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
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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

The product of the identity map and a quotient map

Posted on April 21, 2023 by Dan Ma

1

The Cartesian product of the identity map and a quotient map can be a quotient map under one circumstance. We prove the following theorem.

Theorem 1
Let $X$ be a locally compact space. Let $q:Y \rightarrow Z$ be a quotient map. Let the map $f:X \times Y \rightarrow X \times Z$ be defined by $f(x,y)=(x,q(y))$ for each $(x,y) \in X \times Y$ . Then the map $f$ is a quotient map from $X \times Y$ to $X \times Z$ .

This is Theorem 3.3.17 in the Engelking topology text [1]. The theorem is attributed to J. H. C. Whitehead. The mapping $f$ defined in Theorem 1 is the Cartesian product of the identity map from $X$ to $X$ and the quotient map from $Y$ onto $Z$ . The theorem gives one circumstance in which the Cartesian product is also a quotient map. That is, taking the product of the identity map from a locally compact space to itself and a quotient map produces a quotient map. Potentially this gives us information about the product of the locally compact space in question and the space that is the quotient image. We give two natural applications of this theorem. Sequential spaces are precisely spaces that are quotient images of metric spaces (see here). The spaces called k-spaces are precisely the quotient images of locally compact spaces (see here). As corollary of Theorem 1, we show that the product of a locally compact metric space and a sequential space is a sequential space. In another corollary, we show that the product of a locally compact space and a k-space is a k-space. We have the following corollaries.

Corollary 2
Let $X$ be a locally compact metric space. Let $Y$ be a sequential space. Then $X \times Y$ is a sequential space.

Corollary 3
Let $X$ be a compact metric space. Let $Y$ be a sequential space. Then $X \times Y$ is a sequential space.

Corollary 4
Let $X$ be a locally compact space. Let $Y$ be a k-space. Then $X \times Y$ is a k-space.

We give a proof of Theorem 1 and discuss the corollaries. We also give some examples.

Proof of Theorem 1

Let $X$ , $Y$ and $Z$ be the spaces described in the statement of Theorem 1, and let $q$ and $f$ be the mappings described in Theorem 1. To show that the map $f$ is a quotient map, we need to show that for any set $O \subset X \times Z$ , $O$ is an open set in $X \times Z$ if and only if $f^{-1}(O)$ is an open set in $X \times Y$ . Because the mapping $f$ is continuous, if $O$ is open in $X \times Z$ , we know that $f^{-1}(O)$ is an open set in $X \times Y$ . We only need to prove the other direction: if $f^{-1}(O)$ is an open set in $X \times Y$ , then $O$ is an open set in $X \times Z$ . To this end, let $f^{-1}(O)$ be an open set in $X \times Y$ and $(a,b) \in O$ . We proceed to find some open set $U \times V \subset X \times Z$ such that $(a,b) \in U \times V \subset O$ .

We choose $c \in q^{-1}(b)$ and an open set $U \subset X$ with $a \in U$ such that $\overline{U}$ is compact and $\overline{U} \times \{ c \} \subset f^{-1}(O)$ . We make the following observation,

(1) for any $y \in Y$ , $\overline{U} \times \{ y \} \subset f^{-1}(O)$ if and only if $\overline{U} \times q^{-1} q(y) \subset f^{-1}(O)$

Let $V=\{ z \in Z: \overline{U} \times q^{-1}(z) \subset f^{-1}(O) \}$ . By observation (1), we have $\overline{U} \times q^{-1}q(c) \subset f^{-1}(O)$ . Note that $q(c)=b$ . Thus, $b \in V$ . As a result, we have $(a,b) \in U \times V \subset O$ . We now need to show $V$ is an open subset of $Z$ . Since $q$ is a quotient mapping, we know $V$ is open in $Z$ if we can show $q^{-1}(V)$ is open in $Y$ . The set $q^{-1}(V)$ is described as follows:

$\displaystyle \begin{aligned} q^{-1}(V)&=\{ y \in Y: q(y) \in V \} \\&=\{y \in Y: \overline{U} \times q^{-1}q(y) \subset f^{-1}(O) \} \\&=\{y \in Y: \overline{U} \times \{ y \} \subset f^{-1}(O) \} \end{aligned}$

The last equality is due to Observation (1). Let $\pi: \overline{U} \times Y \rightarrow Y$ be the projection map. Since $\overline{U}$ is compact, the projection map $\pi$ is a closed map according to the Kuratowski Theorem (see here for its proof). Since $(\overline{U} \times Y) \backslash f^{-1}(O)$ is closed in $\overline{U} \times Y$ , $C=\pi(\overline{U} \times Y \backslash f^{-1}(O))$ is closed in $Y$ and $Y \backslash C$ is open in $Y$ . It can be verified that $q^{-1}(V)=Y \backslash C$ . Thus, $q^{-1}(V)$ is open in $Y$ . As a result, $V$ is open in $Y$ . Furthermore, we have $(a,b) \in U \times V \subset O$ . This establishes that $O$ is open in $X \times Z$ . With that, the mapping $f$ is shown to be a quotient map. $\square$

Corollaries

Proof of Corollary 2
Let $X$ be a locally compact metric space and $Y$ be a sequential space. According to the theorem shown here, $Y$ is the quotient space of a metric space. There exists a metric space $M$ such that $Y$ is the quotient image of $M$ . Let $q:M \rightarrow Y$ be a quotient map from $M$ onto $Y$ . Consider the mapping $f:X \times M \rightarrow X \times Y$ defined by $f(x,y)=(x,q(y))$ for all $(x,y) \in X \times M$ . By Theorem 1, $f$ is a quotient map. Since $X \times Y$ is the quotient image of the metric space $X \times M$ , we establish that $X \times Y$ is a sequential space. $\square$

Corollary 3 follows from Corollary 2 since any compact space is a locally compact space.

Proof of Corollary 4
Let $X$ be a locally compact space and let $Y$ be a k-space. According to the theorem shown here, there is a locally compact space $W$ such that $Y$ is the quotient image of $W$ . Let $q:W \rightarrow Y$ be a quotient map from $W$ onto $Y$ . Define $f:X \times W \rightarrow X \times Y$ by letting $f(x,y)=(x,q(y))$ for all $(x,y) \in X \times W$ . According to Theorem 1, $f$ is a quotient map. Since $X \times Y$ is the quotient image of the locally compact space $X \times W$ , we establish that $X \times Y$ is a k-space. $\square$

Sequential Fans

We illustrate the above corollaries using sequential fans. Sequential fans are sequential spaces. Products of sequential fans may no longer be sequential, in fact, may no longer be countably tight. In some cases, the tightness of a product of two sequential fans is dependent of your favorite set theory axiom (see here). However, the product of a sequential fan and a compact metric space is sequential.

Let $S$ be a convergent sequence including its limit. For convenience, denote $S=\{ q_0,q_1,q_2,\cdots \} \cup \{ q \}$ such that each $q_n$ is isolated and an open neighborhood of $q$ consists of the point $q$ and all but finitely many $q_n$ . To make things more concrete, we can also let $S=\{ 1,\frac{1}{2},\frac{1}{3},\cdots \} \cup \{ 0 \}$ with the usual Euclidean topology. Let $\kappa$ be an infinite cardinal number. Let $M(\kappa)$ be the topological sum of $\kappa$ many copies of $S$ . The space $S(\kappa)$ is defined as $M(\kappa)$ with all the sequential limit points identified as one point called $\infty$ . The space $S(\kappa)$ is called the sequential fan with $\kappa$ many spines. In $S(\kappa)$ , there are $\kappa$ many copies of $S \cup \{ \infty \}$ , which is called a spine.

Note that $M(\kappa)$ is a metric space. Because $S(\kappa)$ is the quotient image of $M(\kappa)$ , the sequential fan $S(\kappa)$ is a sequential space. In fact, $S(\kappa)$ is a Frechet space since it is a sequential space that does not contain a copy of the Arens’ space (see here). For the discussion of the Arens’ space, see here.

According to Corollary 3, the product of the sequential fan $S(\kappa)$ and a compact metric space is a sequential space. In particular, the product $S(\kappa) \times S$ is always a sequential space. According to Corollary 4, $S(\kappa) \times S$ is a k-space. The fact that the product is both a sequential space and a k-space is not surprising. Whenever the spaces $X$ and $Y$ are sequential spaces, the product $X \times Y$ is a sequential space if and only if it is a k-space (see Theorem 2.2 [2]).

Reference

Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.

Tanaka Y., On Quasi-k-Spaces, Proc. Japan Acad., 46, 1074-1079, Berlin, 1970.

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Posted in Mappings | Tagged k-space, Quotient Map, Quotient space, Sequential fan, Sequential space | 1 Reply

Defining Arens’ space using diagrams

Posted on April 16, 2023 by Dan Ma

2

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

Perfect preimages of Lindelof spaces

Posted on April 7, 2023 by Dan Ma

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Let $f:X \rightarrow Y$ be a mapping from a topological space $X$ onto a topological space $Y$ . If $f$ is a perfect map and $Y$ is a Lindelof space, then so is $X$ . If $f$ is a closed map and $Y$ is a paracompact space, then so is $X$ . In other words, the pre-image of a Lindelof space under a perfect map is always a Lindelof space. Likewise, the pre-image of a paracompact space under a closed map is always a paracompact space. After proving these two facts, we show that for any compact space $Y$ , the product $X \times Y$ is Lindelof (paracompact) for any Lindelof (paracompact) space $X$ . All spaces under consideration are Hausdorff.

Another way to state the above two facts is that Lindelofness is an inverse invariance under perfect maps and that paracompactness is an inverse invariance of the closed maps. In general, a topological property is an inverse invariance of a class of mappings $\mathcal{M}$ if the following holds: for any mapping $f:X \rightarrow Y$ belonging to $\mathcal{M}$ , if $Y$ has the property, then so does $X$ . In contrast, a topological property is an invariance of a class of mappings $\mathcal{M}$ if for any mapping $f:X \rightarrow Y$ belonging to $\mathcal{M}$ , if $X$ has the property, then so does $Y$ .

All mappings under consideration are continuous maps. A mapping $f:X \rightarrow Y$ , where $f(X)=Y$ , is a closed map if for any closed subset $A$ of $X$ , $f(A)$ is closed in $Y$ . A mapping $f:X \rightarrow Y$ , where $f(X)=Y$ , is a perfect map if $f$ is a closed map and that the point inverse $f^{-1}(y)$ is compact for each $y \in Y$ .

Perfect mappings and closed mappings are objects with strong properties. Such a map places a restriction on what topological properties the “domain” space or the “range” space can have. The theorems below indicate that it is not possible to map a non-Lindelof space onto a Lindelof space using a perfect map and that it is not possible to map a non-paracompact space onto a paracompact space using a closed map. On the other hand, it is not possible to map a separable metric space onto a separable but non-metric space using a perfect map (see here). We prove the following theorems.

Theorem 1…. Lindelofness (or the Lindelof property) is an inverse invariant of the perfect maps.

Theorem 2 ….Paracompactness is an inverse invariant of the closed maps.

Lemma 3 …. Let $f:X \longrightarrow Y$ be a closed map with $f(X)=Y$ . Let $V$ be an open subset of $X$ . Define $f_*(V)=\{ y \in Y: f^{-1}(y) \subset V \}$ . Then the set $f_*(V)$ is open in $Y$ and that $f_*(V) \subset f(V)$ .

For the proof of Lemma 3, see Lemma 2 here.

Proof of Theorem 1
Let $f:X \longrightarrow Y$ be a perfect map with $f(X)=Y$ . Suppose $Y$ is Lindelof. Let $\mathcal{U}$ be an open cover of $X$ . Without loss of generality, we can assume that $\mathcal{U}$ is closed under finite unions. For each $U \in \mathcal{U}$ , define $f_*(U)=\{ y \in Y: f^{-1}(y) \subset U \}$ . By Lemma 3, each $f_*(U)$ is an open subset of $Y$ . We claim that $\mathcal{V}=\{ f_*(U): U \in \mathcal{U} \}$ is an open cover of $Y$ . To this end, let $y \in Y$ . Since $f$ is a perfect map, the point inverse $f^{-1}(y)$ is compact. As a result, we can find a finite $\mathcal{F} \subset \mathcal{U}$ such that $f^{-1}(y) \subset \bigcup \mathcal{F}=W$ . Since $\mathcal{U}$ is closed under finite unions, $W \in \mathcal{U}$ . It follows that $y \in f_*(W)$ . Since $Y$ is Lindelof, there exists a countable $\{W_0,W_1,W_2,\cdots \} \subset \mathcal{V}$ such that $Y = \bigcup_{n=0}^\infty W_n$ . For each $n$ , $W_n=f_*(U_n)$ for some $U_n \in \mathcal{U}$ . We claim that $\{U_0,U_1,U_2,\cdots \}$ is a cover of $X$ . To this end, let $x \in X$ . Then for some $n$ , $y=f(x) \in W_n=f_*(U_n)$ . This implies that $x \in f^{-1}(y) \subset U_n$ . Thus, the open cover $\mathcal{U}$ has a countable subcover. This concludes the proof of Theorem 1. $\square$

Proof of Theorem 2
Let $f:X \longrightarrow Y$ be a perfect map with $f(X)=Y$ . Suppose $Y$ is paracompact. Let $\mathcal{U}$ be an open cover of $X$ . For each $U \in \mathcal{U}$ , define $f_*(U)$ as in Lemma 3. By Lemma 3, each $f_*(U)$ is an open subset of $Y$ . Let $\mathcal{V}=\{ f_*(U): U \in \mathcal{U} \}$ . As shown in the proof of Theorem 1, $\mathcal{V}$ is an open cover of $Y$ . Since $Y$ is paracompact, there exists a locally finite open refinement $\mathcal{W}$ of $\mathcal{V}$ . Let $\mathcal{W}_0=\{ f^{-1}(W): W \in \mathcal{W} \}$ .

We show three facts about $\mathcal{W}_0$ . (1) It is an open cover of $X$ . (2) It is a locally finite collection in $X$ . (3) It is a refinement of $\mathcal{U}$ . To see (1), note that $\mathcal{W}$ is an open cover of $Y$ . As a result, $\mathcal{W}_0$ is an open cover of $X$ . To see (2), let $x \in X$ . We find an open $O \subset X$ such that $x \in O$ and such that $O$ intersects only finitely many elements of $\mathcal{W}_0$ . Since $\mathcal{W}$ is locally finite in $Y$ , there exists an open $B \subset Y$ such that $y=f(x) \in B$ and such that $B$ intersects only finitely many elements of $\mathcal{W}$ , say, $W_0,W_1,\cdots,W_n$ . Let $O=f^{-1}(B)$ . Clearly, $x \in O$ . It can be verified that the only elements of $\mathcal{W}_0$ having non-empty intersections with $O$ are $f^{-1}(W_0)$ , $f^{-1}(W_1),\cdots$ , $f^{-1}(W_n)$ . To see (3), let $f^{-1}(W) \in \mathcal{W}_0$ where $W \in \mathcal{W}$ . Then $W \subset V=f_*(U)$ for some $V \in \mathcal{V}$ and some $U \in \mathcal{U}$ . We claim that $f^{-1}(W) \subset U$ . Let $x \in f^{-1}(W)$ . Then $y=f(x) \in W \subset V=f_*(U)$ . This implies that $x \in f^{-1}(y) \subset U$ . It follows that $\mathcal{W}_0$ is a locally finite open refinement of the open cover $\mathcal{U}$ . This completes the proof of Theorem 2. $\square$

Productively Paracompact Spaces

A space $X$ is productively paramcompact if $X \times Y$ is paracompact for every paracompact space $Y$ . The definition for productively Lindelof can be stated in a similar way. For some reason, the term “productively paracompact” is not used in the literature but is a topic that had been extensively studied. It is also a topic found in this site. The following four classes of spaces are productively paracompact (see here and here).

Compact spaces

$\sigma$ -compact spaces

Locally compact spaces

$\sigma$ -locally compact spaces

The proof for compact spaces being productively paracompact given here uses the Tube Lemma (see here). As applications of Theorem 1 and Theorem 2, we use the two theorems to show that compact spaces are both productively Lindelof and productive paracompact.

Theorem 4…. Let $Y$ any compact space. Then $X \times Y$ is Lindelof for every Lindelof space $X$ .

Theorem 5 ….Let $Y$ any compact space. Then $X \times Y$ is paracompact for every paracompact space $X$ .

Theorems 4 and 5 are corollaries to the Kuratowski theorem (see here) and Theorems 1 and 2 above. Suppose $Y$ is compact. Then the projection map from $X \times Y$ onto $X$ is a closed map. The paracompactness of $X \times Y$ follows whenever $X$ is paracompact. The projection map is also perfect since the point inverses are compact due to the compactness of the factor $Y$ . Then the Lindelofness of $X \times Y$ follows whenever $X$ is Lindelof.

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Posted in Compact space, Lindelof space, Paracompact space, Product space | Tagged Compact space, Lindelof space, Paracompact space, Perfect mapping | Leave a reply

Perfect images of separable metric spaces

Posted on March 28, 2023 by Dan Ma

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The Bow-Tie space is the continuous image of a separable metric space and yet is not metrizable (see here). Though taking continuous image can fail to preserve separable metrizability, we show that the perfect image of a separable metric space is a separable metric space. We prove the following theorem, which says that under a perfect map the weight will not increase. The result about separable metric space is a corollary of Theorem 1.

Theorem 1
Let $f:X \longrightarrow Y$ be a perfect map onto the space $Y$ . Then $w(Y) \le w(X)$ , i.e., the weight of $Y$ is no greater than the weight of $X$ .

Proof of Theorem 1

All spaces under consideration are Hausdorff. Let $X$ and $Y$ be spaces. Let $f: \longrightarrow Y$ be a map (or function) from $X$ onto $Y$ . The map $f$ is said to be a closed map if $f(C)$ is closed in $Y$ for any closed subset $C$ of $X$ . The map $f$ is a perfect map if $f$ is continuous, $f$ is a closed map, and $f^{-1}(y)$ is compact for every $y \in Y$ . In words, the last condition is that every point inverse is compact. A point inverse $f^{-1}(y)$ is also referred to as a fiber. Thus, we can say that a perfect map is a continuous closed surjective map with compact fibers. The following lemma is helpful for proving Theorem 1.

Lemma 2
Let $f:X \longrightarrow Y$ be a closed map such that $f(X)=Y$ . Let $V \subset X$ be open. Let $f_*(V)=\{ y \in Y: f^{-1}(y) \subset V \}$ . Then $f_*(V)$ is open in $Y$ and $f_*(V) \subset f(V)$ .

Proof of Lemma 2
We show that $Y \backslash f_*(V)$ is closed in $Y$ . To this end, we show $f(X \backslash V)=Y \backslash f_*(V)$ . Note that $f(X \backslash V)$ is closed in $Y$ since $f$ is a closed map. First, we show $f(X \backslash V) \subset Y \backslash f_*(V)$ . Let $t \in f(X \backslash V)$ . Then $t=f(x)$ for some $x \in X \backslash V$ . Since $x \notin V$ and $x \in f^{-1}(t)$ , we have $f^{-1}(t) \not \subset V$ . This implies that $t \notin f_*(V)$ .

We now show that $Y \backslash f_*(V) \subset f(X \backslash V)$ . Let $z \in Y \backslash f_*(V)$ . Since $z \notin f_*(V)$ , $f^{-1}(z) \not \subset V$ . Choose $x \in f^{-1}(z) \backslash V$ , which implies that $x \in X \backslash V$ . Thus, $z=f(x) \in f(X \backslash V)$ .

To complete the proof of the lemma, we show $f_*(V) \subset f(V)$ . Let $w \in f_*(V)$ . We have $f^{-1}(w) \subset V$ . As a result, $w=f(x)$ for some $x \in V$ . $\square$

Proof of Theorem 1
Let $\mathcal{B}$ be a base for $X$ . We derive a base $\mathcal{B}_1$ for $Y$ such that $\lvert \mathcal{B}_1 \lvert \le \lvert \mathcal{B} \lvert$ , i.e., the cardinality of $\mathcal{B}_1$ is no more than the cardinality of $\mathcal{B}$ . This implies that the minimal cardinality of a base in $Y$ is no more than the minimal cardinality of a base in $X$ , i.e., $w(Y) \le w(X)$ .

We assume that the base $\mathcal{B}$ is closed under finite unions. We show that $\mathcal{B}_1=\{ f_*(B): B \in \mathcal{B} \}$ is a base for $Y$ . Note that $f_*$ is defined in Lemma 2. Let $U \subset Y$ be an open set. Let $y \in U$ . For each $x \in f^{-1}(y)$ , choose $B_x \in \mathcal{B}$ such that $x \in B_x$ and $f(B_x) \subset U$ . Since $f^{-1}(y)$ is compact, there exists finite $F \subset f^{-1}(y)$ such that $f^{-1}(y) \subset \bigcup_{t \in F} B_t=B$ . Note that $B \in \mathcal{B}$ . Since $f^{-1}(y) \subset B$ , we have $y \in f_*(B)$ . We also have $f_*(B) \subset f(B) \subset U$ . Thus, every open subset $U$ of $Y$ is the union of elements of $\mathcal{B}_1$ . This means that $\mathcal{B}_1$ is a base for $Y$ . Theorem 1 is established. $\square$

Corollary 3
Let $f:X \longrightarrow Y$ be a perfect map onto the space $Y$ . Then if $X$ is a separable metric space, then $Y$ is a separable metric space.

Comment About Lemma 2
A perfect map is not necessarily an open map. If the perfect map $f$ in Theorem 1 is an open map, then Lemma 2 is not needed and $\mathcal{B}_1=\{ f(B): B \in \mathcal{B} \}$ would be a base for $Y$ . However, we cannot assume $f$ is an open map simply because it is a perfect map. To see this, let $X=\mathbb{R}$ be the real line with the usual topology. Collapse the closed interval $[1,2]$ to one point called $p$ . The resulting quotient space is $Y$ where $Y=(-\infty,1) \cup \{ p \} \cup (2,\infty)$ . In $Y$ , the open neighborhoods of points in $(-\infty,1) \cup (2,\infty)$ are the usual Euclidean neighborhoods. The open neighborhoods of the point $p$ are the usual Euclidean open sets containing the interval $[1,2]$ . The resulting quotient map $f$ is an identity map on $(-\infty,1) \cup (2,\infty)$ and it maps points in $[1,2]$ to the point $p$ . It can be verified that $f$ is a perfect map. For the open set $V=(1,2)$ , $f(V)=\{ p \}$ , which is not open in $Y$ . For the open set $V=(0,1.5)$ , $f(V)=(0,1) \cup \{ p \}$ , which is not open in $Y$ . Lemma 2 says that for any open $X \subset$ , $f(V)$ may not be open but has an open subset $f_*(V)$ if $f(V)$ has non-empty interior. The interior sets $f_*(V)$ can work as a base in $Y$ .

Invariant and Inverse Invariant

Let $\mathcal{P}$ be a property of topological spaces. We say that $\mathcal{P}$ is an invariant of the perfect maps or that $\mathcal{P}$ is invariant under the perfect maps if the property $\mathcal{P}$ is preserved by perfect maps, i.e., for each perfect map $f:X \longrightarrow Y$ where $Y=f(X)$ , if the space $X$ has $\mathcal{P}$ , so does $Y$ . On the other hand, $\mathcal{P}$ is an inverse invariant of the perfect maps if this holds: for each perfect map $f:X \longrightarrow Y$ where $Y=f(X)$ , if $Y$ has $\mathcal{P}$ , so does $X$ .

The notions invariant and inverse invariant defined here are for perfect maps. In general, the notions are much broader and can be defined in relation to any class of continuous maps. For example, we know that the continuous image of a separable space is separable. We can say that separability is an invariant of the continuous maps or that separability is invariant under continuous maps. We can now restate Theorem 1 and Corollary 3 as follows.

Theorem 4…..Restatement of Theorem 1
The property “weight $\le \mathcal{K}$ ” is an invariant of the perfect maps.

Corollary 5
The second axiom of countability is invariant under the perfect maps, but is not an invariant of the continuous maps.

The Bow-Tie space is the continuous image of a separable space but cannot have a countable base (see here). In light of Theorem 1, the continuous map that maps a separable metric space to the Bow-Tie space (shown here) cannot be a perfect map. With respect to that map, the upper half plane in the domain (the separable metric space) is closed but its continuous image in the Bow-Tie space is open and not closed.

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Revised March 31, 2023
Revised March 12, 2024

Posted in Separable metrizable | Tagged Bow-tie space, Invariant, Inverse Invariant, Perfect mapping, Separable metrizable space | 2 Replies

The Bow-Tie Space

Posted on March 26, 2023 by Dan Ma

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We present the Bow-Tie space, which exhibits many interesting properties. The Bow-Tie space is hereditarily Lindelof, and hereditarily separable. It is also the continuous image of a separable metric space. These properties follow from the fact that the Bow-Tie space has a countable network. Furthermore, the Bow-Tie space is neither metrizable nor a Moore space. Thus, the example demonstrates that the continuous image of a separable metric space does not have to be a separable metric space.

Louis F. McAuley introduced the Bow-Tie space as an example of a regular semimetric space that is hereditarily separable, collectionwise normal, completely normal and paracompact, but is not second countable and is not developable, hence not a Moore space [3]. The Bow-Tie space is also discussed in Counterexamples in Topology [5] (see p. 175 Dover Edition). In these two references, the Bow-Tie space is defined as a semimetric space. The version given here is found in [4]. The Bow-Tie space is given in [4] as an example of a cosmic space (i.e., a space with a countable network) that is not an $\aleph_0$ -space.

All spaces under consideration are Hausdorff. Let $X$ be a space. A collection $\mathcal{N}$ of subsets of $X$ is said to be a network for $X$ if for each $x \in X$ and for each open set $O$ containing $x$ , there exists $A \in \mathcal{N}$ such that $x \in A \subset O$ . A network behaves like a base but the elements of the network do not have to be open sets. Of interest are the spaces with a countable network. Compact spaces with a countable network is metrizable. Any space with a countable network is both hereditarily separable and hereditarily Lindelof. The space $X$ has a countable network if and only if $X$ is the continuous image of a separable metric space. Having a countable network is a strong property. See here for a discussion of these facts about spaces with countable network.

The Bow-Tie Space

Let $Y$ be the upper half plane, which is the set of all pairs of real numbers $(x,y)$ with $y>0$ . Let $H$ be the x-axis, which is the set of all pairs of real numbers $(x,y)$ with $y=0$ . The bow-tie space is the set $X=Y \cup H$ with the topology defined as follows.

Open neighborhoods of points in the upper half plane $Y$ are the Euclidean open subsets of $Y$ .

An open neighborhood of a point $p$ in the x-axis $H$ is of the form $B(p,c)$ with $0<c \le 1$ . Each set $B(p,c)$ consists of the point $p$ and all points $q \in X$ having Euclidean distance less than $c$ from $p$ and lying underneath either one of the two straight lines emanating from $p$ with slopes $c$ and $-c$ , respectively.

In the following diagram, $B(p,c)$ is represented by the area in the upper half plane shaded in green plus the points in the x-axis having a distance less than $c$ from $p$ and below the two lines with slopes $c$ and $-c$ .

It is straightforward to verify that the open neighborhoods produce a Hausdorff and regular space. The relative Bow-Tie topologies on $Y$ and $H$ coincide with the Euclidean topologies on $Y$ and $H$ , respectively. Let $\mathcal{B}_1$ and $\mathcal{B}_2$ be countable bases for $Y$ and $H$ in their respective relative Euclidean topologies. Then $\mathcal{N}=\mathcal{B}_1 \cup \mathcal{B}_2$ is a network for the Bow-Tie space $X$ . Thus, the Bow-Tie space has a countable network. Any space with a countable network is Lindelof and separable. The property of having a countable netowrk is hereditary. Thus, the Bow-Tie space is hereditarily Lindelof and hereditarily separable. See here for a discussion of these facts about spaces with countable network.

Any space with a countable network is the continuous image of a separable metric space. In the case of the Bow-Tie space, we can see this directly. Let $Y_1$ be the upper half plane $Y$ with the Euclidean topology. Let $H_1$ be the x-axis $H$ with the Euclidean topology. Let $X_1=Y_1 \bigoplus H_1$ , the free sum or free union. This means that $U \subset X_1$ is open if and only if both $U \cap Y_1$ and $U \cap H_1$ are open. It follows that the identity map from $X_1$ onto the Bow-Tie space $X$ is continuous.

The Bow-Tie space $X$ is separable but not metrizable. We show that $X$ does not have a countable base. Suppose it does. Let $\mathcal{B}$ be a countable base for the Bow-Tie space. We can assume that the elements of $\mathcal{B}$ that contain points of the x-axis $H$ are of the form $B(p,c)$ defined above. Since $\mathcal{B}$ is countable, there can only be countably many $B(p,c)$ in $\mathcal{B}$ , say, $B(p_0,c_0)$ , $B(p_1,c_1)$ , $B(p_2,c_2)$ , $B(p_3,c_3),\cdots$ . Pick $p \in H$ such that $p \ne p_i$ for all $i$ . Consider $B(p,1)$ . Since $\mathcal{B}$ is a base, there must exist some $i$ such that $p \in B(p_i,c_i) \subset B(p,1)$ . This means that both the left side and the right side of the bow-tie in $B(p_i,c_i)$ are within $B(p,1)$ . On the other hand, one side of the bow-tie of $B(p_i,c_i)$ (either the left side or the right side) is above the point $p$ . The points on that side of the bow-tie of $B(p_i,c_i)$ right above point $p$ cannot be part of $B(p,1)$ , a contradiction. Thus, the Bow-Tie space $X$ cannot have a countable base and hence not metrizable. The Bow-Tie space cannot be a Moore space since any Lindelof Moore space must have a countable base.

Not only the Bow-Tie space cannot have a countable base, it also cannot have a point-countable base. For any space, a base is a point-countable base if every point in the space belongs to only countably many elements of the base. In [3] and [5], the Bow-Tie space is defined using a semimetric. Heath [2] showed that every semimetric space with a point-countable base is developable, hence a Moore space if the space is regular. The Bow-Tie space cannot have a point-countable base. If it does, it would be a Moore space.

We mention two more facts about the Bow-Tie space. One is that the Bow-Tie space is a Lindelof $\Sigma$ -space. It is well known that any space with a countable network is a Lindelof $\Sigma$ -space [6]. Secondly, $C_p(X)$ , the function space with the pointwise convergence topology on the Bow-Tie space $X$ is a hereditarily D-space. Gruenhage [1] showed that if $L$ is a Lindelof $\Sigma$ -space, then $C_p(L)$ is a hereditarily D-space.

Reference

Gruenhage, G., A note on D-spaces, Topology and Appl. 152, 2229-2240, 2006.

Heath, R. W., On spaces with point-countable bases, Bull. Acad. Polon. Sci. 13, 393-395, 1965.

McAuley, L. F., A relation between perfect separability, completeness, and normality in semimetric spaces, Pacific J. Math. 6, 315-326, 1956.

Michael, E., $\aleph_0$ -spaces, J. Math. Mech., 15, 983-1002, 1966.

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

Tkachuk, V. V., Lindelof $\Sigma$ -spaces: an omnipresent class, RACSAM, 104 (2), 221-244, 2010.

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Posted in Famous Examples, Separable metrizable | Tagged Bow-tie space, Countable network | 1 Reply

Kuratowski Theorem

Posted on February 6, 2023 by Dan Ma

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We give a proof of Kuratowski theorem. This is the Kuratowki Theorem involving the projection map from a product space $X \times Y$ onto one of the factors. All spaces in consideration are at least Hausdorff.

A map $f:L \rightarrow M$ from the space $L$ into the space $M$ is said to be a closed map if $f(C)$ is closed in $M$ for every closed subset $C$ of $L$ . Here’s the statement of the theorem.

Kuratowski Theorem
The following conditions are equivalent.

The space $Y$ is compact.

For every space $X$ , the projection $\pi:X \times Y \rightarrow X$ is a closed map.

For every normal space $X$ , the projection $\pi:X \times Y \rightarrow X$ is a closed map.

Before going to the proof, note that the theorem is not true when none of the factors is compact. Consider the product space $\mathbb{R} \times \mathbb{R}$ where $\mathbb{R}$ is the real number line. The graph of the curve $\frac{1}{x}$ with $x$ ranges over the non-zero real numbers is a closed subset of $\mathbb{R} \times \mathbb{R}$ . The projection of graph of $\frac{1}{x}$ into the x-axis is the set $(- \infty,0) \cup (0,\infty)$ , which is not closed in $\mathbb{R}$ .

The Proof

The direction $2 \rightarrow 3$ is clear. We prove $1 \rightarrow 2$ and $3 \rightarrow 1$ .

$1 \rightarrow 2$
Let $Y$ be compact. Let $C$ be a closed subset of $X \times Y$ . We show that $X \backslash \pi(C)$ is open in $X$ . To this end, let $x \in X$ and $x \notin \pi(C)$ . Note that $\{x \} \times Y \subset (X \times Y) \backslash C$ . By the Tube Lemma (see here), there exists an open set $O \subset X$ with $x \in O$ such that $\{x \} \times Y \subset O \times Y \subset (X \times Y) \backslash C$ . It follows that $O \cap \pi(C)=\varnothing$ . This means that $X \backslash \pi(C)$ is open and $\pi(C)$ is closed in $X$ .

$3 \rightarrow 1$
Suppose condition 3 holds. We show that $Y$ is compact. Suppose not and let $\mathcal{C}$ be a collection of closed subsets of $Y$ with the finite intersection property such that $\bigcap \mathcal{C}=\varnothing$ (see here for the characterization of compactness using the finite intersection property). Choose a point $p \notin Y$ . Let $X=\{ p \} \cup Y$ . Define a topology on $X$ using the following base:

$\displaystyle \mathcal{B}=\{ \{ y \}: y \in Y \} \cup \mathcal{B}_0$

where sets in $\mathcal{B}_0$ are of the form $\{ p \} \cup (\bigcap_{C \in T} C) \cup A$ where $T \subset \mathcal{C}$ is finite and $A \subset Y$ . In words, we can describe the topology on $X$ in this way. The set $Y$ is a discrete subspace of $X$ and each basic open set containing the point $p$ is the union of three sets – the singleton set $\{ p \}$ . the intersection of finitely many sets in $\mathcal{C}$ and a subset of $Y$ . It follows that the space $X$ is Hausdorff. To show this, we only need to separate $p$ and any point in $Y$ by disjoint open sets. Let $y \in Y$ . Note that $\bigcap \mathcal{C}=\varnothing$ . Thus, $y \notin C$ for some $C \in \mathcal{C}$ . As a result, $\{ y \}$ and $\{p \} \cup C$ are disjoint open sets separating $y$ and $p$ , respectively. It also follows that the space $X$ is normal. To see this, note that any subset of $Y$ is open in $X$ . Let $H$ and $K$ be disjoint closed subsets of $X$ . If the point $p$ is not in either $H$ and $K$ , then $H$ and $K$ are disjoint open sets. Suppose that $p \in H$ . Then $X \backslash K$ and $K$ are open sets separating $H$ and $K$ , respectively.

Let $E=\overline{D}$ where $D=\{ (x,x): x \in Y \} \subset X \times Y$ . By assumption, $\pi(E)$ is closed in $X$ . Note that $Y \subset \pi(E)$ . Since $\overline{Y}=X$ , we have $X=\pi(E)$ . This means that $p \in \pi(E)$ . Furthermore, there exists $t \in Y$ such that $(p,t) \in E$ . We claim that $t \in C$ for all $C \in \mathcal{C}$ . This contradicts that the intersection of $\mathcal{C}$ is empty. Let $C \in \mathcal{C}$ . Let $U \subset Y$ be open such that $t \in U$ . Consider the open set $O=(\{p \} \cup C) \times U$ , which contains the point $(p,t)$ . Since $(p,t) \in E$ , and since $(p,t) \in O$ , the open set $O$ contains points of $D$ . Thus $(x,x) \in O$ for some $x \in Y$ . This means that $x \in C$ and $x \in U$ . What we have just shown is that for every open set $U$ containing $t$ , $U \cap C \ne \varnothing$ for all $C \in \mathcal{C}$ . It follows that $t \in C$ for all $C \in \mathcal{C}$ , producing that contradiction indicated earlier. Thus, every collection of closed subsets of $Y$ with the finite intersection property must have non-empty intersection. This implies that $Y$ is compact. With this observation, the proof of Kuratowski Theorem is complete. $\square$

Perfect Mapping

The function $f:L \rightarrow M$ is a perfect map if $f$ is continuous and a closed map such that each point inverse is compact. When each point inverse under the mapping $f$ is compact, the map is said to have compact fibers. A corollary of Kuratowski Theorem is that the projection map in Kuratowski Theorem is also a perfect map.

Theorem 1
Let $X$ be a space and $Y$ be a compact space. Then the projection $\pi:X \times Y \rightarrow X$ is a perfect map.

The projection map $\pi:X \times Y \rightarrow X$ is always continuous. According to Kuratowski Theorem, it is a closed map. For each $x \in X$ , $\pi^{-1}(x)=\{ x \} \times Y$ , which is compact. Thus, each point inverse of the projection map is compact. $\square$

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Revised April 5, 2023

Posted in Product space | Tagged Closed Mapping, Perfect mapping | 2 Replies

A Pixley-Roy Meta-Lindelof Space

Posted on November 1, 2022 by Dan Ma

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The discussion in the preceding post pointed to a natural question of whether we have a theorem of CCC + meta-Lindelof $\rightarrow$ Lindelof or there is a counterexample. In this post, we present a counterexample. The example is a Pixley-Roy hyperspace $\mathcal{F}[X]$ , which is always metacompact, hence meta-Lindelof. We then make sure the ground $X$ is chosen appropriately to ensure that the Pixley-Roy space has the CCC and is not Lindelof.

The following diagram is shown in the preceding post.

$\displaystyle \begin{aligned} & \\& \bold P \bold a \bold r \bold a \bold c \bold o \bold m \bold p \bold a \bold c \bold t \ \ \ \ \ \leftarrow \ \ \ \ \ \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f \\&\ \ \ \ \ \ \ \ \ \downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow\\& \bold M \bold e \bold t \bold a \bold c \bold o \bold m \bold p \bold a \bold c \bold t \ \ \ \ \rightarrow \ \ \ \bold M \bold e \bold t \bold a \bold - \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f \end{aligned}$

In the diagram, if there is an up arrow from meta-Lindelof to Lindelof, then all 4 notions in the diagram are equivalent. But we know Lindelofness and paracompactness are not equivalent. Thus, there must be meta-compact spaces that are not Lindelof. Such examples can be found in the preceding post. On the other hand, among separable spaces, meta-Lindelof implies Lindelof. Thus, among separable spaces, the diagram has a closed loop, implying all 4 notions are equivalent. Naturally, we would like to weaken the separability to the countable chain condition (CCC). Would the diagram be a closed loop for CCC spaces? The answer is no. The counterexample is a Pixley-Roy space as discussed below.

Pixley-Roy Spaces

All spaces are at least Hausdorff. Let $X$ be a space. Let $\mathcal{F}[X]$ be the set of all non-empty finite subsets of $X$ . For any $F \in \mathcal{F}[X]$ and any open $U \subset X$ with $F \subset U$ , define $[F,U]$ as follows:

$[F,U]=\{ D \in \mathcal{F}[X]: F \subset D \subset U \}$

The set of all possible $[F,U]$ is a base for a topology on $\mathcal{F}[X]$ . This topology is called the Pixley-Roy topology (or Pixley-Roy hyperspace topology). The set $\mathcal{F}[X]$ with this topology is called a Pixley-Roy space or Pixley-Roy hyperspace. The space $X$ that defines the hyperspace is called the ground space.

We are interested in a Pixley-Roy space $\mathcal{F}[X]$ that has the CCC, is meta-Lindelof and not Lindelof. The space $\mathcal{F}[X]$ is always metacompact, hence meta-Lindelof whenever the ground space $X$ is Hausdorff. To obtain the desired example, we only need to use a gound space $X$ that is uncountable (to ensure that the hyperspace is not Lindelof) and has a countable network (to ensure that the hyperspace has the CCC). This preceding post discusses the needed facts as well as other basic facts of $\mathcal{F}[X]$ . In the remainder of this post, we show that if the ground space $X$ has a countable network, then the Pixley-Roy space has the CCC.

Spaces with Countable Networks

A collection $\mathcal{N}$ of subsets of the space $X$ is said to be a network if for every $x \in X$ , and for every open $U \subset X$ with $x \in U$ , there exists $N \in \mathcal{N}$ such that $x \in N \subset U$ . This sounds like the definition of a base for a topology. Note that the sets in the network $\mathcal{N}$ do not have to be open. Thus, a network is not necessarily a base. Though not as strong as having a countable base, a space having a countable network is a strong property. For example, a space with a countable network is both hereditarily separable and hereditarily Lindelof. The Lindelof property can fail to be preserved when taking product. The property of having a countable network is preserved by taking countable product.See here for a discussion of spaces with countable network.

We now show that if the ground space $X$ has a countable network, then the Pixley-Roy hyperspace $\mathcal{F}[X]$ has the countable chain condition (CCC), which means that every pairwise disjoint collection of open sets must be countable. Let $\mathcal{N}$ be a countable network for the ground space $X$ . Let $\{ [F_\alpha,U_\alpha]: \alpha \in \omega_1 \}$ be an uncountable collection of open sets in $\mathcal{F}[X]$ . We show that there must exist 2 sets from this collection having non-empty intersection.

To make the argument clear, we can assume that the network $\mathcal{N}$ is closed under finite intersection. If not, we can just make sure that $\mathcal{N}$ include all finite intersections its elements. For each $\alpha$ , there exists $B_\alpha \in \mathcal{N}$ such that $F_\alpha \subset B_\alpha \subset U_\alpha$ . Since $\mathcal{N}$ is countable, there must exists some $B \in \mathcal{N}$ such that $B=B_\alpha$ for uncountably many $\alpha \in \omega_1$ . Consider two such, say $\beta$ and $\gamma$ . Then we have $F_\beta \subset B \subset U_\beta$ and $F_\gamma \subset B \subset U_\gamma$ . Note that the finite set $F_\beta \cup F_\gamma$ belongs to both $[F_\beta,U_\beta]$ and $[F_\gamma,U_\gamma]$ . This completes the proof that the Pixley-Roy space $\mathcal{F}[X]$ satisfies the CCC if the ground space has a countable network.

Counterexamples

Based on the above discussion, for any uncountable Hausdorff space $X$ with a countable network, the Pixley-Roy space $\mathcal{F}[X]$ is a CCC meta-Lindelof space that is not Lindelof. In fact, we can simply one such space that has a countable base (any base is a network). As a concrete example, we can use the real line as the ground space $X$ . Thus, the Pixley-Roy space $\mathcal{F}[\mathbb{R}]$ is a CCC meta-Lindelof space that is not Lindelof.

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Posted 11/1/2022
Updated 11/4/2022

Posted in Lindelof space | Tagged Lindelof space, Meta-Lindelof Space | 1 Reply

Examples of Meta-Lindelof Spaces

Posted on October 29, 2022 by Dan Ma

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Paracompact spaces and Lindelof spaces are familiar notions of covering properties. The covering properties resulting from replacing the para with meta in the case of paracompact and adding meta to Lindelof are also spaces that had been extensively studied. This is an introduction of meta-Lindelof spaces by way of examples.

Definitions

All spaces under discussion are Hausdorff and regular. A space $X$ is paracompact if every open cover of $X$ has a locally finite open refinement. A space $X$ is metacompact if every open cover of $X$ has a point-finite open refinement. A space $X$ is Lindelof if every open cover of $X$ has a countable subcover, i.e., every open cover of $X$ has a countable subcollection that is also a cover of $X$ . A space $X$ is meta-Lindelof if every open cover of $X$ has a point-countable open refinement, i.e., every open cover $\mathcal{U}$ of $X$ has a subcollection $\mathcal{V}$ such that $\mathcal{V}$ is a refinement of $\mathcal{U}$ and that $\mathcal{V}$ is a point-countable collection. From the definitions, we have the implications shown in the following diagram.

$\displaystyle \begin{aligned} & \\& \bold P \bold a \bold r \bold a \bold c \bold o \bold m \bold p \bold a \bold c \bold t \ \ \ \ \ \leftarrow \ \ \ \ \ \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f \\&\ \ \ \ \ \ \ \ \ \downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow\\& \bold M \bold e \bold t \bold a \bold c \bold o \bold m \bold p \bold a \bold c \bold t \ \ \ \ \rightarrow \ \ \ \bold M \bold e \bold t \bold a \bold - \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f \end{aligned}$

It is well known that any regular Lindelof space is paracompact (the left arrow at the top of the diagram). The two down arrows and the right arrow at the bottom in the diagram follow from the definitions. If there is an up arrow from meta-Lindelof to Lindelof, the diagram would be a closed loop, which would mean all 4 notions are equivalent. At minimum, there are paramcompact spaces that are not Lindelof. For example, any non-separable metric space is paracompact and not Lindelof. Thus, in general it is impossible for the diagram to be a closed loop. In other words, there must be meta-Lindelof spaces that are not Lindelof. However, with additional assumptions, an up arrow from meta-Lindelof to Lindelof is possible. Theorem 1 below shows that the diagram is a closed loop for separable spaces. It is well known that paracompact spaces with the countable chain condition (CCC) are Lindelof (see here). Perhaps an up arrow from meta-Lindelof to Lindelof is possible for spaces with CCC. We discuss some partial results at the end of this article.

Spaces that are not meta-Lindelof

We present two spaces that are not meta-Lindelof (Example 1 and Example 2). The following theorem will make the examples clear.

Theorem 1
Let $X$ be a separable space. If $X$ is meta-Lindelof, then $X$ is Lindelof.

Proof
Let $\mathcal{U}$ be an open cover of $X$ . Since $X$ is meta-Lindelof, there is a point-countable open refinement $\mathcal{V}$ of $\mathcal{U}$ . Let $A$ be a countable dense subset of $X$ . Let $\mathcal{W}$ be defined by $\mathcal{W}= \{ V \in \mathcal{V}: V \cap A \not = \varnothing \}$ . Since $\mathcal{V}$ is point-countable, each point of $A$ can belong to only countably many $V$ . It follows that $\mathcal{W}$ is countable. Furthermore, $\mathcal{W}=\mathcal{V}$ . The set inclusion $\mathcal{W} \subset \mathcal{V}$ is clear. The set inclusion $\mathcal{V} \subset \mathcal{W}$ follows from the fact that $A$ is a dense subset. To complete the proof, for each $V \in \mathcal{V}$ , choose $U(V) \in \mathcal{U}$ such that $V \subset U(V)$ . Then $\{ U(V): V \in \mathcal{V} \}$ is a countable subcover of $\mathcal{U}$ . This completes the proof that $X$ is Lindelof. $\square$

In the above proof, the countable dense set forces the point-countable open refinement to be countable, thus leading to a countable subcover of the original open cover. We now look at examples.

Example 1
The idea for this example is that any separable space that is known to be non-Lindelof must not be meta-Lindelof due to the above theorem. A handy example is the Sorgenfrey plane $S \times S$ . Recall that the Sorgenfrey line $S$ is the real number line topologized by the base consisting of the half open intervals of the form $[a, b)$ . It is well known that $S$ is hereditarily Lindelof, separable and not metrizable. The Sorgenfrey line $S$ is a classic example of a Lindelof space whose square is not even normal, thus not Lindelof (see here). By Theorem 1, the Sorgenfrey plane is not meta-Lindelof.

Example 2
This is another example that is separable and non-Lindelof, hence not meta-Lindelof. The space is the Mrowka space (or Psi-space), which is defined here. To define the space, let $\mathcal{A}$ be uncountable almost disjoint family of subsets of $\omega$ , that is, for any $A, B \in \mathcal{A}$ with $A \ne B$ , $A \cap B$ is finite. The underlying set is $X=\omega \cup \mathcal{A}$ . The points in $\omega$ are isolated. An open neighborhood of $A \in \mathcal{A}$ is of the form $\{ A \} \cup (A \backslash F)$ where $F \subset \omega$ is a finite set. The space $X$ is not Lindelof since $\mathcal{A}$ is an uncountable closed and discrete subset. It is separable since the set of integers, $\omega$ , is a dense subset. By Theorem 1, the space $X$ must not be meta-Lindelof. This space is a classic example of a space that has a $G_\delta$ -diagonal but is not submetrizable.

Example 3
Example 1 is not normal. Example 2 is not normal if the almost disjoint family $\mathcal{A}$ has cardinality continuum (due to Jones’ Lemma). For Example 2 to be normal, additional set theory axiom is required. Example 3 is a normal example of a space that is not meta-Lindelof.

Consider $X=\omega_1$ , the space of the countable ordinals with the ordered topology. This space is not Lindelof since the open cover consisting of $[0,\alpha]$ , where $\alpha \in \omega_1$ , is an open cover that has no countable subcover. The space $X$ is also not paracompact since some open covers cannot be locally finite. The same idea can show that certain open covers cannot be point-countable open cover. This is due to the pressing down lemma (see here). To see this, for each limit ordinal $\alpha \in \omega_1$ , let $O_\alpha=(f(\alpha), \alpha]$ be an open set containing $\alpha$ . Then the function $f$ is a so called pressing down function. By the pressing down lemma, there exist some $\delta$ such that the set $\{ \alpha: f(\alpha)=\delta \}$ is a stationary subset of $X=\omega_1$ . This implies that the point $\delta$ would belong to uncountably many open sets $O_\alpha$ . Thus, any open cover of $X=\omega_1$ containing the sets $O_\alpha$ cannot be point-countable.

To wrap up the example, let $\mathcal{V}$ be any open refinement of the open cover consisting of the open sets $[0, \alpha]$ . We choose $O_\alpha$ described above such that each $O_\alpha$ is contained in some $V_\alpha \in \mathcal{V}$ . This means that $\mathcal{V}$ cannot be point-countable. Thus, one open cover of $X$ has no point-countable open refinement, showing that the space of all countable ordinals, $\omega_1$ , cannot be meta-Lindelof.

Example 4
This example makes use of Example 3. Consider $X$ be the product space of $\omega_1$ many copies of the real line $\mathbb{R}$ . Each $x \in X$ is a function $x: \omega_1 \rightarrow \mathbb{R}$ . Let $x_\alpha$ denote $x(\alpha)$ for each $\alpha \in \omega_1$ . Let $Y$ be the set of all points $x$ in $X$ where $x_\alpha$ is non-zero for at most countably many $\alpha \in \omega_1$ . This space is called the $\Sigma$ -product of lines. The topology of $Y$ is the inherited product topology from the product space $X$ . The $\Sigma$ -product of separable metric spaces is collectionwise normal (see here). The $\Sigma$ -product of uncountably many spaces, each of which has at least 2 points, always contains a closed copy of the space $\omega_1$ (see here).

Observe that closed subspaces of a meta-Lindelof space are always meta-Lindelof. Thus the $\Sigma$ -product $Y$ defined here cannot be a meta-Lindelof space. This example is revisited in Example 6 below.

Meta-Lindelof but not Lindelof

Example 5
This example is the Michael line $\mathbb{M}$ , which is the real line topologized by making the irrational numbers isolated and letting the rational numbers retain the usual Euclidean open sets (a basic introduction is found here). The usual Euclidean open sets of the real line are also open in the Michael line. Thus, $\mathbb{M}$ is a submetrizable space.

To see that it is meta-Lindelof, let $\mathcal{U}$ be an open cover of the Michael line $\mathbb{M}$ . Choose $\mathcal{N}=\{ U_0,U_1,U_2, \cdots \} \subset \mathcal{U}$ such that $\mathcal{N}$ covers the rational numbers. Let $\mathcal{V}=\{ \{ y \}: y \in \mathbb{R} \backslash \cup \mathcal{N} \} \cup \mathcal{N}$ . Thus $\mathcal{V}$ is a point-countable open refinement of $\mathcal{U}$ . This shows that the Michael line $\mathbb{M}$ is meta-Lindelof. This fact can also be seen from the above diagram, which shows that paracompact $\rightarrow$ metacompact $\rightarrow$ meta-Lindelof. Note that the Michael line is a paracompact space that is not Lindelof.

The Michael line as an example of a meta-Lindelof space is quite informative. Note that in any Lindelof space, all closed and discrete subsets must be countable (such a space is said to have countable extent). However, a meta-Lindelof space can have uncountable closed and discrete subset. There is a closed and discrete subset of the Michael line of cardinality continuum. To see this, note that there is a Cantor set in the real line consisting entirely of irrational numbers. For example, we can construct a Cantor set within the interval $A=[\sqrt{2},\sqrt{8}]$ . Let $r_0,r_1,r_2,r_4,\cdots$ enumerate all rational numbers within this interval. In the first step, we remove a middle part of the interval $A$ such that the two remaining closed intervals have irrational endpoints and will miss $r_0$ . In the $n$ th step of the construction, we remove the middle part of each of the remaining intervals such that all remaining intervals have irrational endpoints and will miss the rational number $r_n$ . Let $C$ be the resulting Cantor set, which consists only of irrational numbers since it misses all rational numbers in $A$ . The set $C$ , as a subset of the Michael line, is closed and discrete.

The above paragraph shows one direction in which the two notions of “Lindelof” diverge. Lindelof spaces have countable extent. On the other hand, meta-Lindelof space can have uncountable closed and discrete subsets.

Example 5 actually presents a template for producing meta-Lindelof spaces. In Example 5, start with the real line with the usual topology. Identify a “Lindelof” part (the rationals) and make the remainder discrete (the irrationals). The template is further described in the following theorem.

Theorem 2
Let $Y$ be a space and let $W$ be a Lindelof subspace of $Y$ . Define a new space $Z$ such that the underlying set is $Y$ and the new topology is defined as follows. Let points $y \in Y \backslash W$ be isolated and let points $x \in W$ retain the original open sets in $Y$ .

Then the space $Z$ is a meta-Lindelof space.

If the original space $Y$ is a non-Lindelof space, then $Z$ is also not Lindelof.

To see $Y$ with the new topology is meta-Lindelof, follow the same proof in showing the Michael line is meta-Lindelof. To prove part 2, let $\mathcal{U}$ be an open cover of $Y$ (in the original topology) such that no countable subcollection of $\mathcal{U}$ is a cover of $Y$ . Let $\mathcal{C}$ be a countable subcollection of $\mathcal{U}$ such that $\mathcal{C}$ covers $W$ . There must be uncountably many points of $Y$ not covered by $\mathcal{C}$ . Then $\mathcal{Q}=\{ \{ y \}: y \in Z \backslash \cup \mathcal{C} \} \cup \mathcal{C} \}$ is an open cover of $Z$ that has no countable subcover.

Example 6
This example starts with the space $Y$ in Example 4. To define $Y$ , let $X=\Pi_{\alpha \in \omega_1} \mathbb{R}$ , the product space of $\omega_1$ many copies of the real line $\mathbb{R}$ . Let $Y$ and $W$ be the subspaces of $X$ defined as follows:

$\displaystyle Y=\{ y \in X: \lvert \{ \alpha \in \omega_1: x(\alpha) \ne 0 \} \lvert \le \omega \}$
$\displaystyle W=\{ y \in X: \lvert \{ \alpha \in \omega_1: x(\alpha) \ne 0 \} \lvert < \omega \}$

The space $Y$ is called the $\Sigma$ -product of lines and is collectionwise normal. Note that the $\Sigma$ -product of separable metric spaces is collectionwise normal (see here). The space $W$ is called the $\sigma$ -product of lines, which consists of all points in the product space $X$ with non-zero values in at most finitely many coordinates. Note that the $\sigma$ -product of separable metric spaces is a Lindelof space (see here).

The $\Sigma$ -product $Y$ is not Lindelof because it contains a closed copy of $\omega_1$ (see here). Thus, $Y$ is a non-Lindelof space with a Lindelof subspace $W$ , which is exactly what is required in Theorem 2. Define the space $Z$ as described in Theorem 2. The resulting $Z$ is meta-Lindelof but not Lindelof.

Note that the Lindelof $W$ is a dense subset of $\Sigma$ -product $Y$ . Thus, though $Y$ is not Lindelof, it has a dense Lindelof subspace. In Example 4, we see that $Y$ is not meta-Lindelof since it contains a non-meta-Lindelof space as a closed subspace. However, re-topologizing it by making the points not in $W$ isolated and making the points in $W$ retain the open sets in the $\Sigma$ -product topology produces a meta-Lindelof space.

Concluding Remarks

A few observations can be made from the six examples discussed above.

Among the separable spaces, meta-Lindelofness and Lindelofness are the same (Theorem 1). This gives a handy way to show that any separable space that is not Lindelof is not meta-Lindelof.

It is possible that the product of Lindelof spaces is not meta-Lindelof (Example 1).

A question can be asked for Example 2, which is a Moore space. Any Lindelof Morre space is second countable. Example 2 is not meta-Lindelof. Must a meta-Lindelof Moore space be second countable?

The first uncountable ordinal $\omega_1$ with the order topology is not paracompact since it is not possible to cover the limit ordinals with a locally finite collection of open sets. The same idea shows that $\omega_1$ does not even satisfy the weaker property of meta-Lindelof (Example 3).

One take away from Example 4 is that meta-Lindelof spaces resemble Lindelof spaces in one respect, that is, meta-Lindelofness is hereditary with respect to closed subspaces. Because the $\Sigma$ -product of the real lines contains a closed copy of $\omega_1$ , it is not meta-Lindelof.

The Michael line (Example 5) demonstrates one critical difference between meta-Lindelof and Lindelof. Any Lindelof space has countable extent. It is different for meta-Lindelof spaces. In the Michael line, there is an uncountable closed and discrete subset. One way to see this is to construct a Cantor set in the real line consisting of only irrational numbers.

The Michael line (Example 5) gives the hint of a recipe for producing meta-Lindelof spaces. The recipe is described in Theorem 2. Example 6 is a demonstration of the recipe. The $\Sigma$ -product discussed inExample 4 is non-Lindelof with a dense Lindelof subspace $W$ . By letting $W$ retain the original $\Sigma$ -product open sets and making the complement of $W$ discrete, we produce another meta-Lindelof space that is not Lindelof.

The rest of the remark focuses on the diagram given earlier (repeated belo

$\displaystyle \begin{aligned} & \\& \bold P \bold a \bold r \bold a \bold c \bold o \bold m \bold p \bold a \bold c \bold t \ \ \ \ \ \leftarrow \ \ \ \ \ \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f \\&\ \ \ \ \ \ \ \ \ \downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow\\& \bold M \bold e \bold t \bold a \bold c \bold o \bold m \bold p \bold a \bold c \bold t \ \ \ \ \rightarrow \ \ \ \bold M \bold e \bold t \bold a \bold - \bold L \bold i \bold n \bold d \bold e \bold l \bold o \bold f \end{aligned}$

Theorem 1 shows that for separable spaces, meta-Lindelof $\rightarrow$ Lindelof. As a result, the diagram becomes a closed loop, meaning all 4 notions in the diagram are equivalent for separable spaces. As pointed out earlier, CCC may be a good candidate for replacing separable since among CCC spaces, paracompactness equates Lindelofness. It turns out that for CCC spaces, meta-Lindelof does not imply Lindelof. A handy example is the Pixley-Roy space $\mathcal{F}[\mathbb{R}]$ , which is non-Lindelof, metacompact (hence meta-Lindelof) and satisfies the CCC (see the follow up discussion in the next post). We found the following partial result in p. 971 of the Handbook of Set-Theoretic Topology (Chapter 22 on Borel Measures by Gardner and Pfeffer).

Theorem 3 (MA + not CH)
Each locally compact meta-Lindelof space satisfying the CCC is Lindelof.

Theorem 3 is Corollary 4.9 in the article in the Handbook. It implies that for locally compact CCC space, the above diagram is a closed loop but only under Martin’s axiom and the negation of CH.

The Handbook was published in 1984. We wonder if there is any update since then. For any reader who has updated information regarding the consistency result in Theorem 3, please kindly comment in the space below.

For more information on meta-Lindelof and other covering properties, see Chapter 9 in the Handbook of Set-Theoretic Topology (the chapter by D. Burke on covering properties).

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Dan Ma topology
Daniel Ma topology
Dan Ma Lindelof spaces
Daniel Ma Lindelof spaces
Dan Ma meta-Lindelof spaces
Daniel Ma meta-Lindelof spaces

$\copyright$ 2022 – Dan Ma

Posted: 10/29/2022
Updated: 10/30/2022
Updated: 11/1/2022

Posted in Lindelof space | Tagged Lindelof space, Meta-Lindelof Space | 1 Reply

Divergence of the sum of the reciprocals of the primes

Posted on May 29, 2022 by Dan Ma

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The sum of the reciprocals of the prime numbers diverges, i.e., the sum $\sum \limits_{p} \frac{1}{p}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\cdots$ , where $p$ ranges over all the primes, diverges. Facts about prime numbers are always interesting, especially a fundamental fact such as this one. With the divergence of the sum of the reciprocals, a corollary is that there are infinitely many primes. The first proof of this fundamental fact about prime numbers was given by Euler in 1737. The proof discussed below is due to Paul Erdős. It is an elegant and nifty proof that earned a place in the book called Proofs from THE BOOK [1]. Erdős’ proof is by way of a contradiction. It is also a simple proof. After assuming the convergence of the sum, the rest of the proof only utilizes basic information about numbers (integers in particular). There is a Wikipedia entry on the divergence of the sum of the reciprocal of primes. We also briefly mention a conjecture of Erdős and the Green-Tao Theorem.

Erdős’ Proof

Let $p_1,p_2,p_3,\cdots$ be the listing of all the prime numbers. Suppose that $\sum \limits_{p_i} \frac{1}{p_i}$ converges. There is some positive integer $m$ such that $\sum_{i \ge m+1} \frac{1}{p_i}<\frac{1}{2}$ . In the remainder of the proof, the prime numbers $p_1,p_2, \cdots,p_m$ are called small primes and $p_{m+1},p_{m+2},\cdots$ are called big primes. For any positive integer $N$ , we have the following inequality:

(1) ……………. $\displaystyle \sum \limits_{i \ge m+1} \frac{N}{p_i}<\frac{N}{2}$

For each $N$ indicated above, let $N_s$ be the number of all positive integers $n \le N$ such that the prime divisors of $n$ are all small primes. Let $N_b$ be the number of all positive integers $n \le N$ such that at least one prime divisor of $n$ is a big prime. By definition, $N_s+N_b=N$ for all $N$ . We show that $N_s+N_b<N$ for some $N$ , a contradiction. To obtain the contradiction, upper bounds are established for both $N_s$ and $N_b$ .

The symbol $\left\lfloor y \right\rfloor$ refers to the floor function, which is greatest integer less than or equal to $y$ . Consider $\left\lfloor \frac{N}{p_i} \right\rfloor$ , which is identical to the number of positive integers $n \le N$ that are multiples of $p_i$ . We now have the following upper bound for $N_b$ .

(2) ……………. $\displaystyle N_b \le \sum \limits_{i \ge m+1} \left\lfloor \frac{N}{p_i} \right\rfloor \le \sum \limits_{i \ge m+1} \frac{N}{p_i} <\frac{N}{2}$

A positive integer $n$ is square-free if its prime decomposition contains no repeated prime factors. For example, $15=3 \times 5$ is square-free while $18=2 \times 3^2$ is not square-free. Note that any positive integer $n$ can be expressed as the product of two parts, one of which is square-free and the other is a square, i.e., $n=a \times b^2$ with $a$ being square-free. If a positive integer is itself square-free, the square part is 1. Otherwise, we can always rearrange the prime decomposition to obtain $n=a \times b^2$ .

We are now ready to examine $N_s$ . For each $n \le N$ that has only small prime divisors, write $n=a_n \times b_n^2$ where $a_n$ is square-free part. Each $a_n$ is a product of small primes. There are only $m$ small primes. Thus, there can be at most $2^m$ many different square-free parts $a_n$ . On the other hand, since $b_n^2 \le n$ , we have $b_n \le \sqrt{n} \le \sqrt{N}$ . The following is an upper bound on $N_s$ .

(3) ……………. $\displaystyle N_s \le 2^m \times \sqrt{N}$

The inequality (3) holds for all positive integers $N$ . We would like to find an $N$ such that $2^m \times \sqrt{N} \le \frac{N}{2}$ or $2^{m+1} \le \sqrt{N}$ . A good choice is $N=2^{2 m+2}$ . With this particular $N$ , inequality (3) becomes:

(4) ……………. $\displaystyle N_s \le 2^m \times \sqrt{N} \le \frac{N}{2}$

Summing (2) and (4) produces $N_b+N_s <N$ , which contradicts the fact that $N_s+N_b=N$ . Thus, we can conclude that the sum of the reciprocals of the primes must diverge.

Erdős conjecture on arithmetic progressions

A brief mention of the connection with a conjecture posed by Erdős. Erdős posed many mathematical problems. One problem related to the topic at hand is the conjecture on arithmetic progressions. It states that if the sum of the reciprocals of a sequence of positive integers diverges, then the sequence contains arbitrarily long arithmetic progressions. The sequence of primes satisfies the hypothesis of the conjecture. Are there arbitrarily long arithmetic progressions of primes? For example, 199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089 is a 10-term arithmetic progression of primes with difference 210. For each positive integer $k$ , are there arithmetic progression of primes of length $k$ ? This problem about primes would be a worthy conjecture on its own. This is actually an old problem that dates back to 1770. In the 20th century, incremental progresses included “there are infinitely many triples of primes in arithmetic progression” (1939) and “there are infinitely many four-term progressions consisting of three primes and a number that is either a prime or semiprime” (1981), according to Wolfram Math World. This conjecture was settled in 2004 by Ben Green and Terrence Tao. The Green_Tao Theorem states that the prime numbers contain arbitrary long arithmetic progressions.

Reference

Aigner, M., Gunter, M.Z., Proofs from THE BOOK, third edition, Springer-Verlag, Berlin, 2004.

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