# Alexandroff Double Circle

We discuss the Alexandroff double circle, which is a compact and non-metrizable space. A theorem about the hereditarily normality of a product space $Y_1 \times Y_2$ is also discussed.

Let $C_1$ and $C_2$ be the two concentric circles centered at the origin with radii 1 and 2, respectively. Specifically $C_i=\left\{(x,y) \in \mathbb{R}^2: x^2 + y^2 =i \right\}$ where $i=1,2$. Let $X=C_1 \cup C_2$. Furthermore let $f:C_1 \rightarrow C_2$ be the natural homeomorphism. Figure 1 below shows the underlying set.

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Figure 1 – Underlying Set

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We define a topology on $X$ as follows:

• Points in $C_2$ are isolated.
• For each $x \in C_1$ and for each positive integer $j$, let $O(x,j)$ be the open arc in $C_1$ whose center contains $x$ and has length $\frac{1}{j}$ (in the Euclidean topology on $C_1$). For each $x \in C_1$, an open neighborhood is of the form $B(x,j)$ where
$\text{ }$

$B(x,j)=O(x,j) \cup (f(O(x,j))-\left\{f(x) \right\}$).

The following figure shows an open neighborhood at point in $C_1$.

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Figure 2 – Open Neighborhood

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A List of Results

It can be verified that the open neighborhoods defined above form a base for a topology on $X$. We discuss the following points about the Alexandroff double circle.

1. $X$ is a Hausdorff space.
2. $X$ is not separable.
3. $X$ is not hereditarily Lindelof.
4. $X$ is compact.
5. $X$ is sequentially compact.
6. $X$ is not metrizable.
7. $X$ is not perfectly normal.
8. $X$ is completely normal (and thus hereditarily normal).
9. $X \times X$ is not hereditarily normal.

The proof that $X \times X$ is not hereditarily normal can be generalized. We discuss this theorem after presenting the proof of Result 9.
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Results 1, 2, 3

It is clear that the Alexandroff double circle is a Hausdorff space. It is not separable since the outer circle $C_2$ consists of uncountably many singleton open subsets. For the same reason, $C_2$ is a non-Lindelof subspace, making the Alexandroff double circle not hereditarily Lindelof. $\blacksquare$

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Result 4

The property that $X$ is compact is closely tied to the compactness of the inner circle $C_1$ in the Euclidean topology. Note that the subspace topology of the Alexandroff double circle on $C_1$ is simply the Euclidean topology. Let $\mathcal{U}$ be an open cover of $X$ consisting of open sets as defined above. Then there are finitely many basic open sets $B(x_1,j_1)$, $B(x_2,j_2)$, $\cdots$, $B(x_n,j_n)$ from $\mathcal{U}$ covering $C_1$. These open sets cover the entire space except for the points $f(x_1), f(x_2), \cdots,f(x_n)$, which can be covered by finitely many open sets in $\mathcal{U}$. $\blacksquare$

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Result 5

A space $W$ is sequentially compact if every sequence of points of $W$ has a subsequence that converges to a point in $W$. The notion of sequentially compactness and compactness coincide for the class of metric spaces. However, in general these two notions are distinct.

The sequentially compactness of the Alexandroff double circle $X$ hinges on the sequentially compactness of $C_1$ and $C_2$ in the Euclidean topology. Let $\left\{x_n \right\}$ be a sequence of points in $X$. If the set $\left\{x_n: n=1,2,3,\cdots \right\}$ is a finite set, then $\left\{x_n: n>m \right\}$ is a constant sequence for some large enough integer $m$. So assume that $A=\left\{x_n: n=1,2,3,\cdots \right\}$ is an infinite set. Either $A \cap C_1$ is infinite or $A \cap C_2$ is infinite. If $A \cap C_1$ is infinite, then some subsequence of $\left\{x_n \right\}$ converges in $C_1$ in the Euclidean topology (hence in the Alexandroff double circle topology). If $A \cap C_2$ is infinite, then some subsequence of $\left\{x_n \right\}$ converges to $x \in C_2$ in the Euclidean topology. Then this same subsequence converges to $f^{-1}(x)$ in the Alexandroff double circle topology. $\blacksquare$

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Result 6

Note that any compact metrizable space satisfies a long list of properties, which include separable, Lindelof, hereditarily Lindelof. $\blacksquare$

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Result 7

A space is perfectly normal if it is normal with the additional property that every closed set is a $G_\delta$-set. For the Alexandroff double circle, the inner circle $C_1$ is not a $G_\delta$-set, or equivalently the outer circle $C_2$ is not an $F_\sigma$-set. To see this, suppose that $C_2$ is the union of countably many sets, we show that the closure of at least one of the sets goes across to the inner circle $C_1$. Let $C_2=\bigcup \limits_{i=1}^\infty T_n$. At least one of the sets is uncountable. Let $T_j$ be one such. Consider $f^{-1}(T_j)$, which is also uncountable and has a limit point in $C_1$ (in the Euclidean topology). Let $t$ be one such point (i.e. every Euclidean open set containing $t$ contains points of $f^{-1}(T_j)$). Then the point $t$ is a member of the closure of $T_j$ (Alexandroff double circle topology). $\blacksquare$

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Result 8

We first discuss the notion of separated sets. Let $T$ be a Hausdorff space. Let $E \subset T$ and $F \subset T$. The sets $E$ and $F$ are said to be separated (are separated sets) if $E \cap \overline{F}=\varnothing$ and $F \cap \overline{E}=\varnothing$. In other words, two sets are separated if each one does not meet the closure of the other set. In particular, any two disjoint closed sets are separated. The space $T$ is said to be completely normal if $T$ satisfies the property that for any two sets $E$ and $F$ that are separated, there are disjoint open sets $U$ and $V$ with $E \subset U$ and $F \subset V$. Thus completely normality implies normality.

It is a well know fact that if a space is completely normal, it is hereditarily normal (actually the two notions are equivalent). Note that any metric space is completely normal. In particular, any Euclidean space is completely normal.

To show that the Alexandroff double circle $X$ is completely normal, let $E \subset X$ and $F \subset X$ be separated sets. Thus we have $E \cap \overline{F}=\varnothing$ and $F \cap \overline{E}=\varnothing$. Note that $E \cap C_1$ and $F \cap C_1$ are separated sets in the Euclidean space $C_1$. Let $G_1$ and $G_2$ be disjoint Euclidean open subsets of $C_1$ with $E \cap C_1 \subset G_1$ and $F \cap C_1 \subset G_2$.

For each $x \in E \cap C_1$, choose open $U_x$ (Alexandroff double circle open) with $x \in U_x$, $U_x \cap C_1 \subset G_1$ and $U_x \cap \overline{F}=\varnothing$. Likewise, for each $y \in F \cap C_1$, choose open $V_y$ (Alexandroff double circle open) with $y \in V_y$, $V_y \cap C_1 \subset G_2$ and $V_y \cap \overline{E}=\varnothing$. Then let $U$ and $V$ be defined by the following:

$U=\biggl(\bigcup \limits_{x \in E \cap C_1} U_x \biggr) \cup \biggl(E \cap C_2 \biggr)$

$\text{ }$

$V= \biggl(\bigcup \limits_{y \in F \cap C_1} V_y \biggr) \cup \biggl(F \cap C_2 \biggr)$

Because $G_1 \cap G_2 =\varnothing$, the open sets $U_x$ and $V_y$ are disjoint. As a result, $U$ and $V$ are disjoint open sets in the Alexandroff double circle with $E \subset U$ and $F \subset V$.

For the sake of completeness, we show that any completely normal space is hereditarily normal. Let $T$ be completely normal. Let $Y \subset T$. Let $H \subset Y$ and $K \subset Y$ be disjoint closed subsets of $Y$. Then in the space $T$, $H$ and $K$ are separated. Note that $H \cap cl_T(K)=\varnothing$ and $K \cap cl_T(H)=\varnothing$ (where $cl_T$ gives the closure in $T$). Then there are disjoint open subsets $O_1$ and $O_2$ of $T$ such that $H \subset O_1$ and $K \subset O_2$. Now, $O_1 \cap Y$ and $O_2 \cap Y$ are disjoint open sets in $Y$ such that $H \subset O_1 \cap Y$ and $K \subset O_2 \cap Y$.

Thus we have established that the Alexandroff double circle is hereditarily normal. $\blacksquare$

For the proof that a space is completely normal if and only if it is hereditarily normal, see Theorem 2.1.7 in page 69 of [1],
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Result 9

We produce a subspace $Y \subset X \times X$ that is not normal. To this end, let $D=\left\{d_n:n=1,2,3,\cdots \right\}$ be a countable subset of $X$ such that $\overline{D}-D\ne \varnothing$. Let $y \in \overline{D}-D$. Let $Y=X \times X-C_1 \times \left\{y \right\}$. We show that $Y$ is not normal.

Let $H=C_1 \times (X-\left\{y \right\})$ and $K=C_2 \times \left\{y \right\}$. These are two disjoint closed sets in $Y$. Let $U$ and $V$ be open in $Y$ such that $H \subset U$ and $K \subset V$. We show that $U \cap V \ne \varnothing$.

For each integer $j$, let $U_j=\left\{x \in X: (x,d_j) \in U \right\}$. We claim that each $U_j$ is open in $X$. To see this, pick $x \in U_j$. We know $(x,d_j) \in U$. There exist open $A$ and $B$ (open in $X$) such that $(x,d_j) \in A \times B \subset U$. It is clear that $x \in A \subset U_j$. Thus each $U_j$ is open.

Furthermore, we have $C_1 \subset U_j$ for each $j$. Based in Result 7, $C_1$ is not a $G_\delta$-set. So we have $C_1 \subset \bigcap \limits_{j=1}^\infty U_j$ but $C_1 \ne \bigcap \limits_{j=1}^\infty U_j$. There exists $t \in \bigcap \limits_{j=1}^\infty U_j$ but $t \notin C_1$. Thus $t \in C_2$ and $\left\{t \right\}$ is open.

Since $(t,y) \in K$, we have $(t,y) \in V$. Choose an open neighborhood $B(y,k)$ of $y$ such that $\left\{t \right\} \times B(y,k) \subset V$. since $y \in \overline{D}$, there exists some $d_j$ such that $(t,d_j) \in \left\{t \right\} \times B(y,k)$. Hence $(t,d_j) \in V$. Since $t \in U_j$, $(t,d_j) \in U$. Thus $U \cap V \ne \varnothing$. $\blacksquare$

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Generalizing the Proof of Result 9

The proof of Result 9 requires that one of the factors has a countable set that is not discrete and the other factor has a closed set that is not a $G_\delta$-set. Once these two requirements are in place, we can walk through the same proof and show that the cross product is not hereditarily normal. Thus, the statement that is proved in Result 9 is the following.

Theorem
If $Y_1$ has a countable subset that is not closed and discrete and if $Y_2$ has a closed set that is not a $G_\delta$-set then $Y_1 \times Y_2$ has a subspace that is not normal.

The theorem can be restated as:

Theorem
If $Y_1 \times Y_2$ is hereditarily normal, then either every countable subset of $Y_1$ is closed and discrete or $Y_2$ is perfectly normal.

The above theorem is due to Katetov and can be found in [2]. It shows that the hereditarily normality of a cross product imposes quite strong restrictions on the factors. As a quick example, if both $Y_1$ and $Y_2$ are compact, for $Y_1 \times Y_2$ to be hereditarily normal, both $Y_1$ and $Y_2$ must be perfectly normal.

Another example. Let $W=\omega_1+1$, the succesor of the first uncountable ordinal with the order topology. Note that $W$ is not perfectly normal since the point $\omega_1$ is not a $G_\delta$ point. Then for any compact space $Y$, $W \times Y$ is not hereditarily normal. Let $C=\omega+1$, the successor of the first infinite ordinal with the order topology (essentially a convergent sequence with the limit point). The product $W \times C$ is the Tychonoff plank and based on the discussion here is not hereditarily normal. Usually the Tychonoff plank is shown to be not hereditarily normal by removing the cornor point $(\omega_1,\omega)$. The resulting space is the deleted Tychonoff plank and is not normal (see The Tychonoff Plank).

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Przymusinski, T. C., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

# Sorgenfrey Line is not a Moore Space

We found an incorrect statement about the Sorgenfrey line in an entry in Wikipedia about Moore space (link). This statement opens up a discussion on the question of whether the Sorgenfrey line is a Moore space as well as a discussion on Moore space. The following is the incorrect statement found in Wikipedia by the author.

The Sorgenfrey line is the space whose underlying set is the real line $S=\mathbb{R}$ where the topology is generated by a base consisting the half open intervals of the form $[a,b)$. The Sorgenfrey plane is the square $S \times S$.

Even though the Sorgenfrey line is normal, the Sorgenfrey plane is not normal. In fact, the Sorgenfrey line is the classic example of a normal space whose square is not normal. Both the Sorgenfrey line and the Sorgenfrey plane are not Moore space but not for the reason given. The statement seems to suggest that any normal Moore space is second countable. But this flies in the face of all the profound mathematics surrounding the normal Moore space conjecture, which is also discussed in the Wikipedia entry.

The statement indicated above is only a lead-in to a discussion of Moore space. We are certain that it will be corrected. We always appreciate readers who kindly alert us to errors found in this blog.

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Moore Spaces

Let $X$ be a regular space. A development for $X$ is a sequence $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$ of open covers of $X$ such that for each $x \in X$, and for each open subset $U$ of $X$ with $x \in U$, there exists one cover $\mathcal{G}_n$ satisfying the condition that for any open set $V \in \mathcal{G}_n$, $x \in V \Rightarrow V \subset U$. When $X$ has a development, $X$ is said to be a Moore space (also called developable space). A Note On The Sorgenfrey Line is an introductory note on the Sorgenfrey line.

Moore spaces can be viewed as a generalization of metrizable spaces. Moore spaces are first countable (having a countable base at each point). For a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$, the open sets in $\mathcal{G}_n$ are considered “smaller” as the index $n$ increases. In fact, this is how a development is defined for a metric space, where $\mathcal{G}_n$ consists of all open balls with diameters less than $\frac{1}{n}$. Thus metric spaces are developable. There are plenty of non-metrizable Moore space. One example is the Niemytzki’s Tangent Disc space.

In a Moore space, every closed set is a $G_\delta$-set. Thus if a Moore space is normal, it is perfectly normal. Any Moore space has a $G_\delta$-diagonal (the diagonal $\Delta=\left\{(x,x): x \in X \right\}$ is a $G_\delta$-set in $X \times X$). It is a well known theorem that every compact space with a $G_\delta$-diagonal is metrizable. Thus any compact Moore space is metrizable.

The last statement can be shown more directly. Suppose that $X$ is compact and has a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$. Then each $\mathcal{G}_n$ has a finite subcover $\mathcal{H}_n$. Then $\bigcup_{n=1}^\infty \mathcal{H}_n$ is a countable base for $X$. Thus any compact Moore space is second countable and hence metrizable.

What about paracompact Moore space? Suppose that $X$ is paracompact and has a development $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3,\cdots$. Then each $\mathcal{G}_n$ has a locally finite open refinement $\mathcal{H}_n$. Then $\bigcup_{n=1}^\infty \mathcal{H}_n$ is a $\sigma$-locally finite base for $X$. The Smirnov-Nagata metrization theorem states that a space is metrizable if and only if it has a $\sigma$-locally finite base (see Theorem 23.9 on page 170 of [2]). Thus any paracompact Moore space has a $\sigma$-locally finite base and is thus metrizable (after using the big gun of the Smirnov-Nagata metrization theorem).

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Sorgenfrey Line

The Sorgenfrey line is regular and Lindelof. Hence it is paracompact. Since the Sorgenfrey line is not metrizable, by the above discussion it cannot be a Moore space. The Sorgenfrey plane is also not a Moore space. Note that being a Moore space is a hereditary property. So if the Sorgenfrey plane is a Moore space, then every subspace of the Sorgenfrey plane (including the Sorgenfrey line) is a Moore space.

The following theorem is another way to show that the Sorgenfrey line is not a Moore space.

Bing’s Metrization Theorem
A topological space is metrizable if and only if it is a collectionwise normal Moore space.

Every paracompact space is collectionwise normal (see Theorem 5.1.18, p.305 of [1]). Thus the Sorgenfrey line is collectionwise normal and hence cannot be a Moore space. A space $X$ is said to be collectionwise normal if $X$ is a $T_1$-space and for every discrete collection $\left\{W_\alpha: \alpha \in A \right\}$ of closed sets in $X$, there exists a discrete collection $\left\{V_\alpha: \alpha \in A \right\}$ of open subsets of $X$ such that $W_\alpha \subset V_\alpha$. For a proof of Bing’s metrization theorem, see page 329 of [1].

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Remark

The normal Moore space conjecture is the statement that every normal Moore space is metrizable. This conjecture had been one of the key motivating questions for many set theorists and topologists during a large part of the twentieth century. The bottom line is that this statement cannot not be decided just on the basis of the set of generally accepted axioms called Zermelo–Fraenkel set theory with the axiom of choice, commonly abbreviated ZFC. But Bing’s metrization theorem states that if we strengthen normality to collectionwise normality, we have a definite answer.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

# A Space with G-delta Diagonal that is not Submetrizable

The property of being submetrizable implies having a $G_\delta$-diagonal. There are several other properties lying between these two properties (see [1]). Before diving into these other properties, it may be helpful to investigate a classic example of a space with a $G_\delta$-diagonal that is not submetrizable.

The diagonal of a space $X$ is the set $\Delta=\left\{(x,x): x \in X \right\}$, a subset of the square $X \times X$. An interesting property is when the diagonal of a space is a $G_\delta$-set in $X \times X$ (the space is said to have a $G_\delta$-diagonal). Any compact space or a countably compact space with this property must be metrizable (see compact and countably compact space). A space $(X,\tau)$ is said to be submetrizable if there is a topology $\tau^*$ that can be defined on $X$ such that $(X,\tau^*)$ is a metrizable space and $\tau^* \subset \tau$. In other words, a submetrizable space is a space that has a coarser (weaker) metrizable topology. Every submetrizable space has a $G_\delta$-diagonal. Note that when $X$ has a weaker metric topology, the diagonal $\Delta$ is always a $G_\delta$-set in the metric square $X \times X$ (hence in the square in the original topology). The property of having a $G_\delta$-diagonal is strictly weaker than the property of having a weaker metric topology. In this post, we discuss the Mrowka space, which is a classic example of a space with a $G_\delta$-diagonal that is not submetrizable.

The Mrowka space (also called Psi space) was discussed previously in this blog (see this post). For the sake of completeness, the example is defined here.

First, we define some basic notions. Let $\omega$ be the first infinite ordinal (or more conveniently the set of all nonnegative integers). Let $\mathcal{A}$ be a family of infinite subsets of $\omega$. The family $\mathcal{A}$ is said to be an almost disjoint family if for each two distinct $A,B \in \mathcal{A}$, $A \cap B$ is finite. An almost disjoint family $\mathcal{A}$ is said to be a maximal almost disjoint family if $B$ is an infinite subset of $\omega$ such that $B \notin \mathcal{A}$, then $B \cap A$ is infinite for some $A \in \mathcal{A}$. In other words, if you put one more set into a maximal almost disjoint family, it ceases to be almost disjoint.

A natural question is whether there is an uncountable almost disjoint family of subsets of $\omega$. In fact, there is one whose cardinality is continuum (the cardinality of the real line). To see this, identify $\omega$ with $\mathbb{Q}=\lbrace{r_0,r_1,r_2,...}\rbrace$ (the set of all rational numbers). Let $\mathbb{P}=\mathbb{R}-\mathbb{Q}$ be the set of all irrational numbers. For each $x \in \mathbb{P}$, choose a subsequence of $\mathbb{Q}$ consisting of distinct elements that converges to $x$ (in the Euclidean topology). Then the family of all such sequences of rational numbers would be an almost disjoint family. By a Zorn’s Lemma argument, this almost disjoint family is contained within a maximal almost disjoint family. Thus we also have a maximal almost disjoint family of cardinality continuum. On the other hand, there is no countably infinite maximal almost disjoint family of subsets of $\omega$ (see this post).

Let $\mathcal{A}$ be an infinite almost disjoint family of subsets of $\omega$. We now define a Mrowka space (or $\Psi$-space), denoted by $\Psi(\mathcal{A})$. The underlying set is $\Psi(\mathcal{A})=\mathcal{A} \cup \omega$. Points in $\omega$ are isolated. For $A \in \mathcal{A}$, a basic open set is of the form $\lbrace{A}\rbrace \cup (A-F)$ where $F \subset \omega$ is finite. It is straightforward to verify that $\Psi(\mathcal{A})$ is Hausdorff, first countable and locally compact. It has a countable dense set of isolated points. Note that $\mathcal{A}$ is an infinite discrete and closed set in the space $\Psi(\mathcal{A})$. Thus $\Psi(\mathcal{A})$ is not countably compact.

We would like to point out that the definition of a Mrowka space $\Psi(\mathcal{A})$ only requires that the family $\mathcal{A}$ is an almost disjoint family and does not necessarily have to be maximal. For the example discribed in the title, $\mathcal{A}$ needs to be a maximal almost disjoint family of subsets of $\omega$.

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Example
Let $\mathcal{A}$ be a maximal almost disjoint family of subsets of $\omega$. Then $\Psi(\mathcal{A})$ as defined above is a space in which there is a $G_\delta$-diagonal that is not submetrizable.

Note that $\Psi(\mathcal{A})$ is pseudocompact (proved in this post). Because there is no countable maximal almost disjoint family of subsets of $\omega$, $\mathcal{A}$ must be an uncountable in addition to being a closed and discrete subspace of $\Psi(\mathcal{A})$ (thus the space is not Lindelof). Since $\Psi(\mathcal{A})$ is separable and is not Lindelof, $\Psi(\mathcal{A})$ is not metrizable. Any psuedocompact submetrizable space is metrizable (see Theorem 4 in this post). Thus $\Psi(\mathcal{A})$ must not be submetrizable.

On the other hand, any $\Psi$-space $\Psi(\mathcal{A})$ (even if $\mathcal{A}$ is not maximal) is a Moore space. It is well known that any Moore space has a $G_\delta$-diagonal. The remainder of this post has a brief discussion of Moore space.

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Moore Space

A sequence $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ of open covers of a space $X$ is a development for $X$ if for each $x \in X$ and each open set $U \subset X$ with $x \in U$, there is some $n$ such that any open set in $\mathcal{D}_n$ containing the point $x$ is contained in $U$. A developable space is one that has a development. A Moore space is a regular developable space.

Suppose that $X$ is a Moore space. We show that $X$ has a $G_\delta$-diagonal. That is, we wish to show that $\Delta=\left\{(x,x): x \in X \right\}$ is a $G_\delta$-set in $X \times X$.

Let $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ be a development. For each $n$, let $U_n=\bigcup \lbrace{V \times V:V \in \mathcal{D}_n}\rbrace$. Clearly $\Delta \subset \bigcap_{n<\omega} U_n$. Let $(x,y) \in \bigcap_{n<\omega} U_n$. For each $n$, $(x,y) \in V_n \times V_n$ for some $V_n \in \mathcal{D}_n$. We claim that $x=y$. Suppose that $x \ne y$. By the definition of development, there exists some $m$ such that every open set in $\mathcal{D}_m$ containing the point $x$ has to be a subset of $X-\left\{y \right\}$. Then $V_m \subset X-\left\{y \right\}$, which contradicts $y \in V_m$. Thus we have $\Delta = \bigcap_{n<\omega} U_n$.

The remaining thing to show is that $\Psi(\mathcal{A})$ is a Moore space. For each positive integer $n$, let $F_n=\left\{0,1,\cdots,n-1 \right\}$ and let $F_0=\varnothing$. The development is defined by $\lbrace{\mathcal{E}_n}\rbrace_{n<\omega}$, where for each $n$, $\mathcal{E}_n$ consists of open sets of the form $\lbrace{A}\rbrace \cup (A-F_n)$ where $A \in \mathcal{A}$ plus any singleton $\left\{j \right\}$ ($j \in \omega$) that has not been covered by the sets $\lbrace{A}\rbrace \cup (A-F_n)$.

Reference

1. Arhangel’skii, A. V., Buzyakova, R. Z., The rank of the diagonal and submetrizability, Commentationes Mathematicae Universitatis Carolinae, Vol. 47 (2006), No. 4, 585-597.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

# When is a Pseudocompact Space Metrizable?

Compactness, countably compactness and pseudocompactness are three successively weaker properties. It follows easily from definitions that

$(A) \ \ \ \ \ \text{compact} \Rightarrow \text{countably compact} \Rightarrow \text{pseudocompact}$

None of these arrows can be reversed. It is well known that either compactness or countably complactness plus having a $G_\delta$-diagonal implies metrizability. We have:

$(B) \ \ \ \ \ \text{compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable}$

$(C) \ \ \ \ \ \text{countably compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable}$

A question can be asked whether these results can be extended to pseudocompactness.

Question $(D) \ \ \ \ \ \text{pseudocompact compact} + \text{having a } G_\delta \text{-diagonal} \Rightarrow \text{metrizable?}$

The answer to this question is no. The space defined using a maximal almost disjoint family of subsets of $\omega$ is an example of a non-metrizable pseudocompact space with a $G_\delta$-diagonal (discussed in this post). In this post we show that if we strengthen “having a $G_\delta$-diagonal” to being submetrizable, we have a theorem. Specifically, we show:

$(E) \ \ \ \ \ \text{pseudocompact} + \text{submetrizable} \Rightarrow \text{metrizable}$

For the result of $(B)$, see this post. For the result of $(C)$, see this post. In this post, we discuss the basic properties of pseudocompactness that build up to the result of $(E)$. All spaces considered here are at least Tychonoff (i.e. completely regular). For any basic notions not defined here, see [1] or [2].

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Pseudocompact Spaces

A space $X$ is said to be pseudocompact if every real-valued continuous function defined on $X$ is a bounded function. Any real-valued continuous function defined on a compact space must be bounded (and is thus pseudocomppact). If there were an unbounded real-valued continuous function defined on a space $X$, then $X$ would have a countably infinite discrete set (thus not countably compact). Thus countably compact implies pseudocompact, as indicated by $(A)$.

A space $X$ is submetrizable if there is a coarser (i.e. weaker) topology that is a metrizable topology. Specifically the topological space $(X,\tau)$ is submetrizable if there is another topology $\tau^*$ that can be defined on $X$ such that $\tau^* \subset \tau$ and $(X,\tau^*)$ is metrizable. The Sorgenfrey line is non-metrizable and yet the Sorgenfrey topology has a weaker topology that is metrizable, namely the Euclidean topology of the real line.

The following two theorems characterizes pseudocompact spaces in terms of locally finite open family of open sets (Theorem 1) and the finite intersection property (Theorem 2). Both theorems are found in Engelking (Theorem 3.10.22 and Theorem 3.10.23 in page 207 of [1]). Theorem 3 states that in a pseudocompact space, closed domains are pseudocompact (the definition of closed domain is stated before the theorem). Theorem 4 is the main theorem (result $E$ stated above).

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

1. The space $X$ is pseudocompact.
2. If $\mathcal{V}$ is a locally finite family of non-empty open subsets of $X$, then $\mathcal{V}$ is finite.
3. If $\mathcal{V}$ is a locally finite open cover of $X$, then $\mathcal{V}$ is finite.
4. If $\mathcal{V}$ is a locally finite open cover of $X$, then $\mathcal{V}$ has a finite subcover.

Proof
$1 \Rightarrow 2$
Suppose that condition $2$ does not hold. Then there is an infinite locally finite family of non-empty open sets $\mathcal{V}$ such that $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$. We wish to define an unbounded continuous function using $\mathcal{V}$.

This is where we need to invoke the assumption of complete regularity. For each $n$ choose a point $x_n \in V_n$. Then for each $n$, there is a continuous function $f_n:X \rightarrow [0,n]$ such that $f_n(x_n)=n$ and $f_n(X-V_n) \subset \left\{ 0 \right\}$. Define $f:X \rightarrow [0,\infty)$ by $f(x)=f_1(x)+f_2(x)+f_3(x)+\cdots$.

Because $\mathcal{V}$ is locally finite, the function $f$ is essentially pointwise the sum of finitely many $f_n$. In other words, for each $x \in X$, for some positive integer $N$, $f_j(x)=0$ for all $j \ge N$. Thus the function $f$ is well defined and is continuous at each $x \in X$. Note that for each $x_n$, $f(x_n) \ge n$, showing that it is unbounded.

The directions $2 \Rightarrow 3$ and $3 \Rightarrow 4$ are clear.

$4 \Rightarrow 1$
Let $g:X \rightarrow \mathbb{R}$ be a continuous function. We want to show that $g$ is a bounded function. Consider the open family $\mathcal{O}=\left\{\cdots,O_{-3},O_{-2},O_{-1},O_0,O_1,O_2,O_3,\cdots \right\}$ where each $O_n=g^{-1}((n,n+2))$. Note that $\mathcal{O}$ is a locally finite family in $X$ since its members $O_n=g^{-1}((n,n+2))$ are inverse images of members of a locally finite family in the range space $\mathbb{R}$. By condition $4$, $\mathcal{O}$ has a finite subcover, leading to the conclusion that $g$ is a bounded function. $\blacksquare$

Theorem 2
Let $X$ be a space. The following conditions are equivalent:

1. The space $X$ is pseudocompact.
2. If $\mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $O_n \supset O_{n+1}$ for each $n$, then $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$.
3. If $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is a family of non-empty open subsets of $X$ such that $\mathcal{V}$ has the finite intersection property, then $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$.

Proof
$1 \Rightarrow 2$
Suppose that $X$ is pseudocompact. Suppose $\mathcal{O}=\left\{O_1,O_2,O_3,\cdots \right\}$ satisfies the hypothesis of condition $2$. If there is some positive integer $m$ such that $O_n=O_m$ for all $n \ge m$, then we are done. So assume that $O_n$ are distinct for infinitely many $n$. According to condition $2$ in Theorem 1, $\mathcal{O}$ must not be a locally finite family. Then there exists a point $x \in X$ such that every open set containing $x$ must meet infinitely many $O_n$. This implies that $x \in \overline{O_n}$ for infinitely many $n$. Thus $x \in \bigcap \limits_{n=1}^\infty \overline{O_n}$.

$2 \Rightarrow 3$
Suppose $\mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\}$ is a family of non-empty open sets with the finite intersection property as in the hypothesis of $3$. Then let $O_1=V_1$, $O_2=V_1 \cap V_2$, $O_3=V_1 \cap V_2 \cap V_3$, and so on. By condition $2$, we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$, which implies $\bigcap \limits_{n=1}^\infty \overline{V_n} \ne \varnothing$.

$3 \Rightarrow 1$
Let $g:X \rightarrow \mathbb{R}$ be a continuous function such that $g$ is unbounded. For each positive integer $n$, let $V_n=\left\{x \in X: \lvert g(x) \lvert > n \right\}$. Clearly the open sets $V_n$ have the finite intersection property. Because $g$ is unbounded, it follows that $\bigcap \limits_{n=1}^\infty \overline{V_n} = \varnothing$. $\blacksquare$

Let $X$ be a space. Let $A \subset X$. The interior of $A$, denoted by $\text{int}(A)$, is the set of all points $x \in X$ such that there exists an open set $O$ with $x \in O \subset A$. Points of $\text{int}(A)$ are called the interior points of $A$. A subset $C \subset X$ is said to be a closed domain if $C=\overline{\text{int}(C)}$. It is clear that $C$ is a closed domain if and only if $C$ is the closure of an open set.

Theorem 3
The property of being a pseudocompact space is hereditary with respect to subsets that are closed domains.

Proof
Let $X$ be a pseudocompact space. We show that $\overline{U}$ is pseudocompact for any nonempty open set $U \subset X$. Let $Y=\overline{U}$ where $U$ is a non-empty open subset of $X$. Let $S_1 \supset S_2 \supset S_3 \supset \cdots$ be a decreasing sequence of open subsets of $Y$. Note that each $S_i$ contains points of the open set $U$. Let $O_i=S_i \cap U$ for each $i$. Note that the open sets $O_i$ form a decreasing sequence of open sets in the pseudocompact space $X$. By Theorem 2, we have $\bigcap \limits_{n=1}^\infty \overline{O_n} \ne \varnothing$ (closure here is with respect to $X$). Note that points in $\bigcap \limits_{n=1}^\infty \overline{O_n}$ are also points in $\bigcap \limits_{n=1}^\infty \overline{S_n}$ (closure with respect to $Y$). By Theorem 2, $Y=\overline{U}$ is pseudocompact. $\blacksquare$

Theorem 4 (Statement $E$ above)
Let $X$ be a pseudocompact submetrizable space. Then $X$ is metrizable.

Proof
Let $(X,\tau)$ be a pseudocompact submetrizable space. Then there exists topology $\tau^*$ on $X$ such $(X,\tau^*)$ is metrizable and $\tau^* \subset \tau$. We show that $\tau \subset \tau^*$, leading to the conclusion that $(X,\tau)$ is also metrizable. If $A \subset X$, we denote the closure of $A$ in $(X,\tau)$ by $cl_{\tau}(A)$ and the closure of $A$ in $(X,\tau^*)$ by $cl_{\tau^*}(A)$.

To show that $\tau \subset \tau^*$, we show any closed set with respect to the topology $\tau$ is also a closed set with respect to the topology $\tau^*$. Let $C$ be a closed set in $(X,\tau)$. Consider the family $\mathcal{W}=\left\{cl_{\tau}(U): U \in \tau \text{ and } C \subset U \right\}$. We make the following claims.

Claim 1. $C=\bigcap \left\{W: W \in \mathcal{W} \right\}$.

Claim 2. Each $W \in \mathcal{W}$ is pseudocompact in $(X,\tau)$.

Claim 3. Each $W \in \mathcal{W}$ is pseudocompact in $(X,\tau^*)$.

Claim 4. Each $W \in \mathcal{W}$ is compact in $(X,\tau^*)$.

We now discuss each of these four claims. For Claim 1, it is clear that $C \subset \bigcap \left\{W: W \in \mathcal{W} \right\}$. The reverse set inclusion follows from the fact that $X$ is a regular space. Claim 2 follows from Theorem 3. Note that sets in $\mathcal{W}$ are closed domains in the pseudocompact space $(X,\tau)$.

If sets in $\mathcal{W}$ are pseudocompact in the larger topology $\tau$, they would be pseudocompact in the weaker topology $\tau^*$ too. Thus Claim 3 is established. In a metrizable space, compactness and weaker notions such as countably compactness and pseudocompactness coincide. Because they are pseudocompact subsets, sets in $\mathcal{W}$ are compact in the metrizable space $(X,\tau^*)$. Thus Claim 4 is established.

It follows that $C$ is closed in $(X,\tau^*)$ since it is the intersection of compact sets in $(X,\tau^*)$. Thus $(X,\tau)$ is identical to $(X,\tau^*)$, implying that $(X,\tau)$ is metrizable. $\blacksquare$

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.
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# Thinking about the Space of Irrationals Topologically

Let $\mathbb{R}$ denote the real number line and $\mathbb{P}$ denote the set of all irrational numbers. The irrational numbers and the set $\mathbb{P}$ occupy an important place in mathematics. The set $\mathbb{P}$ with the Euclidean topology inherited from the full real line is a topological space in its own right. Thus the space $\mathbb{P}$ has some of the same properties inherited from the Euclidean real line, e.g., just to name a few, it is hereditarily separable, hereditarily Lindelof, paracompact and metrizable. The space of the irrational numbers $\mathbb{P}$ has many properties apart from the full real line (e.g. $\mathbb{P}$ is totally disconnected). In this post, we look at a topological description of the space $\mathbb{P}$.

Let $\omega$ be the set of all nonnegative integers. Then the space $\mathbb{P}$ of irrational numbers is topologically equivalent (i.e. homeomorphic to) the product space $\prod \limits_{i=0}^\infty X_i$ where each $X_i=\omega$ has the discrete topology. We will also denote the product space $\prod \limits_{i=0}^\infty X_i$ by $\omega^\omega$. We have the following theorem.

Theorem
The space $\mathbb{P}$ of irrational numbers is homeomorphic to the product space $\omega^\omega$.

Because of this theorem, we can look at irrational numbers as sequences of nonnegative integers. Specifically each irrational number can be identified by a unique sequence of nonnegative integers. We can think of each such unique sequence as an address to locate an irrational number. In the remainder of the post, we describe at a high level how to define the unique addresses, which will also give us a homeomorphic map between the space $\mathbb{P}$ and the product space $\omega^\omega$.

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Step 0
Put the rational numbers in a sequence $r_0,r_1,r_2,r_3,\cdots$ such that $r_0=0$. We divide the real line into countably many non-overlapping intervals. Specifically, let $A_0=[0,1]$, $A_1=[-1,0]$, $A_2=[1,2]$, etc (see the following figure).

To facilitate the remaining construction, we denote the left endpoint of the interval $A_i$ by $L_i$ and denote the right endpoint by $R_i$.

Step 1
In each of the interval $A_i$, we further divide it into non-overlapping intervals. The left endpoint and the right endpoint of the interval $A_i$ are $L_i$ and $R_i$, respectively. We choose a sequence $x_{i,0}, x_{i,1}, x_{i,2},\cdots$ of rational numbers converging to the right endpoint $R_i$. Then let $A_{i,0}=[L_i,x_{i,0}]$, $A_{i,1}=[x_{i,0},x_{i,1}]$, $A_{i,2}=[x_{i,1},x_{i,2}]$, etc (see the following figure).

Two important points to consider in Step $1$. One is that we make sure the rational number $r_1$ is chosen as an endpoint of some interval in Step 1. The second is that the length of each $A_{i,j}$ is less than $\displaystyle \frac{1}{2^1}$.

Step 2
In each of the interval $A_{i,j}$, we further divide it into non-overlapping intervals. The left endpoint and the right endpoint of the interval $A_{i,j}$ are $L_{i,j}$ and $R_{i,j}$, respectively. We choose a sequence $x_{i,j,0}, x_{i,j,1}, x_{i,j,2},\cdots$ of rational numbers converging to the left endpoint $L_{i,j}$. Then let $A_{i,j,0}=[x_{i,j,0},R_{i,j}]$, $A_{i,j,1}=[x_{i,j,1},x_{i,j,0}]$, $A_{i,j,2}=[x_{i,j,2},x_{i,j,1}]$, etc (see the following figure).

As in the previous step, two important points to consider in Step $2$. One is that we make sure the rational number $r_2$ is chosen as an endpoint of some interval in Step 2. The second is that the length of each $A_{i,j,k}$ is less than $\displaystyle \frac{1}{2^2}$.

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Remark

In the process described above, the endpoints of the intervals $A_{f(0),\cdots,f(n)}$ are rational numbers and we make sure that all the rational numbers are used as endpoints. We also make sure that the intervals from the successive steps are nested closed intervals with lengths $\rightarrow 0$. The consequence of this point is that the nested decreasing closed intervals will collapse to one single point (since the real line is a complete metric space) and this single point must be an irrational number (since all the rational numbers are used up as endpoints of the nested closed intervals).

In Step $i$ where $i$ is an odd integer, we make the endpoints of the new intervals converge to the right. In Step $i$ where $i>1$ is an even integer, we make the endpoints of the new intervals converge to the left. This manipulation is to ensure that the nested closed intervals will never share the same endpoint from one step all the way to the end of the process.

Thus if we have $f \in \omega^\omega$, then $\bigcap \limits_{n=0}^\infty A_{f(0),\cdots,f(n)} \ne \varnothing$ and in fact has only one point that is an irrational number.

On the other hand, for each point $x \in \mathbb{P}$, we can locate inductively a sequence of intervals, $A_{f(0)}, A_{f(0),f(1)}, A_{f(0),f(1),f(2)}, \cdots$, containing the point $x$. This sequence of nested closed intervals must collapse to a single point and the single point must be the irrational number $x$.

The process described above gives us a one-to-one mapping from $\mathbb{P}$ onto the product space $\omega^\omega$. This mapping is also continuous in both directions, making it a homeomorphism. the nested intervals defined above form a base for the Euclidean topology on $\mathbb{P}$. These basic open sets have a natural correspondance with basic open sets in the product space $\omega^\omega$.

For example, for $f \in \omega^\omega$, $\left\{ A_{f(0),\cdots,f(n)} \cap \mathbb{P}: n \in \omega \right\}$ is a local base at the point $x \in \bigcap \limits_{n=0}^\infty A_{f(0),\cdots,f(n)}$. One the other hand, each $A_{f(0),\cdots,f(n)} \cap \mathbb{P}$ has a natural counterpart in a basic open set in the product space, namely the following set:

$\displaystyle . \ \ \ \ \left\{g \in \omega^\omega: g \restriction n = f \restriction n \right\}$

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The above process establishes that the countable product of the integers, $\omega^\omega$, is equivalent topologically to the Euclidean space $\mathbb{P}$.

# Elementary Examples of Lindelof Spaces and Separable Spaces

The Euclidean spaces $\mathbb{R}$ and $\mathbb{R}^n$ are both Lindelof and separable. In fact these two properties are equivalent in the class of metrizable spaces. A space is metrizable if its topology can be induced by a metric. In a metrizable space, having one of these properties implies the other one. Any students in beginning topology courses who study basic notions such as the Lindelof property and separability must venture outside the confine of Euclidean spaces or metric spaces. The goal of this post is to present some elementary examples showing that these two notions are not equivalent.

All topological spaces under consideration are Hausdorff. Let $X$ be a space. Let $D \subset X$. The set $D$ is said to be dense in $X$ if every nonempty open subset of $X$ contains some point of $D$. The space $X$ is said to be separable if there is countable subset of $X$ that is also dense in $X$. All Euclidean spaces are separable. For example, in the real line $\mathbb{R}$, every open interval contains a rational number. Thus the set of all rational numbers $\mathbb{Q}$ is dense in $\mathbb{R}$.

Let $\mathcal{U}$ be a collection of subsets of the space $X$. The collection $\mathcal{U}$ is said to be a cover of $X$ if every point of $X$ is contained in some element of $\mathcal{U}$. The collection $\mathcal{U}$ is said to be an open cover of $X$ if, in addition it being a cover, $\mathcal{U}$ consists of open sets in $X$.

Let $\mathcal{U}$ be a cover of the space $X$. Let $\mathcal{V} \subset \mathcal{U}$. If the collection $\mathcal{V}$ is also a cover of $X$, we say that $\mathcal{V}$ is a subcover of $\mathcal{U}$. The space $X$ is a Lindelof space (or has the Lindelof property) if every open cover of $X$ has a countable subcover.

The real $\mathbb{R}$ is Lindelof. Both the Lindelof property and the separability of $\mathbb{R}$ follows from the fact that the Euclidean topology on $\mathbb{R}$ can be generated by a countable base (e.g. one countable base consists of all open intervals with rational endpoints). Now some non-Euclidean (and non-metrizable) examples.

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Example 1 – A Lindelof space that is not separable
Let $X$ be any uncountable set. Let $p$ be a point that is not in $X$, e.g., let $p=\left\{ X \right\}$. Define the space $Y = \left\{p\right\} \cup X$ as follows. Let every point in $X$ be isolated, meaning any singleton set $\left\{ x \right\}$ is declared open for any $x \in X$. An open neighborhood of the point $p$ is of the form $\left\{p\right\} \cup W$ where $X-W$ is a countable subset of $X$.

It is clear that the resulting space $Y$ is Lindelof since every open set containing $p$ contains all but countably many points of $X$. It is also clear that no countable set can be dense in $Y$.

Even though this example $Y$ is Lindelof, it is not hereditarily Lindelof since the subspace $X$ is uncountable discrete space.

In a previous post, we showed that the space $Y$ defined in this example is a productively Lindelof space (meaning that its product with every Lindelof space is Lindelof).

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Remark

A space is said to have the countable chain condition (CCC) if there are no uncountable family of pairwise disjoint open subsets. It is clear that any separable space has the CCC. It follows that the space $Y$ in Example 1 does not have the CCC, since the singleton sets $\left\{ x \right\}$ (with $x \in X$) forms a pairwise disjoint collection of open sets, showing that the Lindelof property does not even imply the weaker property of having the CCC.

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Example 2 – A separable space that is not Lindelof
The example here is the Tangent Disc Space (Niemytzki’s Tangent Disc Topology in [2]). The underlying set is the upper half plane (the x-axis and the plane above the x-axis). In other words, consider the following set:

$\displaystyle . \ \ \ \ \ X=\left\{(x,y) \in \mathbb{R}^2: y \ge 0 \right\}$

Let $\displaystyle X_u=\left\{(x,y) \in \mathbb{R}^2: y>0 \right\}$ and $T=\left\{(x,0): x \in \mathbb{R} \right\}$. The line $T$ is the x-axis and $X_u$ is the upper plane without the x-axis. We define a topology on $X$ such that $X_u$ as a subspace in this topology is Euclidean. The open neighborhoods of a point $p=(x,0) \in T$ are of the form $\left\{p \right\} \cup D$ where $D$ is an open disc tangent to the x-axis at the point $p$. The figure below illustrates how open neighborhoods at the x-axis are defined.

It is clear that the points with rational coordinates in the upper half plane $X_u$ form a dense set in the tangent disc topology. Thus $X$ is separable. In any Lindelof space, there are no uncountable closed and discrete subsets. Note that the x-axis $T$ is a closed and discrete subspace in the tangent disc space. Thus $X$ is not Lindelof.

Though separable, the Tangent Disc Space is not hereditarily separable since the x-axis $T$ is uncountable and discrete.

The Tangent Disc Space is an interesting example. For example, it is a completely regular space that is an example of a Moore space that is not normal. For these and other interesting facts about the Tangent Disc Space, see [2].

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For the Lindelof property and the property of being separable, there are plenty of examples of spaces that possess only one of the properties. All three references indicated below are excellent places to look. The book by Steen and Seebach ([2]) is an excellent catalog of interesting spaces (many of them are elementary).
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Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Steen, L. A., Seebach, J. A.,Counterexamples in Topology, 1995, Dover Edition, Dover Publications, New York.
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

# The product of first countable spaces

All spaces under consideration are Hausdorff. First countable spaces are those spaces where there is a countable local base at every point in the space. This is quite a strong property. For example, every first countable space that is also compact has a cap on its cardinality and the cap is the cardinality of the real line (the continuum). See The cardinality of compact first countable spaces, I in this blog. In fact, if the compact and first countable space is uncountable, it has cardinality continuum (see The cardinality of compact first countable spaces, III). Any metric space (or metrizable space) is first countable. In this post, we discuss the product of first countable spaces. In this regard, first countable spaces and metrizable spaces behave similarly. We show that the product of countably many first countable spaces is first countable while the product of uncountably many first countable is not first countable. For more information on the product topology, see [2].

The Product Space
Consider a collection of sets $A_\alpha$ where $\alpha \in S$. Let $W=\bigcup \limits_{\alpha \in S} A_\alpha$. The product $\prod \limits_{\alpha \in S} A_\alpha$ is the set of all functions $f:S \mapsto W$ such that for each $\alpha \in S$, $f(\alpha) \in A_\alpha$. If the index set $S=\left\{1,2,\cdots,n\right\}$ is finite, the functions $f$ can be regarded as sequences $(f_1,f_2,\cdots,f_n)$ where each $f_i \in A_i$. If the index set $S=\mathbb{N}$, we can think of elements $f$ of the product as the sequence $(f_1,f_2,\cdots)$ where each $f_i \in A_i$. In general we can regard $f \in \prod \limits_{\alpha \in S} A_\alpha$ as functions $f:S \mapsto W$ or as sequences $f=(f_\alpha)_{\alpha \in S}$.

Consider the topological spaces $X_\alpha$ where $\alpha \in S$. Let $X=\prod \limits_{\alpha \in S} X_\alpha$ be the product as defined above. The product space of the spaces $X_\alpha$ is $X$ with the topology defined in the following paragraph.

Let $\tau_\alpha$ be the topology of each space $X_\alpha$, $\alpha \in S$. Consider $Y=\prod \limits_{\alpha \in S} O_\alpha$ where for each $\alpha \in S$, $O_\alpha \in \tau_\alpha$ (i.e. $O_\alpha$ is open in $X_\alpha$) and $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in S$. The set of all such sets $Y$ is a base for a topology on the product $X=\prod \limits_{\alpha \in S} X_\alpha$. This topology is called the product topology of the spaces $X_\alpha$, $\alpha \in S$.

To more effectively work with product spaces, we consider a couple of equivalent bases that we can define for the product topology. Let $\mathcal{B}_\alpha$ be a base for the space $X_\alpha$. Consider $B=\prod \limits_{\alpha \in S} B_\alpha$ such that there is a finite set $F \subset S$ where $B_\alpha \in \mathcal{B}_\alpha$ for each $\alpha \in F$ and $B_\alpha=X_\alpha$ for all $\alpha \in S-F$. The set of all such sets $B$ is an equivalent base for the product topology.

Another equivalent base is defined using the projection maps. For each $\alpha \in S$, consider the map $\pi_\alpha:\prod \limits_{\beta \in S} X_\beta \mapsto X_\alpha$ such that $\pi_\alpha(f)=f_\alpha$ for each $f$ in the product. In words, the function $\pi_\alpha$ maps each point in the product space to its $\alpha^{th}$ coordinate. The mapping $\pi_\alpha$ is called the $\alpha^{th}$ projection map. For each set $U \subset X_\alpha$, $\pi_\alpha^{-1}(U)$ is the following set:

$\pi_\alpha^{-1}(U)=\left\{f \in \prod \limits_{\beta \in S} X_\beta: \pi_\alpha(f)=f_\alpha \in U\right\}$

Consider sets of the form $\bigcap \limits_{\alpha \in F} \pi_\alpha^{-1}(U_\alpha)$ where $F \subset S$ is finite and $U_\alpha$ is open in $X_\alpha$ for each $\alpha \in F$. The set of all such sets is another equivalent base for the product topology. If we only require that each $U_\alpha \in \mathcal{B}_\alpha$, a predetermined base for the coordinate space $X_\alpha$, we also obtain an equivalent base for the product topology.

Countable Product of First Countable Spaces
For $i=1,2,3,\cdots$, let $X_i$ be a first countable space. We show that $X=\prod \limits_{i=1}^{\infty}X_i$ is a first countable space.

Let $\mathbb{N}$ be the set of positive integers. For each $n \in \mathbb{N}$, let $[n]=\left\{1,2,\cdots,n\right\}$ and let $\mathbb{N}^{[n]}$ be the set of all functions $t:[n] \mapsto \mathbb{N}$. For each $i$ and each $x \in X_i$, let $\mathcal{B}_x(i)=\left\{B_x(i,j): j \in \mathbb{N}\right\}$ be a countable local base at $x$.

Let $f \in X=\prod \limits_{i=1}^{\infty}X_i$. We wish to define a countable local base at $f$. For each $n \in \mathbb{N}$, define $W_n$ to be:

$W_n=\left\{\prod \limits_{i=1}^n B_{f(i)}(i,t(i)):t \in \mathbb{N}^{[n]}\right\}$

Let $W$ be the set of all subsets of the product space $X$ of the following form:

$\prod \limits_{j=1}^{\infty}V_j$ where there is some $n \in \mathbb{N}$ such that $\prod \limits_{j=1}^{n}V_j \in W_n$ and for all $j>n$, $V_j=X_j$.

Each $W_n$ is countable and $W$ is essentially the union of all the $W_n$. Thus $W$ is countable. We claim that $W$ is a local base at $f$. Let $O \subset X$ be an open set containing $f$. We can assume that $O=\prod \limits_{i=1}^\infty O_i$ where there is some $n \in \mathbb{N}$ such that for each $i \le n$, $O_i$ is open in $X_i$ and for $i>n$, $O_i=X_i$.

For each $i \le n$, $f(i) \in O_i$. Choose some $B_{f(i)}(i,t(i))$ such that $f(i) \in B_{f(i)}(i,t(i)) \subset O_i$. Let $V=\prod \limits_{i=1}^{\infty} V_i$ such that $\prod \limits_{i=1}^{n} V_i=\prod \limits_{i=1}^n B_{f(i)}(i,t(i))$ and $V_i=X_i$ for all $i>n$. Then $V \in W$ and $f \in V \subset O$. This completes the proof that $X=\prod \limits_{i=1}^{\infty}X_i$ is a first countable space.

Uncountable Product
Let $S$ be an uncountable index set. For $\alpha \in S$, let $X_\alpha$. We want to avoid the situation that all but countably many $X_\alpha$ are one-point space. So we assume each coordinate space $X_\alpha$ has at least two points, say, $p_\alpha$ and $q_\alpha$ with $p_\alpha \ne q_\alpha$. We show that $X=\prod \limits_{\alpha \in S}X_\alpha$ is not first countable.

Let $f \in \prod \limits_{\alpha \in S}X_\alpha$. Let $U_1,U_2, \cdots$ be open subsets of the product space such that for each $i$, $f \in U_i$. We show that there is some open set $O$ such that $f \in O$ and each $U_i \nsubseteq O$. For each $i$, there is a basic open set $B_i=\bigcap \limits_{\alpha \in F_i} \pi_\alpha^{-1}(U_{\alpha,i})$ such that $f \in B_i \subset U_i$.

Let $F=F_1 \cup F_2 \cup \cdots$. Since $S$ is uncountable and $F$ is countable, choose $\gamma \in S-F$. Since $X_\gamma$ has at least two points $p_\gamma$ and $q_\gamma$, choose one of them that is different from $f_\gamma$, say, $p_\gamma$. Choose two disjoint open subsets $M_1$ and $M_2$ of $X_\gamma$ such that $f_\gamma \in M_1$ and $p_\gamma \in M_2$ of $X_\gamma$. Let $O=\prod \limits_{\alpha \in S}O_\alpha$ such that $O_\gamma=M_1$ and $O_\alpha=X_\alpha$ for all $\alpha \ne \gamma$. We have $f \in O$. For each $i$, there is $g_i \in B_i \subset U_i$ such that $g_i(\gamma)=p_\gamma$. Thus each $g_i \notin O$. Thus there is no countable local base at $f$. Thus any product space with uncountably many factors, each of which has at least two points, is never first countable.

Reference

1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.