Sequential fan and the dominating number

Posted on March 11, 2024 by Dan Ma

The sequential fan $S(\kappa)$ is the quotient space obtained by identifying the limit points of a topological sum of $\kappa$ many convergent sequences. We focus on $S(\omega)$ , the sequential fan derived from countably and infinitely many convergent sequences. Because only countably many convergent sequences are used, $S(\omega)$ is intimately connected to the combinatorics in $\omega^\omega$ , which is the family of all functions from $\omega$ into $\omega$ . In particular, we show that the character at the limit point $\infty$ in $S(\omega)$ equals to the dominating number $\mathfrak{d}$ . The dominating number $\mathfrak{d}$ and the bounding number $\mathfrak{b}$ , introduced below, are quite sensitive to set theoretic assumptions. As a result, pinpointing the precise cardinality of the character of the point $\infty$ in the sequential fan $S(\omega)$ requires set theory beyond ZFC. The fact that the character at $\infty$ in the sequential fan $S(\omega)$ is identical to the dominating number $\mathfrak{d}$ is mentioned in page 13 in chapter a-3 of the Encyclopedia of General Topology [1].

Sequential fans had been discussed previously (see here). See here, here, here, here, and here for previous discussion on the bounding number and the dominating number.

Sequential Fans

As mentioned above, a sequential fan is the quotient space on a disjoint union of convergent sequences with all the limit points of the sequences collapsed to one point called $\infty$ . We first give a working definition. To further provide intuition, we also show that our sequential fan of interest $S(\omega)$ is the quotient space of a subspace of the Euclidean plane (i.e., the countably many convergent sequences can be situated in the plane).

In the discussion that follows, $\omega$ is the set of all non-negative integers. The set $\omega^\omega$ is the family of all functions from $\omega$ into $\omega$ . Let $\kappa$ be an infinite cardinal number. The sequential fan $S(\kappa)$ with $\kappa$ many spines is the set $S(\kappa)=\{ \infty \} \cup (\kappa \times \omega)$ with the topology defined as follows:

Every point in $\kappa \times \omega$ is made an isolated point.
An open neighborhood of the point $\infty$ is of the following form:

$B_f=\{ \infty \} \cup \{ (\alpha,n) \in \kappa \times \omega: n \ge f(\alpha) \}$

$f \in \omega^\omega$

In this formulation of the sequential fan, the set $\{(\alpha, n): n \in \omega \}$ , where $\alpha < \kappa$ , is a sequence converging to $\infty$ . For each such convergent sequence, the open neighborhood $B_f$ contains all but finitely many points.

Our focus is $S(\omega)$ , where $S(\omega)=\{ \infty \} \cup (\omega \times \omega)$ .

A View From the Euclidean Plane

The formulation of the sequential fan $S(\kappa)$ given above is a good working formulation. We now describe how $S(\omega)$ can be derived from the Euclidean plane. Consider the following diagram.

$\displaystyle \begin{aligned} & \\& \text{ } & \text{ } & S_1 & \text{ } & S_2 & \text{ } & S_3 & \text{ } & S_4 & \text{ } & S_5 & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ }& \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \downarrow &\text{ } & \downarrow & \text{ } & \downarrow &\text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & p_1 &\text{ } & p_2 & \text{ } & p_3 &\text{ } & p_4 & \text{ } & p_5 & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \end{aligned}$

In the above diagram, the thick bullets are the points in the space and the tiny dots represent “dot dot dot”, indicating that the points or the sequences are continuing countably infinitely. The arrows in the diagrams point to the direction to which the points are converging. There are countably and infinitely many convergent sequences, named $S_1,S_2,S_3,\cdots$ , with $p_n$ being the limit of the sequence $S_n$ . For convenience, we can let $p_n$ be the point $(n,0)$ in the plane and $S_n$ be a sequence converging downward to $p_n$ . Let $S=S_1 \cup S_2 \cup S_3 \cup \cdots$ and let $P=\{p_1,p_2,p_3,\cdots \}$ . Consider the space $X=S \cup P$ with the topology inherited from the Euclidean plane. Any point in any one of the convergent sequence $S_n$ is an isolated point. An open neighborhood of the limit point $p_n$ consists of $p_n$ and all but finitely many points in $S_n$ .

The diagram and the preceding paragraph set up the scene. We are now ready to collapse points (or define the quotient map). We collapse the set of all limit points $P$ to one point called $\infty$ . The resulting quotient space is $Y=S \cup \{ \infty \}$ . In this quotient space, $S$ is the set of all points in the countably many convergent sequences with each point isolated. An open neighborhood at $\infty$ consists of $\infty$ and all but finitely many points in each convergent sequence. This formulation is clearly equivalent to the sequential fan $S(\omega)$ formulated earlier.

When $\kappa$ is uncountable, the topological sum of $\kappa$ many convergent sequences can no longer viewed in a Euclidean space. However, the topological sum is still a metric space (just not a separable one). We can still collapse the limit points into one point called $\infty$ . The resulting quotient space is identical to $S(\kappa)$ formulated above.

The Combinatorics on the Integers

We begin the combinatorics by defining the order $\le^*$ . Recall that $\omega^\omega$ is the family of all functions from $\omega$ into $\omega$ . For $f,g \in \omega^\omega$ , declare $f \le^* g$ if $f(n) \le g(n)$ for all but finitely many $n \in \omega$ . We write $f \not \le^* g$ if the negation of $f \le^* g$ is true, i.e., $f(n)>g(n)$ for infinitely many $n \in \omega$ . The order $\le^*$ is a reflexive and transitive relation.

A set $F \subset \omega^\omega$ is said to be bounded if $F$ has an upper bound according to the order $\le^*$ , i.e., there exists $g \in \omega^\omega$ such that $f \le^* g$ for all $f \in F$ (Here, $g$ is the upper bound of $F$ ). The set $F$ is said to be unbounded if it is not bounded according to $\le^*$ . That is, $F$ is unbounded if for each $g \in \omega^\omega$ , there exists $f \in F$ such that $f \not \le^* g$ . A set $F \subset \omega^\omega$ is said to be a dominating set if $F$ is cofinal in $\le^*$ , i.e., for each $f \in \omega^\omega$ , there exists $g \in F$ such that $f \le^* g$ . We now define two cardinal numbers as follows:

$\mathfrak{b}=\text{min} \{ \lvert F \lvert: F \subset \omega^\omega \text{ is unbounded} \}$

$\mathfrak{d}=\text{min} \{ \lvert F \lvert: F \subset \omega^\omega \text{ is dominating} \}$

The first number $\mathfrak{b}$ is called the bounding number and the second one $\mathfrak{d}$ is called the dominating number. Both are upper bounded by the continuum $\mathfrak{c}$ , i.e., $\mathfrak{b} \le \mathfrak{c}$ and $\mathfrak{d} \le \mathfrak{c}$ . Using a diagonal argument, we can show that both of these cardinal numbers are not countable. Thus, we have $\omega_1 \le \mathfrak{b},\mathfrak{d} \le \mathfrak{c}$ . How do $\mathfrak{b}$ and $\mathfrak{d}$ relate? We have $\mathfrak{b} \le \mathfrak{d}$ since every dominating set is also an unbounded set.

The Character at Infinity

The sequential fan $S(\omega)$ is not first countable at the point $\infty$ . In other word, there does not exist a countable local base at $\infty$ . To see this, let $\{ B_{f_1},B_{f_2},B_{f_3},\cdots \}$ be a countable collection of open neighborhoods of $\infty$ . Using a diagonal argument, we can find $f \in \omega^\omega$ such that $B_{f_n} \not \subset B_f$ for all $n$ . This shows that no countable collection of open neighborhoods can be a base at $\infty$ . Thus, the character at $\infty$ must be uncountable (the character at a point is the minimum cardinality of a local base at the point). Thus, we have have $\chi(S(\omega),\infty)>\omega$ . Furthermore, we have $\omega_1 \le \chi(S(\omega),\infty) \le \mathfrak{c}$ (character is at least $\omega_1$ but no more than continuum). The range from $\omega_1$ to continuum $\mathfrak{c}$ is a narrow range if continuum hypothesis holds, but can be a large range if continuum hypothesis does not hold. Can we pinpoint the character at $\infty$ more narrowly and more precisely?

Connecting the Dominating Number to the Sequential Fan

We claim the for the sequential fan $S(\omega)$ , the character at the point $\infty$ is the dominating number $\mathfrak{d}$ introduced above. To establish this claim, we set up a different formulation of dominating set. A set $F \subset \omega^\omega$ is said to be a special dominating set if for each $f \in \omega^\omega$ , there exists $g \in F$ such that $f(n) \le g(n)$ for all $n \in \omega$ . We define the cardinal number $\mathfrak{d}_1$ as follows:

$\mathfrak{d}_1=\text{min} \{ \lvert F \lvert: F \subset \omega^\omega \text{ is a special dominating set} \}$

Note that the term “special dominating” is not a standard term. It is simply a definition that facilitates the argument at hand. One key observation is that when $F$ is a special dominating set, the collection $\{B_f: f \in F \}$ becomes a base at the point $\infty$ . Since the cardinal number $\mathfrak{d}_1$ is the minimum cardinality of a base at $\infty$ , we only need to show that $\mathfrak{d}=\mathfrak{d}_1$ . Since every special dominating set is a dominating set, we have $\mathfrak{d} \le$ the cardinality of every special dominating set. Thus, $\mathfrak{d} \le \mathfrak{d}_1$ .

Next we show $\mathfrak{d}_1 \le \mathfrak{d}$ . To this end, we show that $\mathfrak{d}_1 \le$ the cardinality of every dominating set. We claim that for every dominating set $F$ , there exists a special dominating set $F_*$ with $\lvert F_* \lvert=\lvert F \lvert$ . Once this is established, we have $\mathfrak{d}_1 \le$ the cardinality of every dominating set and thus $\mathfrak{d}_1 \le \mathfrak{d}$ .

Let $F$ be a dominating set. For each $n \in \omega$ with $n \ge 1$ , define the following:

$D_n=\{0,1,\cdots,n-1 \}$
$E_n=\{ n,n+1,n+2,\cdots \}$
$A_n=\omega^{D_n}$
$B_n=\omega^{E_n}$

If $h \in A_n$ and $k \in B_n$ , then we take $h \cup k$ to be a function in $\omega^\omega$ . For each $n \ge 1$ and for each $f \in F$ , define the following:

$F_{f,n}=\{h \cup (f \upharpoonright E_n): h \in A_n \}$

with $f \upharpoonright E_n$ representing the function $f$ restricted to the set $E_n$ . Let $F_*=\bigcup \{F_{f,n}: n \ge 1, f \in F \}$ . Note that each $F_{f,n}$ is countable. As a result, $\lvert F_* \lvert=\lvert F \lvert$ . Because $F$ is a dominating set, $F_*$ is a special dominating set. We have just established that $\mathfrak{d}_1 = \mathfrak{d}$ and that the character of the point $\infty$ in the sequential fan $S(\omega)$ is the dominating number $\mathfrak{d}$ .

Remarks

Can we pinpoint the character at $\infty$ ? The answer is a partial yes. We establish that $\chi(S(\omega),\infty)=\mathfrak{d}$ . However, the dominating number and the bounding number as well as other small cardinals are very sensitive to set theory. For example, when continuum hypothesis (CH) holds, The dominating number $\mathfrak{d}$ is continuum. Thus, it is consistent with ZFC that $\chi(S(\omega),\infty)$ is continuum. It is also consistent with ZFC that $\omega_1 \le \mathfrak{b} <\mathfrak{d}<\mathfrak{c}$ . Thus it is consistent that $\chi(S(\omega),\infty)$ is greater than $\omega_1$ and less than continuum. Though the dominating number tells us how big the character at $\infty$ is, we cannot pinpoint precisely where the character is in the range between $\omega_1$ and continuum. For more information about dominating number and other small cardinals, see chapter 3 in the Handbook of Set-Theoretic Topology [2].

The fact that the character at $\infty$ in the sequential fan $S(\omega)$ is identical to the dominating number $\mathfrak{d}$ is mentioned in page 13 in chapter a-3 of the Encyclopedia of General Topology [1].

The sequential fan $S(\omega)$ is a space that has a simple definition. After all, the starting point is a subspace of the Euclidean plane with $S(\omega)$ obtained by collapsing the limit points. Though the space is very accessible, the size of the character at the limit point $\infty$ is unknowable if we work only in ZFC. It is a short “distance” from the definition of the sequential fan $S(\omega)$ to the set-theoretic unknowable statement. This makes the sequential fan $S(\omega)$ an interesting example and an excellent entry point of learning more set-theoretic topology.

Reference

Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
Van Douwen, E. K., The Integers and Topology, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 111-167.

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Dan Ma Sequential Fan
Daniel Ma Sequential Fan

Dan Ma Dominating Number
Daniel Ma Dominating Number

Dan Ma Bounding Number
Daniel Ma Bounding Number

Dan Ma quotient mapping
Daniel Ma quotient mapping

Dan Ma Quotient Space
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$\copyright$ 2024 – Dan Ma

Posted in Convergence Property, Well known examples | Tagged Quotient Map, Quotient space, Sequential fan, Sequential space | Leave a reply

Defining Arens’ space using diagrams

Posted on April 16, 2023 by Dan Ma

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One way to define the Arens’ space is a 2-step approach, which is the quotient space approach. The first step is to identify an Euclidean plane consisting of convergent sequences (usually conveniently situated in the two-dimensional plane). The second step is to collapse certain points to make it a quotient space. Another way is to define the space directly, usually using an appropriate subset of the plane (of course, the resulting space is not an Euclidean space). We demonstrate both approaches using diagrams. In the first approach, we use two diagrams, the first one showing what the Euclidean space should look like, the second showing the resulting Arens’ space after certain points are identified. In the second approach, only one diagram is used (the standalone approach). The two-step approach is actually more informative since the quotient space of a separable metric space is a sequential space.

The following diagrams define the spaces without identifying specific points or locations in the Euclidean plane. The diagrams only indicate how the points relate to one another. For a definition of Arens’ space using the quotient space approach using specific points in the plane, see here. For a definition without connection to quotient space, see here. The red diagram and the blue diagram are for the quotient space approach (two-step). The pink diagram is the standalone approach.

The Arens’ space as discussed here is related to the Arens-Fort space, example 26 in Counterexamples in Topology [2].

The Red Diagram – The Euclidean Space

$\displaystyle \begin{aligned} & \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & S_5 & \text{ } & S_4 & \text{ } & S_3 & \text{ } & S_2 & \text{ } & S_1 \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } &\bullet & \text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet & \text{ } &\bullet & \text{ } & \bullet & \text{ } &\bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot &\text{ } &\cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot &\text{ } & \cdot & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & q_5 &\text{ } & q_4 & \text{ } & q_3 &\text{ } & q_2 & \text{ } & q_1 \\& \text{ } & \text{ } & \text{ } &\text{ } & \text{ } & \text{ } & \text{ } &\text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\& \bullet & \leftarrow & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet \\& p & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & p_5 &\text{ } & p_4 & \text{ } & p_3 &\text{ } & p_2 & \text{ } & p_1 \end{aligned}$

In all 3 diagrams, the thick bullets represent points in the space and the tiny dots represent “dot dot dot”, indicating that the points or the sequences are continuing countably infinitely. The arrows in the diagrams point to the direction to which the points are converging.

The points in the red diagram form a subspace of the Euclidean plane. There are convergent sequences $S_n$ going downward and going across from right to left. The point $q_n$ is the limit of the sequence $S_n$ . The sequence of points $q_n$ converges to a point, which is ignored and not shown in the diagram. The points $p_n$ are situated below the points $q_n$ and converge to the point $p$ . In this Euclidean space, the points in the sequences $S_n$ are isolated points. An open set of the point $q_n$ consists of $q_n$ and all but finitely many points in the sequence $S_n$ . Each point $p_n$ is isolated. An open set of the point $p$ consists of $p$ and all but finitely many $p_n$ .

The Blue Diagram – The Arens’ Space as a Quotient Space

$\displaystyle \begin{aligned} & \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & S_5 & \text{ } & S_4 & \text{ } & S_3 & \text{ } & S_2 & \text{ } & S_1 \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } &\bullet & \text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet & \text{ } &\bullet & \text{ } & \bullet & \text{ } &\bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot &\text{ } &\cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot &\text{ } & \cdot & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow \\& \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet \\& p & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & p_5 &\text{ } & p_4 & \text{ } & p_3 &\text{ } & p_2 & \text{ } & p_1 \end{aligned}$

The blue diagram is established from the red diagram. The blue diagram is obtained by identifying the points $q_n$ and $p_n$ in the red diagram as one point called $p_n$ . The resulting quotient space is the Arens’ space. Let $S$ be the set of all points in the sequences $S_n$ . Let $P$ be the set of all points $p_n$ . The Arens’ space is $X=S \cup P \cup \{ p \}$ with the quotient topology. With the quotient topology, an open set containing the point $p$ is obtained by removing finitely many vertical columns (a column is $S_n$ plus the point $p_n$ ) from $X$ and by removing finitely many points in the remaining columns. An open neighborhood of the point $p_n$ consists of $p_n$ and all but finitely many points in the sequence $S_n$ . Points in the sequences $S_n$ continue to be isolated points.

The Arens’ space is a sequential space since it is the quotient image of a separable metric space.

The Pink Diagram – The Arens’ Space as a Standalone Space

$\displaystyle \begin{aligned} & \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ }& \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & S_1 & \text{ } & S_2 & \text{ } & S_3 & \text{ } & S_4 & \text{ } & S_5 & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \text{ } &\text{ } & \text{ } & \text{ } & \text{ } &\text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\& \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& p & \text{ } & p_1 &\text{ } & p_2 & \text{ } & p_3 &\text{ } & p_4 & \text{ } & p_5 & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \end{aligned}$

In the pink diagram, the bottom rows are the limit points. The point $p_n$ is the sequential limit of the sequence $S_n$ . The sequence $S_n$ is displayed vertically. The convergence of the sequence $S_n$ is not exhibited in the diagram and follows from how the open sets are defined.

Let $S$ be the set of all points in the sequences $S_n$ . Let $P$ be the set of all points $p_n$ . The Arens’ space is $X=S \cup P \cup \{ p \}$ with the topology defined as follows. Each point in $S$ is an isolated point. An open neighborhood of $p_n \in P$ consists of the point $p_n$ and all but finitely many points in $S_n$ . An open neighbood containing the point $p$ is obtained by removing finitely many vertical columns (a column is $S_n$ plus the point $p_n$ ) from $X$ and by removing finitely many points in the remaining columns.

Remarks

The blue diagram and the pink diagram are both representations of the Arens’ space. The space consists of countably many convergent sequences and their limits points plus one additional limit point called $p$ . Both diagrams present the essential ideas without being tied to specific points or sequences in the Euclidean plane. Perhaps the diagrams will make it easier to think about the Arens’ space and remember the definition.

As mentioned earlier, the Arens’ space is a sequential space since it is the quotient space of a metric space (see the theorem here). Recall from above that the Arens’ space is $X=S \cup P \cup \{ p \}$ . Clearly, $p \in \overline{S}$ . Note that no sequence of points in $S$ can converge to the point $p$ . Thus, the Arens’ space is an example of a sequential space that is not a Frechet space. In fact, in order to know whether a sequential space $W$ is Frechet, all we need to do is to determine if $W$ contains a copy of the Arens’ space (see the theorem here). Thus, any space that is sequential but not Frechet contains a copy of the Arens’ space. In this case, Frechetness is characterized the absence of an Arens’ subspace. The Arens’ space is a canonical quotient space that appears in the characterization of other properties. See [1] for an example.

The property of being a sequential space is not hereditary. Consider the subspace of the Arens’ space $Y=S \cup \{ p \}$ . As observed in the preceding paragraph, no sequence of points in $S$ can converge to $p$ . Thus, the set $S$ is a sequentially closed set but not closed in $Y$ . This means that $Y$ is not a sequential space. Thus, the Arens’ space is a sequential that is not hereditarily sequential. In fact, a space is a Frechet space if and only if it is a hereditarily sequential space (see Theorem 1 here).

The subspace $Y=S \cup \{ p \}$ discussed in the preceding paragraph is the Arens-Fort space, which is the example 26 in Steen and Seebach [2].

Reference

Lin, S., A note on the Arens’ space and sequential fan, Topology Appl, 81, 185-196, 1997.

Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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Dan Ma Arens’ space
Daniel Ma Arens’ space

Dan Ma sequential space
Daniel Ma sequential space

Dan Ma Frechet space
Daniel Ma Frechet space

Dan Ma quotient space
Daniel Ma quotient space

Dan Ma topology
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$\copyright$ 2023 – Dan Ma

Posted in Well known examples | Tagged Arens' Space, Quotient space, Sequential space | 2 Replies

Revisiting example 106 from Steen and Seebach

Posted on December 25, 2021 by Dan Ma

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The example 106 from Counterexamples in Topology by Steen and Seebach [3] is the space $\omega_1 \times I^I$ where the first factor $\omega_1$ is the space of countable ordinals with the usual order topology and the second factor $I^I$ is the product of continuum many copies of the unit interval $I=[0,1]$ .

This space was previously discussed in this site. One of the key results from that discussion is that $\omega_1 \times I^I$ is not normal, a result not shown in Steen and Seebach. The proof that was given in this site (see here) is based on an article published in 1976 [1], long before the publication date of the first edition of Steen and Seebach in 1970. It turns out that the non-normality of $\omega_1 \times I^I$ was given as an exercise in Steen and Seebach in the problem section at the end of the book (problem 127 in page 211, Dover edition). Problem 127: Show that $[0, \Omega) \times I^I$ is not normal. This indicates that the result not shown in Steen and Seebach was because it was given as a problem and not because the tool for solving it was not yet available. The fact that it is given as an exercise also means that there is a more basic proof of the non-normality of $\omega_1 \times I^I$ . So, once this is realized, I set out to find a simpler proof or at least one that does not rely on the result from [1]. Interestingly, this proof brings out a broader discussion that is worthwhile and goes beyond the example at hand. The goal here is to examine the more basic proof and the broader discussion.

A Classic Example

Before talking about the promised proof, we consider the product of $\omega_1$ and its immediate successor.

As noted at the beginning, the space $\omega_1$ is the set of all countable ordinals with the order topology. The ordinal $\omega_1+1$ is the immediate successor of $\omega_1$ . It can be regarded as the result of adding one more point to $\omega_1$ . The extra point is $\omega_1$ , i.e., $\omega_1 +1=\omega_1 \cup \{ \omega_1 \}$ with $\omega_1$ greater than all points $\beta < \omega_1$ . The ordinal $\omega_1+1$ with the order topology is a compact space. Using interval notation, $\omega_1=[0, \omega_1)$ and $\omega_1+1=[0, \omega_1]$ . As ordinals, $\omega_1$ is the first uncountable ordinal and $\omega_1+1$ is the first uncountable successor ordinal. For more information, see here.

The product $[0, \omega_1) \times [0, \omega_1]$ is a classic example of a product of a normal space (the first factor) and a compact space (the second factor) that is not normal. This example and others like it show that normality is easily broken upon taking product even if one of the factors is as nice as a compact space. The non-normality of $[0, \omega_1) \times [0, \omega_1]$ is discussed here. In that proof, two disjoint closed sets $H$ and $K$ are given such that they cannot be separated by disjoint open sets. The $H$ and $K$ are:

$H=\{ (\alpha, \alpha): \alpha < \omega_1 \}$
$K=\{(\alpha, \omega_1): \alpha < \omega_1 \}$

The Basic Proof

To show that $\omega_1 \times I^I$ is not normal, we show that one of its closed subspaces is not normal. That closed subspace is $[0, \omega_1) \times [0, \omega_1]$ . To this end, we show that $[0, \omega_1]$ can be embedded in the product space $I^I$ . With a non-normal closed subspace, it follows that $\omega_1 \times I^I$ is not normal. The remainder of the proof is to give the embedding.

We show that $[0, \omega_1]$ can be embedded as a closed subspace of $I^{\omega_1}$ , the product of $\omega_1$ many copies of $I$ . This means that $[0, \omega_1]$ is also a closed subspace of $I^I$ .

For each $\beta < \omega_1$ , define $T_\beta: \omega_1 \rightarrow I$ as follows:

$T_\beta(\gamma) = \begin{cases} 1 & \ \ \ \mbox{if } \gamma < \beta \\ 0 & \ \ \ \mbox{if } \gamma \ge \beta \end{cases}$

Furthermore, define $T: \omega_1 \rightarrow I$ by letting $T(\beta)=1$ for all $\beta < \omega_1$ . Consider the correspondence $\beta \rightarrow T_\beta$ with $\beta < \omega_1$ and $\omega_1 \rightarrow T$ . The mapping is clearly one-to-one from $[0, \omega_1]$ onto $\{ T_\beta: \beta < \omega_1 \} \cup \{ T \}$ . Upon closer inspection, the mapping in each direction is continuous (this is a good exercise to walk through). Thus, the mapping is a homeomorphism. It follows that $[0, \omega_1]$ can be considered a subspace of $I^{\omega_1}$ . Since $[0, \omega_1]$ is compact, it must be a closed subspace. With the cardinality of $\omega_1$ being less than or equal to continuum, it follows that $[0, \omega_1]$ can be embedded as a closed subspace of $I^I$ .

Stone-Cech Compactification

The first broader discussion is that of Stone-Cech compactification. More specifically, $\beta \omega_1=\omega_1+1$ , i.e., the Stone-Cech compactification of the first uncountable ordinal is its immediate successor.

To see that $\beta \omega_1=\omega_1+1$ , note that every continuous function defined on $[0,\omega_1)$ is bounded and is eventually constant (see result B here). As a result, every continuous function defined on $[0,\omega_1)$ can be extended to a continuous function defined on $[0, \omega_1]$ . For any continuous function $f: \omega_1 \rightarrow \mathbb{R}$ , we can simply define $f(\omega_1)$ to be the eventual constant value. A subspace $W$ of a space $Y$ is $C^*$ -embedded in $Y$ if every bounded continuous real-valued function on $W$ can be extended to $Y$ . According to theorem 19.12 in [4], if $Y$ is a compactification of $X$ and if $X$ is $C^*$ -embedded in $Y$ , then $Y$ is the Stone-Cech compactification of $X$ . Thus $[0,\omega_1)$ is $C^*$ -embedded in $[0,\omega_1]$ and $[0,\omega_1]$ is the Stone-Cech compactification of $[0,\omega_1)$ . In this instance, the Stone-Cech compactification agrees with the one-point compactification. Consider the following class theorem about normality in product space. The theorem is Corollary 3.4 in the chapter on products of normal spaces in the handbook of set-theoretic topology [2].

Theorem 1
Let $X$ be a space. The following conditions are equivalent.

The space $X$ is paracompact.

The product space $X \times \beta X$ is normal.

Based on the discussion presented above, the non-normality of $\omega_1 \times I^I$ is due to the non-normality of $[0, \omega_1) \times [0, \omega_1]$ . Based on this theorem, the non-normality of $[0, \omega_1) \times [0, \omega_1]$ is due to the non-paracompactness of $[0, \omega_1)$ . See result G here for a proof that $[0, \omega_1)$ is not paracompact.

The discussion up to this point points to two ways to prove that $\omega_1 \times I^I$ is not normal. One way is the basic proof indicated above. The other way is to use Theorem 1, along with the homeomorphic embedding from $[0, \omega_1]$ into $I^I$ , the fact that $\beta \omega_1=\omega_1+1$ and the fact that $[0, \omega_1)$ is not paracompact. Both are valuable. The first way is basic and is a constructive proof. Because it is more hands-on, it is a better proof to learn from. The second way provides a broader perspective that is informative but requires quoting a couple of fairly deep results. Perhaps it is best used as a second proof for perspective.

Countable Tightness

The essence of the basic proof above goes like this: if the space $Y$ contains a copy of $\omega_1+1=[0, \omega_1]$ , then the product space $[0, \omega_1) \times Y$ is not normal. The contrapositive statement would be the following:

Corollary
Let $Y$ be a space. If the product space $\omega_1 \times Y$ is normal, then $Y$ cannot contain a copy of $\omega_1+1$ .

In the space of $\omega_1+1=[0, \omega_1]$ , note the following about the last point: $\omega_1 \in \overline{[0, \omega_1)}$ but $\omega_1 \notin \overline{C}$ for any countable $C \subset [0, \omega_1)$ , i.e., the last point is the limit point of the set of all the points preceding it but is not in the closure of any countable set. This means that the space $\omega_1+1=[0, \omega_1]$ does not have countable tightness (or is not countably tight). See here for definition. The property of countable tightness is hereditary. If $Y$ contains a copy of $\omega_1+1$ , then $Y$ is not countably tight (or is uncountably tight). This brings us to the following theorem.

Theorem 2
Let $Y$ be an infinite compact space. Then $\omega_1 \times Y$ is normal if and only if $Y$ has countable tightness.

Whenever we consider the normality of a product with the first factor being $\omega_1$ and the second factor being a compact space, the real story is the tightness of that compact space. If the tightness is countable, the product is normal. Otherwise, the product is not normal. The theorem is another reason that $\omega_1 \times I^I$ is not normal. Instead of embedding $[0, \omega_1]$ into $I^I$ , we can actually show that $I^I$ does not have countable tightness. This is the approach that was taken in this previous post.

Theorem 2 is the result from 1976 alluded to earlier [1]. A proof of Theorem 2 is found in this previous post. For results concerning normality in a product space with a compact factor (the other factor does not have to be $\omega_1$ ), see the chapter on products of normal spaces in the handbook of set-theoretic topology [2].

Reference

Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976.

Przymusinski T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.

Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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Dan Ma topology

Daniel Ma topology

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$\copyright$ 2021 – Dan Ma

Posted in First uncountable ordinal, Normal space, Normality in Product, Omega 1, Product of unit interval, Product space, Well known examples | Tagged General topology, Non-normal product, Normal space, Omega 1, Point-set topology, Product space, Topology | Leave a reply

Helly Space

Posted on June 10, 2019 by Dan Ma

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This is a discussion on a compact space called Helly space. The discussion here builds on the facts presented in Counterexample in Topology [2]. Helly space is Example 107 in [2]. The space is named after Eduard Helly.

Let $I=[0,1]$ be the closed unit interval with the usual topology. Let $C$ be the set of all functions $f:I \rightarrow I$ . The set $C$ is endowed with the product space topology. The usual product space notation is $I^I$ or $\prod_{t \in I} W_t$ where each $W_t=I$ . As a product of compact spaces, $C=I^I$ is compact.

Any function $f:I \rightarrow I$ is said to be increasing if $f(x) \le f(y)$ for all $x<y$ (such a function is usually referred to as non-decreasing). Helly space is the subspace $X$ consisting of all increasing functions. This space is Example 107 in Counterexample in Topology [2]. The following facts are discussed in [2].

The space $X$ is compact.

The space $X$ is first countable (having a countable base at each point).

The space $X$ is separable.

The space $X$ has an uncountable discrete subspace.

From the last two facts, Helly space is a compact non-metrizable space. Any separable metric space would have countable spread (all discrete subspaces must be countable).

The compactness of $X$ stems from the fact that $X$ is a closed subspace of the compact space $C$ .

Further Discussion

Additional facts concerning Helly space are discussed.

The product space $\omega_1 \times X$ is normal.

Helly space $X$ contains a copy of the Sorgenfrey line.

Helly space $X$ is not hereditarily normal.

The space $\omega_1$ is the space of all countable ordinals with the order topology. Recall $C$ is the product space $I^I$ . The product space $\omega_1 \times C$ is Example 106 in [2]. This product is not normal. The non-normality of $\omega_1 \times C$ is based on this theorem: for any compact space $Y$ , the product $\omega_1 \times Y$ is normal if and only if the compact space $Y$ is countably tight. The compact product space $C$ is not countably tight (discussed here). Thus $\omega_1 \times C$ is not normal. However, the product $\omega_1 \times X$ is normal since Helly space $X$ is first countable.

To see that $X$ contains a copy of the Sorgenfrey line, consider the functions $h_t:I \rightarrow I$ defined as follows:

$\displaystyle h_t(x) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ 0 \le x \le t \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ t<x \le 1 \\ \end{array} \right.$

for all $0<t<1$ . Let $S=\{ h_t: 0<t<1 \}$ . Consider the mapping $\gamma: (0,1) \rightarrow S$ defined by $\gamma(t)=h_t$ . With the domain $(0,1)$ having the Sorgenfrey topology and with the range $S$ being a subspace of Helly space, it can be shown that $\gamma$ is a homeomorphism.

With the Sorgenfrey line $S$ embedded in $X$ , the square $X \times X$ contains a copy of the Sorgenfrey plane $S \times S$ , which is non-normal (discussed here). Thus the square of Helly space is not hereditarily normal. A more interesting fact is that Helly space is not hereditarily normal. This is discussed in the next section.

Finding a Non-Normal Subspace of Helly Space

As before, $C$ is the product space $I^I$ where $I=[0,1]$ and $X$ is Helly space consisting of all increasing functions in $C$ . Consider the following two subspaces of $X$ .

$Y_{0,1}=\{ f \in X: f(I) \subset \{0, 1 \} \}$

$Y=X - Y_{0,1}$

The subspace $Y_{0,1}$ is a closed subset of $X$ , hence compact. We claim that subspace $Y$ is separable and has a closed and discrete subset of cardinality continuum. This means that the subspace $Y$ is not a normal space.

First, we define a discrete subspace. For each $x$ with $0<x<1$ , define $f_x: I \rightarrow I$ as follows:

$\displaystyle f_x(y) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ 0 \le y < x \\ \text{ } & \text{ } \\ \displaystyle \frac{1}{2} &\ \ \ \ \ y=x \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ x<y \le 1 \\ \end{array} \right.$

Let $H=\{ f_x: 0<x<1 \}$ . The set $H$ as a subspace of $X$ is discrete. Of course it is not discrete in $X$ since $X$ is compact. In fact, for any $f \in Y_{0,1}$ , $f \in \overline{H}$ (closure taken in $X$ ). However, it can be shown that $H$ is closed and discrete as a subset of $Y$ .

We now construct a countable dense subset of $Y$ . To this end, let $\mathcal{B}$ be a countable base for the usual topology on the unit interval $I=[0,1]$ . For example, we can let $\mathcal{B}$ be the set of all open intervals with rational endpoints. Furthermore, let $A$ be a countable dense subset of the open interval $(0,1)$ (in the usual topology). For convenience, we enumerate the elements of $A$ and $\mathcal{B}$ .

$A=\{ a_1,a_2,a_3,\cdots \}$

$\mathcal{B}=\{B_1,B_2,B_3,\cdots \}$

We also need the following collections.

$\mathcal{G}=\{G \subset \mathcal{B}: G \text{ is finite and is pairwise disjoint} \}$

$\mathcal{A}=\{F \subset A: F \text{ is finite} \}$

For each $G \in \mathcal{G}$ and for each $F \in \mathcal{A}$ with $\lvert G \lvert=\lvert F \lvert=n$ , we would like to arrange the elements in increasing order, notated as follow:

$F=\{t_1,t_2,\cdots,t_n \}$

$G=\{E_1,E_2,\cdots,E_n \}$

For the set $F$ , we have $0<t_1<t_2< \cdots <t_n<1$ . For the set $G$ , $E_i$ is to the left of $E_j$ for $i<j$ . Note that elements of $G$ are pairwise disjoint. Furthermore, write $E_i=(p_i,q_i)$ . If $0 \in E_1$ , then $E_1=[p_1,q_1)=[0,q_1)$ . If $1 \in E_n$ , then $E_n=(p_n,q_n]=(p_n,1]$ .

For each $F$ and $G$ as detailed above, we define a function $L(F,G):I \rightarrow I$ as follows:

$\displaystyle L(F,G)(x) = \left\{ \begin{array}{ll} \displaystyle t_1 &\ \ \ \ \ 0 \le x < q_1 \\ \text{ } & \text{ } \\ \displaystyle t_2 &\ \ \ \ \ q_1 \le x < q_2 \\ \text{ } & \text{ } \\ \displaystyle \vdots &\ \ \ \ \ \vdots \\ \text{ } & \text{ } \\ \displaystyle t_{n-1} &\ \ \ \ \ q_{n-2} \le x < q_{n-1} \\ \text{ } & \text{ } \\ \displaystyle t_n &\ \ \ \ \ q_{n-1} \le x \le 1 \\ \end{array} \right.$

The following diagram illustrates the definition of $L(F,G)$ when both $F$ and $G$ have 4 elements.

Figure 1 – Member of a countable dense set

Let $D$ be the set of $L(F,G)$ over all $F \in \mathcal{A}$ and $G \in \mathcal{G}$ . The set $D$ is a countable set. It can be shown that $D$ is dense in the subspace $Y$ . In fact $D$ is dense in the entire Helly space $X$ .

To summarize, the subspace $Y$ is separable and has a closed and discrete subset of cardinality continuum. This means that $Y$ is not normal. Hence Helly space $X$ is not hereditarily normal. According to Jones’ lemma, in any normal separable space, the cardinality of any closed and discrete subspace must be less than continuum (discussed here).

Remarks

The preceding discussion shows that both Helly space and the square of Helly space are not hereditarily normal. This is actually not surprising. According to a theorem of Katetov, for any compact non-metrizable space $V$ , the cube $V^3$ is not hereditarily normal (see Theorem 3 in this post). Thus a non-normal subspace is found in $V$ , $V \times V$ or $V \times V \times V$ . In fact, for any compact non-metric space $V$ , an excellent exercise is to find where a non-normal subspace can be found. Is it in $V$ , the square of $V$ or the cube of $V$ ? In the case of Helly space $X$ , a non-normal subspace can be found in $X$ .

A natural question is: is there a compact non-metric space $V$ such that both $V$ and $V \times V$ are hereditarily normal and $V \times V \times V$ is not hereditarily normal? In other words, is there an example where the hereditarily normality fails at dimension 3? If we do not assume extra set-theoretic axioms beyond ZFC, any compact non-metric space $V$ is likely to fail hereditarily normality in either $V$ or $V \times V$ . See here for a discussion of this set-theoretic question.

Reference

Kelly, J. L., General Topology, Springer-Verlag, New York, 1955.

Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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Dan Ma topology

Daniel Ma topology

Dan Ma math

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$\copyright$ 2019 – Dan Ma

Posted in Well known examples | Tagged Compact space, General topology, Helly Space, Hereditarily normal space, Jones' Lemma, Point-set topology, Topology | Leave a reply

Drawing more Sorgenfrey continuous functions

Posted on January 1, 2018 by Dan Ma

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In this previous post, we draw continuous functions on the Sorgenfrey line $S$ to gain insight about the function $C_p(S)$ . In this post, we draw more continuous functions with the goal of connecting $C_p(S)$ and $C_p(D)$ where $D$ is the double arrow space. For example, $C_p(D)$ can be embedded as a subspace of $C_p(S)$ . More interestingly, both function spaces $C_p(D)$ and $C_p(S)$ share the same closed and discrete subspace of cardinality continuum. As a result, the function space $C_p(D)$ is not normal.

Double Arrow Space

The underlying set for the double arrow space is $D=[0,1] \times \{ 0,1 \}$ , which is a subset in the Euclidean plane.

Figure 1 – The Double Arrow Space

The name of double arrow comes from the fact that an open neighborhood of a point in the upper line segment points to the right while an open neighborhood of a point in the lower line segment points to the left. This is demonstrated in the following diagram.

Figure 2 – Open Neighborhoods in the Double Arrow Space

More specifically, for any $a$ with $0 \le a < 1$ , a basic open set containing the point $(a,1)$ is of the form $\displaystyle \biggl[ [a,b) \times \{ 1 \} \biggr] \cup \biggl[ (a,b) \times \{ 0 \} \biggr]$ , painted red in Figure 2. One the other hand, for any $a$ with $0<a \le 1$ , a basic open set containing the point $(a,0)$ is of the form $\biggl[ (c,a) \times \{ 1 \} \biggr] \cup \biggl[ (c,a] \times \{ 0 \} \biggr]$ , painted blue in Figure 2. The upper right point $(1,1)$ and the lower left point $(0,0)$ are made isolated points.

The double arrow space is a compact space that is perfectly normal and not metrizable. Basic properties of this space, along with those of the lexicographical ordered space, are discussed in this previous post.

The drawing of continuous functions in this post aims to show the following results.

The function space $C_p(D)$ can be embedded as a subspace in the function $C_p(S)$ .

Both function spaces $C_p(D)$ and $C_p(S)$ share the same closed and discrete subspace of cardinality continuum.

The function space $C_p(D)$ is not normal.

Drawing a Map from Sorgenfrey Line onto Double Arrow Space

In order to show that $C_p(D)$ can be embedded into $C_p(S)$ , we draw a continuous map from the Sorgenfrey line $S$ onto the double arrow space $D$ . The following diagram gives the essential idea of the mapping we need.

Figure 3 – Mapping Sorgenfrey Line onto Double Arrow Space

The mapping shown in Figure 3 is to map the interval $[0,1]$ onto the upper line segment of the double arrow space, as demonstrated by the red arrow. Thus $x \mapsto (x,1)$ for any $x$ with $0 \le x \le 1$ . Essentially on the interval $[0,1]$ , the mapping is the identity map.

On the other hand, the mapping is to map the interval $[-1,0)$ onto the lower line segment of the double arrow space less the point $(0, 0)$ , as demonstrated by the blue arrow in Figure 3. Thus $-x \mapsto (x,0)$ for any $-x$ with $0<x \le 1$ . Essentially on the interval $[-1,0)$ , the mapping is the identity map times -1.

The mapping described by Figure 3 only covers the interval $[-1,1]$ in the domain. To complete the mapping, let $x \mapsto (1,1)$ for any $x \in (1, \infty)$ and $x \mapsto (0,0)$ for any $x \in (-\infty, -1)$ .

Let $h$ be the mapping that has been described. It maps the Sorgenfrey line onto the double arrow space. It is straightforward to verify that the map $h: S \rightarrow D$ is continuous.

Embedding

We use the following fact to show that $C_p(D)$ can be embedded into $C_p(S)$ .

Suppose that the space $Y$ is a continuous image of the space $X$ . Then $C_p(Y)$ can be embedded into $C_p(X)$ .

Based on this result, $C_p(D)$ can be embedded into $C_p(S)$ . The embedding that makes this true is $E(f)=f \circ h$ for each $f \in C_p(D)$ . Thus each function $f$ in $C_p(D)$ is identified with the composition $f \circ h$ where $h$ is the map defined in Figure 3. The fact that $E(f)$ is an embedding is shown in this previous post (see Theorem 1).

Same Closed and Discrete Subspace in Both Function Spaces

The following diagram describes a closed and discrete subspace of $C_p(S)$ .

Figure 4 – a family of Sorgenfrey continuous functions

For each $0<a<1$ , let $f_a: S \rightarrow \{0,1 \}$ be the continuous function described in Figure 4. The previous post shows that the set $F=\{ f_a: 0<a<1 \}$ is a closed and discrete subspace of $C_p(S)$ . We claim that $F \subset C_p(D) \subset C_p(S)$ .

To see that $F \subset C_p(D)$ , we define continuous functions $U_a: D \rightarrow \{0,1 \}$ such that $f_a=U_a \circ h$ . We can actually back out the map $U_a$ from $f_a$ in Figure 4 and the mapping $h$ . Here’s how. The function $f_a$ is piecewise constant (0 or 1). Let’s focus on the interval $[-1,1]$ in the domain of $f_a$ .

Consider where the function $f_a$ maps to the value 1. There are two intervals, $[a,1)$ and $[-1,-a)$ , where $f_a$ maps to 1. The mapping $h$ maps $[a,1)$ to the set $[a,1) \times \{ 1 \}$ . So the function $U_a$ must map $[a,1) \times \{ 1 \}$ to the value 1. The mapping $h$ maps $[-1,-a)$ to the set $(a,1] \times \{ 0 \}$ . So $U_a$ must map $(a,1] \times \{ 0 \}$ to the value 1.

Now consider where the function $f_a$ maps to the value 0. There are two intervals, $[0,a)$ and $[-a,0)$ , where $f_a$ maps to 0. The mapping $h$ maps $[0,a)$ to the set $[0,a) \times \{ 1 \}$ . So the function $U_a$ must map $[0,a) \times \{ 1 \}$ to the value 0. The mapping $h$ maps $[-a,0)$ to the set $(0,a] \times \{ 0 \}$ . So $U_a$ must map $(0,a] \times \{ 0 \}$ to the value 0.

To take care of the two isolated points $(1,1)$ and $(0,0)$ of the double arrow space, make sure that $U_a$ maps these two points to the value 0. The following is a precise definition of the function $U_a$ .

$\displaystyle U_a(y) = \left\{ \begin{array}{ll} \displaystyle 1 &\ \ \ \ \ \ y \in [a,1) \times \{ 1 \} \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ y \in (a,1] \times \{ 0 \} \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ y \in (0,a] \times \{ 0 \} \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ y \in [0,a) \times \{ 1 \} \\ \text{ } & \text{ } \\ 0 &\ \ \ \ \ \ y=(0,0) \text{ or } y = (1,1) \end{array} \right.$

The resulting $U_a$ is a translation of $f_a$ . Under the embedding $E$ defined earlier, we see that $E(U_a)=f_a$ . Let $U=\{ U_a: 0<a<1 \}$ . The set $U$ in $C_p(D)$ is homeomorphic to the set $F$ in $C_p(S)$ . Thus $U$ is a closed and discrete subspace of $C_p(D)$ since $F$ is a closed and discrete subspace of $C_p(S)$ .

Remarks

The drawings and the embedding discussed here and in the previous post establish that $C_p(D)$ , the space of continuous functions on the double arrow space, contains a closed and discrete subspace of cardinality continuum. It follows that $C_p(D)$ is not normal. This is due to the fact that if $C_p(X)$ is normal, then $C_p(X)$ must have countable extent (i.e. all closed and discrete subspaces must be countable).

While $C_p(D)$ is embedded in $C_p(S)$ , the function space $C_p(S)$ is not embedded in $C_p(D)$ . Because the double arrow space is compact, $C_p(D)$ has countable tightness. If $C_p(S)$ were to be embedded in $C_p(D)$ , then $C_p(S)$ would be countably tight too. However, $C_p(S)$ is not countably tight due to the fact that $S \times S$ is not Lindelof (see Theorem 1 in this previous post).

Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

Tkachuk V. V., A $C_p$ -Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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Dan Ma math

Daniel Ma mathematics

$\copyright$ 2018 – Dan Ma

Posted in Function space, The Sorgenfrey Line, Well known examples | Tagged Double arrow space, Function space, General topology, Normal space, Point-set topology, The Sorgenfrey Line, Topology | 2 Replies

A stroll in Bing’s Example G

Posted on December 5, 2016 by Dan Ma

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In this post we take a leisurely walk in Bing’s Example G, which is a classic example of a normal but not collectionwise normal space. Hopefully anyone who is new to this topological space can come away with an intuitive feel and further learn about it. Indeed this is a famous space that had been extensively studied. This example has been written about in several posts in this topology blog. In this post, we explain how Example G is defined, focusing on intuitive idea as much as possible. Of course, the intuitive idea is solely the perspective of the author. Any reader who is interested in building his/her own intuition on this example can skip this post and go straight to the previous introduction. Other blog posts on various subspaces of Example G are here, here and here. Bing’s Example H is discussed here.

At the end of the post, we will demonstrate that the product of Bing’s Example G with the closed unit interval, $F \times [0,1]$ , is a normal space.

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The Product Space Angle

The topology in Example G is tweaked from the product space topology. It is thus a good idea to first examine the relevant product space. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$ . In other words, $Q$ is the power set of $P$ . Consider the product of $\lvert Q \lvert$ many copies of the two element set $\left\{0,1 \right\}$ . The usual notation of this product space is $2^Q$ . The elements of $2^Q$ are simply the functions from $Q$ into $\left\{0,1 \right\}$ . An arbitrary element of $2^Q$ is a function $f$ that maps every subset of $P$ to either 0 or 1.

Though the base set $P$ can be any uncountable set, it is a good idea to visualize clearly what $P$ is. In the remainder of this section, think of $P$ as the real line $\mathbb{R}$ . Then $Q$ is simply the collection of all subsets of the real line. The elements of the product space are simply functions that map each set of real numbers to either 0 or 1. Or think of each function as a 2-color labeling of the subsets of the real line, where each subset is either red or green for example. There are $2^c$ many subsets of the real line where $c$ is the cardinality of the continuum.

To further visualize the product space, let’s look at a particular subspace of $2^Q$ . For each real number $p$ , define the function $f_p$ such that $f_p$ always maps any set of real numbers that contains $p$ to 1 and maps any set of real numbers that does not contain $p$ to 0. For example, the following are several values of the function $f_0$ .

$f_0([0,1])=1$

$f_0([1,2])=0$

$f_0(\left\{0 \right\})=1$

$f_0(\mathbb{R}-\left\{0 \right\})=0$

$f_0(\mathbb{R})=1$

$f_0(\varnothing)=0$

$f_0(\mathbb{P})=0$

where $\mathbb{P}$ is the set of all irrational numbers. Consider the subspace $F_P=\left\{f_p: p \in P \right\}$ . Members of $F_P$ are easy to describe. Each function in $F_P$ maps a subset of the real line to 0 or 1 depending on whether the subscript belongs to the given subset. Another reason that $F_P$ is important is that Bing’s Example is defined by declaring all points not in $F_P$ isolated points and by allowing all points in $F_P$ retaining the open sets in the product topology.

Any point $f$ in $F_P$ determines $f(q)=0 \text{ or } 1$ based on membership (whether the reference point belongs to the set $q$ ). Points not in $F_P$ have no easy characterization. It seems that any set can be mapped to 0 or 1. Note that any $f$ in $F_P$ maps equally to 0 or 1. So the constant functions $f(q)=0$ and $f(q)=1$ are not in $F_P$ . Furthermore, any $f$ such that $f(q)=1$ for at most countably many $q$ would not be in $F_P$ .

Let’s continue focusing on the product space for the time being. When $F_P$ is considered as a subspace of the product space $2^Q$ , $F_P$ is a discrete space. For each $p \in P$ , there is an open set $W_p$ containing $f_p$ such that $W_p$ contains no other points of $F_P$ . So $F_P$ is relatively discrete in the product space $2^Q$ . Of course $F_P$ cannot be closed in $2^Q$ since $2^Q$ is a compact space. The open set $W_p$ is defined as follows:

$W_p=\left\{f \in 2^Q: f(\left\{p \right\})=1 \text{ and } f(P-\left\{p \right\})=0 \right\}$

It is clear that $f_p \in W_p$ and that $f_t \notin W_p$ for any real number $t \ne p$ .

Two properties of the product space $2^Q$ would be very relevant for the discussion. By the well known Tychonoff theorem, the product space $2^Q$ is compact. Since $P$ is uncountable, $2^Q$ always has the countable chain condition (CCC) since it is the product of separable spaces. A space having CCC means that there can only be at most countably many pairwise disjoint open sets. As a result, the uncountably many open sets $W_p$ cannot be all pairwise disjoint. So there exist at least a pair of $W_p$ , say $W_{a}$ and $W_{b}$ , with nonempty intersection.

The last observation can be generalized. For each $p \in P$ , let $V_p$ be any open set containing $f_p$ (open in the product topology). We observe that there are at least two $a$ and $b$ from $P$ such that $V_a \cap V_b \ne \varnothing$ . If there are only countably many distinct sets $V_p$ , then there are uncountably many $V_p$ that are identical and the observation is valid. So assume that there are uncountably many distinct $V_p$ . By the CCC in the product space, there are at least two $a$ and $b$ with $V_a \cap V_b \ne \varnothing$ . This observation shows that the discrete points in $F_P$ cannot be separated by disjoint open sets. This means that Bing’s Example G is not collectionwise Hausdorff and hence not collectionwise normal.

Another observation is that any disjoint $A_1, A_2 \subset F_P$ can be separated by disjoint open sets. To see this, define the following two open sets $E_1$ and $E_2$ in the product topology.

$q_1=\left\{p \in P: f_p \in A_1 \right\}$

$q_2=\left\{p \in P: f_p \in A_2 \right\}$

$E_1=\left\{f \in 2^Q: f(q_1)=1 \text{ and } f(q_2)=0 \right\}$

$E_2=\left\{f \in 2^Q: f(q_1)=0 \text{ and } f(q_2)=1 \right\}$

It is clear that $A_1 \subset E_1$ and $A_2 \subset E_2$ . Furthermore, $E_1 \cap E_2=\varnothing$ . This observation will be the basis for showing that Bing’s Example G is normal.

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The Topology of Bing’s Example G

The topology for Bing’s Example G is obtained by tweaking the product topology on $2^Q$ . Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$ . The set $F_P$ is defined as above. Bing’s Example G is $F=2^Q$ with points in $F_P$ retaining the open sets in the product topology and with points not in $F_P$ declared isolated. For some reason, in Bing’s original paper, the notation $F$ is used even though the example is identified by G. We will follow Bing’s notation.

The subspace $F_P$ is discrete but not closed in the product topology. However, $F_P$ is both discrete and closed in Bing’s Example G. Based on the discussion in the previous section, one immediate conclusion we can made is that the space $F$ is not collectionwise Hausdorff. This follows from the fact that points in the uncountable closed and discrete set $F_P$ cannot be separated by disjoint open sets. By declaring points not in $F_P$ isolated, the countable chain condition in the original product space $2^Q$ is destroyed. However, there is still a strong trace of CCC around the points in the set $F_P$ , which is sufficient to prevent collectionwise Hausdorffness, and consequently collectionwise normality.

To show that $F$ is normal, let $H$ and $K$ be disjoint closed subsets of $F$ . To make it easy to follow, let $H=A_1 \cup B_1$ and $K=A_2 \cup B_2$ where

$A_1=H \cap F_P \ \ \ \ B_1=H \cap (F-F_P)$

$A_2=K \cap F_P \ \ \ \ B_2=K \cap (F-F_P)$

In other words, $A$ is the non-isolated part and $B$ is the isolated part of the respective closed set. Based on the observation made in the previous section, obtain the disjoint open sets $E_1$ and $E_2$ where $A_1 \subset E_1$ and $A_2 \subset E_2$ . Set the following open sets.

$O_1=(E_1 \cup B_1) - K$

$O_2=(E_2 \cup B_2) - H$

It follows that $O_1$ and $O_2$ are disjoint open sets and that $A_1 \subset O_1$ and $A_2 \subset O_2$ . Thus Bing’s Example G is a normal space.

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Bing’s Example G is Countably Paracompact

We discuss one more property of Bing’s Example G. A space $X$ is countably paracompact if every countable countable open cover of $X$ has a locally finite open refinement. In other words, such a space satisfies the property of being a paracompact space but just for countable open covers. A space is countably metacompact if every countable open cover has a point-finite open refinement (i.e. replacing locally finite in the paracompact definition with point-finite). It is well known that in the class of normal spaces, the two notions are equivalent (see Corollary 2 here). Since Bing’s Example G is normal, we only need to show that it is countably metacompact. Note that Bing’s Example G is not metacompact (see here).

Let $\mathcal{U}$ be a countable open cover of $F$ . Let $\mathcal{U}^*=\left\{U_1,U_2,U_3,\cdots \right\}$ be the set of all open sets in $\mathcal{U}$ that contain points in $F_P$ . For each $i$ , let $A_i=U_i \cap F_P$ . From the perspective of Bing’s Example G, the sets $A_i$ are discrete closed sets. In any normal space, countably many discrete closed sets can be separated by disjoint open sets (see Lemma 1 here). Let $O_1,O_2,O_3,\cdots$ be disjoint open sets such that $A_i \subset O_i$ for each $i$ .

We now build a point-finite open refinement of $\mathcal{U}$ . For each $i$ , let $V_i=U_i \cap O_i$ . Let $V=\cup_{i=1}^\infty V_i$ . Consider the following.

$\mathcal{V}=\left\{V_i: i=1,2,3,\cdots \right\} \cup \left\{\left\{ x \right\}: x \in F-V \right\}$

It follows that $\mathcal{V}$ is an open cover of $F$ . All points of $F_P$ belong to the open sets $V_i$ . Any point that is not in one of the $V_i$ belongs to a singleton open set. It is also clear that $\mathcal{V}$ is a refinement of $\mathcal{U}$ . For each $i$ , $V_i \subset U_i$ and each singleton set is contained in some member of $\mathcal{U}$ . It follows that each point in $F$ belongs to at most finitely many sets in $\mathcal{V}$ . In fact, each point belongs to exactly one set in $\mathcal{V}$ . Each point in $F_P$ belongs to exactly one $V_i$ since the open sets $O_i$ are disjoint. Any point in $V$ belongs to exactly one singleton open set. What we just show is slightly stronger than countably metacompact. The technical term would be countably 1-bounded metacompact.

Since among normal spaces, countably paracompactness is equivalent to countably metacompact, we can now say that Bing’s Example G is a topological space that is normal and countably paracompact. By Dowker’s Theorem, we can conclude that the product of Bing’s Example G with the closed unit interval, $F \times [0,1]$ , is a normal space.

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Previous Posts

Introduction of Bing’s Example G

Some subspaces of Bing’s Example G

Compact subspaces of Bing’s Example G

A subspace of Bing’s Example G

Bing’s Example H

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$\copyright \ 2016 \text{ by Dan Ma}$

Posted in Normal space, Well known examples | Tagged Bing's G, Collectionwise normal, General topology, Point-set topology, Product space, Topology | Leave a reply

The Sorgenfrey plane is subnormal

Posted on May 7, 2014 by Dan Ma

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The Sorgenfrey line is the real line with the topology generated by the base of half-open intervals of the form $[a,b)$ . The Sorgenfrey line is one of the most important counterexamples in general topology. One of the often recited facts about this counterexample is that the Sorgenfrey plane (the square of the Sorgengfrey line) is not normal. We show that, though far from normal, the Sorgenfrey plane is subnormal.

A subset $M$ of a space $Y$ is a $G_\delta$ subset of $Y$ (or a $G_\delta$ -set in $Y$ ) if $M$ is the intersection of countably many open subsets of $Y$ . A subset $M$ of a space $Y$ is a $F_\sigma$ subset of $Y$ (or a $F_\sigma$ -set in $Y$ ) if $Y-M$ is a $G_\delta$ -set in $Y$ (equivalently if $M$ is the union of countably many closed subsets of $Y$ ).

A space $Y$ is normal if for any disjoint closed subsets $H$ and $K$ of $Y$ , there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$ . A space $Y$ is subnormal if for any disjoint closed subsets $H$ and $K$ of $Y$ , there exist disjoint $G_\delta$ subsets $V_H$ and $V_K$ of $Y$ such that $H \subset V_H$ and $K \subset V_K$ . Clearly any normal space is subnormal. The Sorgenfrey plane is an example of a subnormal space that is not normal.

In the proof of the non-normality of the Sorgenfrey plane in this previous post, one of the two disjoint closed subsets of the Sorgenfrey plane that cannot be separated by disjoint open sets is countable. Thus the Sorgenfrey plane is not only not normal; it is not pseudonormal (also discussed in this previous post). A space $Y$ is pseudonormal if for any disjoint closed subsets $H$ and $K$ of $Y$ (one of which is countable), there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$ . The examples of the Sorgenfrey plane and $\omega_1 \times (\omega_1+1)$ show that these two weak forms of normality (pseudonormal and subnormal) are not equivalent. The space $\omega_1 \times (\omega_1+1)$ is pseudonormal but not subnormal (see this previous post for the non-subnormality).

A space $Y$ is said to be a perfect space if every closed subset of $Y$ is a $G_\delta$ subset of $Y$ (equivalently, every open subset of $Y$ is an $F_\sigma$ -subset of $Y$ ). It is clear that any perfect space is subnormal. We show that the Sorgenfrey plane is perfect. There are subnormal spaces that are not perfect (see the example below).

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The Sorgenfrey plane is perfect

Let $S$ denote the Sorgenfrey line, i.e., the real line $\mathbb{R}$ topologized using the base of half-open intervals of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x <b \right\}$ . The Sorgenfrey plane is the product space $S \times S$ . We show the following:

Proposition 1
The Sorgenfrey line $S$ is perfect.

Proof of Proposition 1
Let $U$ be a non-empty subset of $S$ . We show that $U$ is a $F_\sigma$ -set. Let $U_0$ be the interior of $U$ in the usual topology. In other words, $U_0$ is the following set:

$U_0=\left\{x \in U: \exists \ (a,b) \text{ such that } x \in (a,b) \text{ and } (a,b) \subset U \right\}$

The real line with the usual topology is perfect. Thus $U_0=\bigcup_{n=1}^\infty F_n$ where each $F_n$ is a closed subset of the real line $\mathbb{R}$ . Since the Sorgenfrey topology is finer than the usual topology, each $F_n$ is also closed in the Sorgenfrey line.

Consider $Y=U-U_0$ . We claim that $Y$ is countable. Suppose $Y$ is uncountable. Since the Sorgenfrey line is hereditarily Lindelof, there exists $y \in Y$ such that $y$ is a limit point of $Y$ (see Corollary 2 in this previous post). Since $y \in Y \subset U$ , $[y,t) \subset U$ for some $t$ . Note that $(y,t) \subset U_0$ , which means that no point of the open interval $(y,t)$ can belong to $Y$ . On the other hand, since $y$ is a limit point of $Y$ , $y<w<t$ for some $w \in Y$ , a contradiction. Thus $Y$ must be countable. It follows that $U$ is the union of countably many closed subsets of $S$ . $\blacksquare$

Proposition 2
If $X$ is perfect and $Y$ is metrizable, then $X \times Y$ is perfect.

Proof of Proposition 2
Let $X$ be perfect. Let $Y$ be a space with a base $\mathcal{B}=\bigcup_{n=1}^\infty \mathcal{B}_n$ such that each $\mathcal{B}_n$ , in addition to being a collection of basic open sets, is a discrete collection. The existence of such a base is equivalent to metrizability, a well known result called Bing’s metrization theorem (see Theorem 4.4.8 in [1]). Let $U$ be a non-empty open subset of $X \times Y$ . We show that it is an $F_\sigma$ -set in $X \times Y$ . For each $x \in U$ , there is some open subset $V$ of $X$ and there is some $W \in \mathcal{B}$ such that $x \in V \times W$ and $V \times \overline{W} \subset U$ . Thus $U$ is the union of a collection of sets of the form $V \times \overline{W}$ . Thus we have:

$U=\bigcup \mathcal{O} \text{ where } \mathcal{O}=\left\{ V_\alpha \times \overline{W_\alpha}: \alpha \in A \right\}$

for some index set $A$ . For each positive integer $m$ , let $\mathcal{O}_m$ be defined by

$\mathcal{O}_m=\left\{V_\alpha \times \overline{W_\alpha} \in \mathcal{O}: W_\alpha \in \mathcal{B}_m \right\}$

For each $\alpha \in A$ , let $V_\alpha=\bigcup_{n=1}^\infty V_{\alpha,n}$ where each $V_{\alpha,n}$ is a closed subset of $X$ . For each pair of positive integers $n$ and $m$ , define $\mathcal{O}_{n,m}$ by

$\mathcal{O}_{n,m}=\left\{V_{\alpha,n} \times \overline{W_\alpha}: V_\alpha \times \overline{W_\alpha} \in \mathcal{O}_m \right\}$

We claim that each $\mathcal{O}_{n,m}$ is a discrete collection of sets in the space $X \times Y$ . Let $(a,b) \in X \times Y$ . Since $\mathcal{B}_m$ is discrete, there exists some open subset $H_b$ of $Y$ with $b \in H_b$ such that $H_b$ can intersect at most one $\overline{W}$ where $W \in \mathcal{B}_m$ . Then $X \times H_b$ is an open subset of $X \times Y$ with $(a,b) \in X \times H_b$ such that $X \times H_b$ can intersect at most one set of the form $V_{\alpha,n} \times \overline{W_\alpha}$ . Then $C_{n,m}=\bigcup \mathcal{O}_{n,m}$ is a closed subset of $X \times Y$ . It is clear that $U$ is the union of $C_{n,m}$ over all countably many possible pairs $n,m$ . Thus $U$ is an $F_\sigma$ -set in $X \times Y$ . $\blacksquare$

Proposition 3
The Sorgenfrey plane $S \times S$ is perfect.

Proof of Proposition 3
To get ready for the proof, consider the product spaces $X_1=\mathbb{R} \times S$ and $X_2=S \times \mathbb{R}$ where $\mathbb{R}$ has the usual topology. By both Proposition 1 and Proposition 2, both $X_1$ and $X_2$ are perfect. Also note that the Sorgenfrey plane topology is finer than the topologies for both $X_1$ and $X_2$ . Thus a closed set in $X_1$ (in $X_2$ ) is also a closed set in $S \times S$ . It follows that any $F_\sigma$ -set in $X_1$ (in $X_2$ ) is also an $F_\sigma$ -set in $S \times S$ .

Let $U$ be a non-empty subset of $S \times S$ . We show that $U$ is a $F_\sigma$ -set. We assume that $U$ is the union of basic open sets of the form $[a,b) \times [a,b)$ . Consider the sets $U_1$ and $U_2$ defined by:

$U_1=\left\{x \in U: \exists \ (a,b) \times [a,b) \text{ such that } x \in (a,b) \times [a,b) \text{ and } (a,b) \times [a,b) \subset U \right\}$

$U_2=\left\{x \in U: \exists \ [a,b) \times (a,b) \text{ such that } x \in [a,b) \times (a,b) \text{ and } [a,b) \times (a,b) \subset U \right\}$

Note that $U_1$ is the interior of $U$ when $U$ is considered as a subspace of $X_1$ . Likewise, $U_2$ is the interior of $U$ when $U$ is considered as a subspace of $X_2$ . Since both $X_1$ and $X_2$ are perfect, $U_1$ and $U_2$ are $F_\sigma$ in $X_1$ and $X_2$ , respectively. Hence both $U_1$ and $U_2$ are $F_\sigma$ -sets in $S \times S$ .

Let $Y=U-(U_1 \cup U_2)$ . We claim that $Y$ is an $F_\sigma$ -set in $S \times S$ . Proposition 3 is established when this claim is proved. To get ready to prove this claim, for each $x=(x_1,x_2) \in S \times S$ , and for each positive integer $k$ , let $B_k(x)$ be the half-open square $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$ . Then $\mathcal{B}(x)=\left\{B_k(x): k=1,2,3,\cdots \right\}$ is a local base at the point $x$ . For each positive integer $k$ , define $Y_k$ by

$Y_k=\left\{y=(y_1,y_2) \in Y: B_k(y) \subset U \right\}$

Clearly $Y=\bigcup_{k=1}^\infty Y_k$ . We claim that each $Y_k$ is closed in $S \times S$ . Suppose $x=(x_1,x_2) \in S \times S-Y_k$ . In relation to the point $x$ , $Y_k$ can be broken into several subsets as follows:

$Y_{k,1}=\left\{y=(y_1,y_2) \in Y_k: y_1=x_1 \text{ and } y_2 \ne x_2 \right\}$

$Y_{k,2}=\left\{y=(y_1,y_2) \in Y_k: y_1 \ne x_1 \text{ and } y_2 = x_2 \right\}$

$Y_{k,\varnothing}=\left\{y=(y_1,y_2) \in Y_k: y_1 \ne x_1 \text{ and } y_2 \ne x_2 \right\}$

Since $x \notin Y_k$ , it follows that $Y_k=Y_{k,1} \cup Y_{k,2} \cup Y_{k,\varnothing}$ . We show that for each of these three sets, there is an open set containing the point $x$ that is disjoint from the set.

Consider $Y_{k,1}$ . If $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$ is disjoint from $Y_{k,1}$ , then we are done. So assume $B_k(x) \cap Y_{k,1} \ne \varnothing$ . Let $t=(t_1,t_2) \in B_k(x) \cap Y_{k,1}$ . Note that $t_1=x_1$ and $t_2 > x_2$ . Now consider the following open set:

$G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_2<t_2 \right\}$

The set $G$ is an open set containing the point $x$ . We claim that $G \cap Y_{k,1}=\varnothing$ . Suppose $g \in G \cap Y_{k,1}$ . Then $g_1=x_1$ and $x_2<g_2<t_2$ . Consider the following set:

$H=B_k(g) \cap \left\{h=(h_1,h_2) \in S \times S: g_2<h_2 \right\}$

Note that $H$ is an open subset of $X_2=S \times \mathbb{R}$ . Since $g \in Y_k$ , it follows that $H \subset B_k(g) \subset U$ . Thus $H$ is a subset of the interior of $U$ (as a subspace of $X_2$ ). We have $H \subset U_2$ . It follows that $t \in H$ since

$x_1=g_1=t_1$

$x_2<g_2<t_2<x_2+\frac{1}{k}<g_2+\frac{1}{k}$

On the other hand, $t \in Y_{k,1} \subset Y_k \subset Y$ . Hence $t \notin U_2$ , a contradiction. Thus the claim that $G \cap Y_{k,1}=\varnothing$ must be true.

The case $Y_{k,2}$ is symmetrical to the case $Y_{k,1}$ . Thus by applying a similar argument, there is an open set containing the point $x$ that is disjoint from the set $Y_{k,2}$ .

Now consider the case $Y_{k,\varnothing}$ . If $B_k(x)=[x_1,x_1+\frac{1}{k}) \times [x_2,x_2+\frac{1}{k})$ is disjoint from $Y_{k,\varnothing}$ , then we are done. So assume $B_k(x) \cap Y_{k,\varnothing} \ne \varnothing$ . Let $t=(t_1,t_2) \in B_k(x) \cap Y_{k,\varnothing}$ . Note that $t_1>x_1$ and $t_2 > x_2$ . Now consider the following open set:

$G=B_k(x) \cap \left\{y=(y_1,y_2) \in S \times S: y_1<t_1 \text{ and }y_2<t_2 \right\}$

The set $G$ is an open set containing the point $x$ . We claim that $G \cap Y_{k,\varnothing}=\varnothing$ . Suppose $g \in G \cap Y_{k,\varnothing}$ . Then $x_1<g_1<t_1$ and $x_2<g_2<t_2$ . Consider the following set:

$H=B_k(g) \cap \left\{h=(h_1,h_2) \in S \times S: g_2<h_2 \right\}$

As in the previous case, $H$ is an open subset of $X_2=S \times \mathbb{R}$ . Since $g \in Y_k$ , it follows that $H \subset B_k(g) \subset U$ . As before, $H \subset U_2$ . We also have a contradiction in that $t \in H$ (based on the following)

$x_1<g_1<t_1<x_1+\frac{1}{k}<g_1+\frac{1}{k}$

$x_2<g_2<t_2<x_2+\frac{1}{k}<g_2+\frac{1}{k}$

and on the one hand and $t \in Y_{k,\varnothing} \subset Y=U-(U_1 \cup U_2)$ . Thus the claim that $G \cap Y_{k,\varnothing}=\varnothing$ is true. Take the intersection of the three open sets from the three cases, we have an open set containing $x$ that is disjoint from $Y_k$ . Thus $Y_k$ is closed in $S \times S$ and $Y=\bigcup_{k=1}^\infty Y_k$ is $F_\sigma$ in $S \times S$ . $\blacksquare$

Remarks
The authors of [2] showed that any finite power of the Sorgenfrey line is perfect. The proof in [2] is an inductive proof: if $S^n$ is perfect, then $S^{n+1}$ is perfect. We take the inductive proof in [2] and adapt it for the Sorgenfrey plane. The authors in [2] also proved that for a sequence of spaces $X_1,X_2,X_3,\cdots$ such that the product of any finite number of these spaces is perfect, the product $\prod_{n=1}^\infty X_n$ is perfect. Then $S^\omega$ is perfect.

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A non-perfect example

Any perfect space is subnormal. Subnormal spaces do not have to be perfect. In fact subnormal non-normal spaces do not have to be perfect. From a perfect space that is not normal (e.g. the Sorgenfrey plane), one can generate a subnormal and non-normal space that is not perfect. Let $X$ be a subnormal and non-normal space. Let $Y$ be a normal space that is not perfectly normal. There are many possible choices for $Y$ . If a specific example is needed, one can take $Y=\omega_1$ with the order topology. Let $X \bigoplus Y$ be the disjoint sum (union) of $X$ and $Y$ . The presence of $Y$ destroys the perfectness. It is clear that any two disjoint closed sets can be separated by disjoint $G_\delta$ -sets.

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Reference

Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

Heath, R. W., Michael, E., A property of the Sorgenfrey line, Compositio Math., 23, 185-188, 1971.

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$\copyright \ 2014 \text{ by Dan Ma}$

Posted in Normal space, Normality in Product, Product space, The Sorgenfrey Line, Well known examples | Tagged G-delta sets, General topology, Non-normal product, Point-set topology, Product space, Subnormal space, The Sorgenfrey Line, Topology | Leave a reply

Normal x compact needs not be subnormal

Posted on May 7, 2014 by Dan Ma

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In this post, we revisit a counterexample that was discussed previously in this blog. A previous post called “Normal x compact needs not be normal” shows that the Tychonoff product of two normal spaces needs not be normal even when one of the factors is compact. The example is $\omega_1 \times (\omega_1+1)$ . In this post, we show that $\omega_1 \times (\omega_1+1)$ fails even to be subnormal. Both $\omega_1$ and $\omega_1+1$ are spaces of ordinals. Thus they are completely normal (equivalent to hereditarily normal). The second factor is also a compact space. Yet their product is not only not normal; it is not even subnormal.

A subset $M$ of a space $Y$ is a $G_\delta$ subset of $Y$ (or a $G_\delta$ -set in $Y$ ) if $M$ is the intersection of countably many open subsets of $Y$ . A subset $M$ of a space $Y$ is a $F_\sigma$ subset of $Y$ (or a $F_\sigma$ -set in $Y$ ) if $Y-M$ is a $G_\delta$ -set in $Y$ (equivalently if $M$ is the union of countably many closed subsets of $Y$ ).

A space $Y$ is normal if for any disjoint closed subsets $H$ and $K$ of $Y$ , there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$ . A space $Y$ is subnormal if for any disjoint closed subsets $H$ and $K$ of $Y$ , there exist disjoint $G_\delta$ subsets $V_H$ and $V_K$ of $Y$ such that $H \subset V_H$ and $K \subset V_K$ . Clearly any normal space is subnormal.

A space $Y$ is pseudonormal if for any disjoint closed subsets $H$ and $K$ of $Y$ (one of which is countable), there exist disjoint open subsets $U_H$ and $U_K$ of $Y$ such that $H \subset U_H$ and $K \subset U_K$ . The space $\omega_1 \times (\omega_1+1)$ is pseudonormal (see this previous post). The Sorgenfrey plane is an example of a subnormal space that is not pseudonormal (see here). Thus the two weak forms of normality (pseudonormal and subnormal) are not equivalent.

The same two disjoint closed sets that prove the non-normality of $\omega_1 \times (\omega_1+1)$ are also used for proving non-subnormality. The two closed sets are:

$H=\left\{(\alpha,\alpha): \alpha<\omega_1 \right\}$

$K=\left\{(\alpha,\omega_1): \alpha<\omega_1 \right\}$

The key tool, as in the proof for non-normality, is the Pressing Down Lemma ([1]). The lemma has been used in a few places in this blog, especially for proving facts about $\omega_1$ (e.g. this previous post on the first uncountable ordinal). Lemma 1 below is a lemma that is derived from the Pressing Down Lemma.

Pressing Down Lemma
Let $S$ be a stationary subset of $\omega_1$ . Let $f:S \rightarrow \omega_1$ be a pressing down function, i.e., $f$ satisfies: $\forall \ \alpha \in S, f(\alpha)<\alpha$ . Then there exists $\alpha<\omega_1$ such that $f^{-1}(\alpha)$ is a stationary set.

Lemma 1
Let $L=\left\{(\alpha,\alpha) \in \omega_1 \times \omega_1: \alpha \text{ is a limit ordinal} \right\}$ . Suppose that $L \subset \bigcap_{n=1}^\infty O_n$ where each $O_n$ is an open subset of $\omega_1 \times \omega_1$ . Then $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty O_n$ for some $\gamma<\omega_1$ .

Proof of Lemma 1
For each $n$ and for each $\alpha<\omega_1$ where $\alpha$ is a limit, choose $g_n(\alpha)<\alpha$ such that $[g_n(\alpha),\alpha] \times [g_n(\alpha),\alpha] \subset O_n$ . The function $g_n$ can be chosen since $O_n$ is open in the product $\omega_1 \times \omega_1$ . By the Pressing Down Lemma, for each $n$ , there exists $\gamma_n < \omega_1$ and there exists a stationary set $S_n \subset \omega_1$ such that $g_n(\alpha)=\gamma_n$ for all $\alpha \in S_n$ . It follows that $[\gamma_n,\omega_1) \times [\gamma_n,\omega_1) \subset O_n$ for each $n$ . Choose $\gamma<\omega_1$ such that $\gamma_n<\gamma$ for all $n$ . Then $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset O_n$ for each $n$ . $\blacksquare$

Theorem 2
The product space $\omega_1 \times (\omega_1+1)$ is not subnormal.

Proof of Theorem 2
Let $H$ and $K$ be defined as above. Suppose $H \subset \bigcap_{n=1}^\infty U_n$ and $K \subset \bigcap_{n=1}^\infty V_n$ where each $U_n$ and each $V_n$ are open in $\omega_1 \times (\omega_1+1)$ . Without loss of generality, we can assume that $U_n \cap (\omega_1 \times \left\{\omega_1 \right\})=\varnothing$ , i.e., $U_n$ is open in $\omega_1 \times \omega_1$ for each $n$ . By Lemma 1, $[\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty U_n$ for some $\gamma<\omega_1$ .

Choose $\beta>\gamma$ such that $\beta$ is a successor ordinal. Note that $(\beta,\omega_1) \in \bigcap_{n=1}^\infty V_n$ . For each $n$ , there exists some $\delta_n<\omega_1$ such that $\left\{\beta \right\} \times [\delta_n,\omega_1] \subset V_n$ . Choose $\delta<\omega_1$ such that $\delta >\delta_n$ for all $n$ and that $\delta >\gamma$ . Note that $\left\{\beta \right\} \times [\delta,\omega_1) \subset \bigcap_{n=1}^\infty V_n$ . It follows that $\left\{\beta \right\} \times [\delta,\omega_1) \subset [\gamma,\omega_1) \times [\gamma,\omega_1) \subset \bigcap_{n=1}^\infty U_n$ . Thus there are no disjoint $G_\delta$ sets separating $H$ and $K$ . $\blacksquare$

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Reference

Kunen, K., Set Theory, An Introduction to Independence Proofs, First Edition, North-Holland, New York, 1980.

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$\copyright \ 2014 \text{ by Dan Ma}$

Posted in First uncountable ordinal, Normal space, Normality in Product, Omega 1, Well known examples | Tagged G-delta sets, General topology, Non-normal product, Normal space, Omega 1, Point-set topology, Pressing Down Lemma, Product space, Topology | Leave a reply

An exercise involving non-normal spaces

Posted on February 22, 2014 by Dan Ma

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A space is normal if any two disjoint closed subsets of the space can be separated by disjoint open sets. A space is pseudonormal if any two disjoint closed subsets of the space, one of which is countable, can be separated by disjoint open sets. In this post, we present an interesting exercise that deals with non-normal spaces:

Take a space that is not normal. Then determine whether it is pseudonormal. You can supply your own examples or you can start with several non-normal spaces listed below. Once you have a list, determine which ones are psuedonormal and which ones are not.

To make the exercise more interesting, we propose that the focus is on spaces that are $T_1$ (i.e. singleton sets are closed) and regular. Since regular Lindelof spaces are normal, we will be certain that any non-normal (and regular) space is not Lindelof.

In the previous post called Pseudonormal spaces, we identify four spaces that are known to be non-normal. Three of these spaces are not normal because one countable closed set and another closed set cannot be separated, hence not pseudonormal (one is the Sorgenfrey plane and one is the Niemmytzkis’ plane). The fourth non-normal space is pseudonormal.

Here’s a list of several other non-normal spaces previously discussed in this blog.

The Tychonoff Plank.

See this post.

The sigma-product of $\omega_1$ many copies of $\omega_1+1$ .

See this post.

The product space $\omega^{\omega_1}$ .

See this post.

The product of the Michael line and the space of irrationals.

See this post.

The product of countably many copies of the Michael line.

See this post.

The product of a Lindelof space and a Bernstein set.

See this post.

The Pixley-Roy space $\mathcal{F}[\mathbb{R}]$ .

See this post.

Mrowka space, defined on a maximal almost disjoint family of subsets of $\omega$ .

See this post.

Readers are welcome to submit other examples of non-normal spaces. Submit examples by entering a comment below. Submitted examples that are different from the ones listed above will be appended to this post.

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$\copyright \ 2014 \text{ by Dan Ma}$

Posted in Basic Topology, Normal space, Well known examples | Tagged General topology, Normal space, Point-set topology, Pseudonormal space, Topology | Leave a reply

Pseudonormal spaces

Posted on February 17, 2014 by Dan Ma

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When two disjoint closed sets in a topological space cannot be separated by disjoint open sets, the space fails to be a normal space. When one of the two closed sets is countable, the space fails to satisfy a weaker property than normality. A space $X$ is said to be a pseudonormal space if $H$ and $K$ can always be separated by two disjoint open sets whenever $H$ and $K$ are disjoint closed subsets of $X$ and one of them is countable. In this post, we discuss several non-normal spaces that actually fail to be pseudonormal. We also give an example of a pseudonormal space that is not normal.

We work with spaces that are at minimum $T_1$ spaces, i.e., spaces in which singleton sets are closed. Then any pseudonormal space is regular. To see this, let $X$ be $T_1$ and pseudonormal. For any closed subset $C$ of $X$ and for any point $x \in X-C$ , we can always separate the disjoint closed sets $\left\{ x \right\}$ and $C$ by disjoint open sets. This is one reason why we insist on having $T_1$ separation axiom as a starting point. We now show some examples of spaces that fail to be pseudonormal.

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Some Non-Pseudonormal Examples

All three examples in this section are spaces where the failure of normality is exhibited by the inability of separating a countable closed set and another disjoint closed set.

Example 1
This example of a non-normal space that fails to be pseudonormal is defined in the previous post called An Example of a Completely Regular Space that is not Normal. This is an example of a Hausdorff, locally compact, zero-dimensional (having a base consisting of closed and open sets), metacompact, completely regular space that is not normal. We state the definition of the space and present a proof that it is not pseudonormal.

Let $E$ be the set of all points $(x,y) \in \mathbb{R} \times \mathbb{R}$ such that $y \ge 0$ . For each real number $x$ , define the following sets:

$V_x=\left\{(x,y) \in E: 0 \le y \le 2 \right\}$

$D_x=\left\{(s,s-x) \in E: x \le s \le x+2 \right\}$

$O_x=V_x \cup D_x$

The set $V_x$ is the vertical line of height 2 at the point $(x,0)$ . The set $D_x$ is the line originating at $(x,0)$ and going in the Northeast direction reaching the same vertical height as $V_x$ as shown in the following figure.

The topology on $E$ is defined by the following:

Each point $(x,y) \in E$ where $y>0$ is isolated.

For each point $(x,0) \in E$ , a basic open set is of the form $O_x - F$ where $(x,0) \notin F$ and $F$ is a finite subset of $O_x$ .

The x-axis in this example is a closed and discrete set of cardinality continuum. Amy two disjoint subsets of the x-axis are disjoint closed sets. The two closed sets that cannot be separated are:

$H=\left\{(x,0) \in E: x \text{ is rational} \right\}$

$K=\left\{(x,0) \in E: x \text{ is irrational} \right\}$

For each $(x,0)$ , let $W_x=O_x-F_x$ where $F_x \subset O_x$ is finite and $(x,0) \notin F_x$ . Furthermore, break up $F_x$ by letting $F_{x,d}=F_x \cap D_x$ and $F_{x,v}=F_x \cap V_x$ . Let $U$ and $V$ be defined by:

$U_H=\bigcup \limits_{(x,0) \in H} W_x$

$U_K=\bigcup \limits_{(x,0) \in K} W_x$

The open sets $U_H$ and $U_K$ are essentially arbitrary open sets containing $H$ and $K$ respectively. We claims that $U_H \cap U_K \ne \varnothing$ .

Define the projection map $\tau_1:\mathbb{R}^2 \rightarrow \mathbb{R}$ by $\tau_1(x,y)=x$ . Let $A$ and $B$ be defined by:

$A=\bigcup \left\{\tau_1(F_{x,d}): (x,0) \in H \right\}$

$B=\left\{(x,0) \in K: (x,0) \notin A \right\}$

The set $A$ is countable. So the set $B$ is uncountable. Choose $(x,0) \in B$ . Choose $(a,0) \in H$ on the left of $(x,0)$ and close enough to $(x,0)$ such that $V_x \cap D_a=\left\{t \right\}$ and $t \notin F_{x,v}$ . This means that

$t \in V_x \cup D_x -F_x=O_x-F_x=W_x$

$t \in V_a \cup D_a -F_a=O_a-F_a=W_a$ .

Thus $U_H \cap U_K \ne \varnothing$ . We have shown that the space $E$ is not pseudonormal and thus not normal.

Example 2
The Sorgenfrey line is the real line $\mathbb{R}$ topologized by the base consisting of half open and half closed intervals of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x < b \right\}$ . In this post, we use $S$ to denote the real line $\mathbb{R}$ with this topology.

The Sorgenfrey line $S$ is a classic example of a normal space whose square $S \times S$ is not normal. In the Sorgenfrey plane $S \times S$ , the set $\left\{(x,-x) \in S \times S: x \in \mathbb{R} \right\}$ is a closed and discrete set and is called the anti-diagonal. The proof presented in this previous post shows that the following two disjoint closed subsets of $S \times S$

$H=\left\{(x,-x) \in S \times S: x \text{ is rational} \right\}$

$K=\left\{(x,-x) \in S \times S: x \text{ is irrational} \right\}$

cannot be separated by disjoint open sets. The argument is based on the fact that the real line with the usual topology is of second category. The key point in the argument is that the set of the irrationals cannot be the union of countably many closed and nowhere dense sets (in the usual topology of the real line).

Thus $S \times S$ fails to be pseudonormal. This example shows that normality can fail to be preserved by taking Cartesian product in such a way that even pseudonormality cannot be achieved in the Cartesian product!

Example 3
Another example of a non-normal space that fails to be pseudonormal is the Niemmytzkis’ plane (Example 2 in in this previous post). The underlying set is $N=\left\{(x,y) \in \mathbb{R} \times \mathbb{R}: y \ge 0 \right\}$ . The points lying above the x-axis have the usual Euclidean open neighborhoods. A point $(x,0)$ in the x-axis has as neighborhoods $\left\{(x,0) \right\}$ together with the interior of a disc in the upper half plane that is tangent at the point $(x,0)$ . Consider the following the two disjoint closed sets on the x-axis:

$H=\left\{(x,0): x \text{ is rational} \right\}$

$K=\left\{(x,0): x \text{ is irrational} \right\}$

The disjoint closed sets $H$ and $K$ cannot be separated by disjoint open sets (see Niemytzki’s Tangent Disc Topology in [2], Example 82). Like Example 2 above, the argument that $H$ and $K$ cannot be separated is also a Baire category argument.

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An Example of Pseudonormal but not Normal

Example 4
One way to find such a space is to look for spaces that are non-normal and see which one is pseudonormal. On the other hand, in a pseudonormal space, countable closed sets are easily separated from other disjoint closed sets. One space in which “countable” is nice is the first uncountable ordinal $\omega_1$ with the order topology. But $\omega_1$ is normal. So we look at the Cartesian product $\omega_1 \times (\omega_1 +1)$ . The second factor is the successor ordinal to $\omega_1$ or as a space that is obtained by tagging one more point to $\omega_1$ that is considered greater than all the points in $\omega_1$ . Let’s use $X \times Y=\omega_1 \times (\omega_1 +1)$ to denote this space.

The space $X \times Y$ is not normal (shown in this previous post). In the previous post, $X \times Y$ is presented as an example showing that the product of a normal space with a compact space needs not be normal. However, in this case at least, the product is pseudonormal.

Let $\alpha < \omega_1$ . Then the square $\alpha \times \alpha$ as a subspace of $X \times Y$ is a countable space and a first countable space. So it has a countable base (second countable) and thus metrizable, and in particular normal. Any countable subset of $X \times Y$ is contained in one of these countable squares, making it easy to separate a countable closed set from another closed set.

Let $H$ and $K$ be disjoint closed sets in $X \times Y$ such that $H$ is countable. Then there is some successor ordinal $\mu < \omega_1$ ( $\mu=\alpha+1$ for some ordinal $\alpha<\omega_1$ ) such that $H \subset \mu \times \mu$ . Based on the discussion in the preceding paragraph, there are disjoint open sets $O_H$ and $O_K$ in $\mu \times \mu$ such that $H \subset O_H$ and $(K \cap (\mu \times \mu)) \subset O_K$ . With $\mu$ being a successor ordinal, the square $\mu \times \mu$ is both closed and open in $X \times Y$ . Then the following sets

$V_H=O_H$

$V_K=O_K \cup (X \times Y-\mu \times \mu)$

are disjoint open sets in $X \times Y$ separating $H$ and $K$ .

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Some Comments about Examples 1 – 3

In each of Examples 1, 2 and 3 discussed above, there is a closed and discrete set of cardinality continuum (the x-axis in Examples 1 and 3 and the anti-diagonal in Example 2). So the extent of each of these three spaces is continuum. Note that the extent of a space is the maximum cardinality of a closed and discrete subset.

In each of these examples, it just so happens that it is not possible to separate the rationals from the irrationals in the x-axis or the anti-diagonal by disjoint open sets, making each example not only not normal but also not pseudonormal.

What if we consider a smaller subset of the x-axis or anti-diagonal? For example, consider an uncountable set of cardinality less than continuum. Then what can we say about the pseudonormality or normality of the resulting subspaces? For Example 1, the picture is clear cut.

In Example 1, the argument that $H$ and $K$ cannot be separated is a “countable vs. uncountable” argument. The argument will work as long as $H$ is a countable dense set in the x-axis (dense in the usual topology) and $K$ is any uncountable set.

For Example 2 and Example 3, the argument that $H$ and $K$ cannot be separated is not a “countable vs. uncountable” argument and instead is a Baire category argument. The fact that one of the closed sets is the irrationals is a crucial point. On the other hand, both Example 2 and Example 3 (especially Example 3) are set-theoretic sensitive examples. For Example 2 and Example 3, the normality of the resulting smaller subspaces is dependent on some extra axioms beyond ZFC. For pseudonormality, it could be set-theoretic sensitive too. We give some indication here why this is so.

Let $S$ be the Sorgenfrey line as in Example 2 above. Assuming Martin’s Axiom and the negation of the continuum hypothesis (abbreviated by MA + not CH), for any uncountable $X \subset S$ with $\lvert X \lvert < c$ , $X \times X$ is normal but not paracompact (see Example 6.3 in [1] and see [3]). Even though $X \times X$ is not exactly a comparable example, this example shows that restricting to a smaller subset on the anti-diagonal seems to make the space normal.

Example 3 has an illustrious history with respect to the normal Moore space conjecture. There is not surprise that extra set-theory axioms are used. For any subset $B$ of the x-axis, let $N(B)$ be the space defined as in Example 3 above except that only points of $B$ are used on the x-axis. Assuming MA + not CH, for any uncountable $B$ that is of cardinality less than continuum, it can be shown that $N(B)$ is normal non-metrizable Moore space (see Example F in [4]). So by assuming extra axiom of MA + not CH, we cannot get a non-pseudonormal example out of Example 3 by restricting to a smaller uncountable subset of the x-axis. Under other set-theoretic axioms, there exists no normal non-metrizable Moore space. Just because this is a set-theoretic sensitive example, it is conceivable that $N(B)$ could be a space that is not pseudonormal under some other axioms.

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Reference

Burke, D. K., Covering Properties, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 347-422, 1984.

Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc, Amsterdam, New York, 1995.

Przymusinski, T. C., A Lindelof space $X$ such that $X \times X$ is normal but not paracompact, Fund. Math., 91, 161-165, 1973.

Tall, F. D., Normality versus Collectionwise Normality, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 685-732, 1984.

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$\copyright \ 2014 \text{ by Dan Ma}$

Posted in Basic Topology, First uncountable ordinal, Normal space, Well known examples | Tagged General topology, Normal space, Omega 1, Point-set topology, Pseudonormal space, The Sorgenfrey Line, Topology | Leave a reply

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