# The product of uncountably many factors is never hereditarily normal

The space $Y=\prod_{\alpha<\omega_1} \left\{0,1 \right\}=\left\{0,1 \right\}^{\omega_1}$ is the product of $\omega_1$ many copies of the two-element set $\left\{0,1 \right\}$ where $\omega_1$ is the first uncountable ordinal. It is a compact space by Tychonoff’s theorem. It is a normal space since every compact Hausdorff space is normal. A space is hereditarily normal if every subspace is normal. Is the space $Y$ hereditarily normal? In this post, we give two proofs that it is not hereditarily normal. It then follows that any product space $\prod X_\alpha$ cannot be hereditarily normal as long as there are uncountably many factors and every factor has at least two point.

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The connection with a theorem of Katetov

It turns out that there is a connection with a theorem of Katetov. For any compact space, knowing hereditary normality of the first several self product spaces can reveal a great deal of information about the compact space. More specifically, for any compact space $X$, knowing whether $X$, $X^2$ and $X^3$ are hereditarily normal can tell us whether $X$ is metrizable. If all three are hereditarily normal, then $X$ is metrizable. If one of the three self products is not hereditarily normal, then $X$ is not metrizable. This fact is based on a theorem of Katetov (see this previous post). The space $Y=\left\{0,1 \right\}^{\omega_1}$ is not metrizable since it is not first countable (see Problem 1 below). Thus one of its first three self products must fail to be hereditarily normal.

These two proofs are not direct proof in the sense that a non-normal subspace is not explicitly produced. Instead the proofs use other theorem or basic but important background results. One of the two proofs (#2) uses a theorem of Katetov on hereditarily normal spaces. The other proof (#1) uses the fact that the product of uncountably many copies of a countable discrete space is not normal. We believe that these two proofs and the required basic facts are an important training ground for topology. We list out these basic facts as exercises. Anyone who wishes to fill in the gaps can do so either by studying the links provided or by consulting other sources.

The theorem of Katetov mentioned earlier provides a great exercise – for any non-metrizable compact space $X$, determine where the hereditary normality fails. Does it fail in $X$, $X^2$ or $X^3$? This previous post examines a small list of compact non-metrizable spaces. In all the examples in this list, the hereditary normality fails in $X$ or $X^2$. The space $Y=\left\{0,1 \right\}^{\omega_1}$ can be added to this list. All the examples in this list are defined using no additional set theory axioms beyond ZFC. A natural question: does there exist an example of compact non-metrizable space $X$ such that the hereditary normality holds in $X^2$ and fails in $X^3$? It turns out that this was a hard problem and the answer is independent of ZFC. This previous post provides a brief discussion and has references for the problem.

All spaces under consideration are Hausdorff spaces.

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Exercises

Problem 1
Let $X$ be a compact space. Show that $X$ is normal.

Problem 2
For each $\alpha<\omega_1$, let $A_\alpha$ be a set with cardinality $\le \omega_1$. Show that $\lvert \bigcup_{\alpha<\omega_1} A_\alpha \lvert \le \omega_1$.

Problem 2 holds for any infinite cardinal, not just $\omega_1$. One reference for Problem 2 is Lemma 10.21 on page 30 of Set Theorey, An Introduction to Independence Proofs by Kenneth Kunen.

Problem 3
For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Show that for every point $p \in \prod_{\alpha<\omega_1} X_\alpha$, there does not exist a countable base at the point $p$. In other words, the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not first countable at every point. It follows that product space $\prod_{\alpha<\omega_1} X_\alpha$ is not metrizable.

Problem 4
In any space, a $G_\delta$-set is a set that is the intersection of countably many open sets. When a singleton set $\left\{ x \right\}$ is a $G_\delta$-set, we say the point $x$ is a $G_\delta$-point. For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Show that every point $p$ in the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not a $G_\delta$-point.

Note that Problem 4 implies Problem 3.

For Problem 3 and Problem 4, use the fact that there are uncountably many factors and that a basic open set in the product space is of the form $\prod_{\alpha<\omega_1} O_\alpha$ and that it has only finitely many coordinates at which $O_\alpha \ne X_\alpha$.

Problem 5
For each $\alpha<\omega_1$, let $X_\alpha=\left\{0,1,2,\cdots \right\}$ be the set of non-negative integers with the discrete topology. Show that the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not normal.

See here for a discussion of Problem 5.

Problem 6
Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. Show that $Y$ has a countably infinite subspace

$W=\left\{y_0,y_1,y_2,y_3\cdots \right\}$

such that $W$ is relatively discrete. In other words, $W$ is discrete in the subspace topology of $W$. However $W$ is not discrete in the product space $Y$ since $Y$ is compact.

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Proof #1

Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. We show that $Y$ is not hereditarily normal.

Note that the product space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$ can be written as the product of $\omega_1$ many copies of itself:

$\displaystyle \left\{0,1 \right\}^{\omega_1} \cong \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

The fact (1) follows from the fact that the union of $\omega_1$ many pairwise disjoint sets, each of which has cardinality $\omega_1$, has cardinality $\omega_1$ (see Problem 2). The space $\left\{0,1 \right\}^{\omega_1}$ has a countably infinite subspace that is relatively discrete (see Problem 6). In other words, it has a subspace that is homemorphic to $\omega=\left\{0,1,2,\cdots \right\}$ where $\omega$ has the discrete topology. Thus the following is homeomorphic to a subspace of $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$.

$\displaystyle \omega^{\omega_1} = \omega \times \omega \times \omega \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

By Problem 5, the space $\omega^{\omega_1}$ is not normal. Hence the compact space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$ contains the non-normal space $\omega^{\omega_1}$ and is thus not hereditarily normal. $\blacksquare$

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Proof #2

Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. We show that $Y$ is not hereditarily normal. This proof uses a theorem of Katetov, discussed in this previous post and stated below.

Theorem 1
If $X_1 \times X_2$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

• The factor $X_1$ is perfectly normal.
• Every countable and infinite subset of the factor $X_2$ is closed.

First, $Y$ can be written as the product of two copies of itself:

$\displaystyle \left\{0,1 \right\}^{\omega_1} \cong \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

This is because the union of two disjoints sets, each of which has cardinality $\omega_1$, has carinality $\omega_1$. Note that the countably infinite subset $W$ from Problem 6 is not a closed subset of $Y$. If it were, the compact space $Y$ would contain an infinite set with no limit point. Thus the second condition of Theorem 1 is not satisfied. If $Y \cong Y \times Y$ were to be hereditarily normal, then the first condition must be satisfied, i.e. $Y$ is perfectly normal (meaning that $Y$ is normal and that every closed subset of it is a $G_\delta$-set). However, Problem 4 indicates that no point in $Y$ can be a $G_\delta$ point. Therefore $Y$ cannot be hereditarily normal. $\blacksquare$

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Corollary

The product of uncountably many spaces, each one of which has at least two points, contains a homeomorphic copy of the space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. Thus such a product space can never be hereditarily normal. We state this more formally below.

Theorem 2
Let $\kappa$ be any uncountable cardinal. For each $\alpha<\kappa$, let $X_\alpha$ be a space with at least two points. Then $\prod_{\alpha<\kappa} X_\alpha$ is not hereditarily normal.

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$\copyright \ 2015 \text{ by Dan Ma}$

# Looking for non-normal subspaces of the square of a compact X

A theorem of Katetov states that if $X$ is compact with a hereditarily normal cube $X^3$, then $X$ is metrizable (discussed in this previous post). This means that for any non-metrizable compact space $X$, Katetov’s theorem guarantees that some subspace of the cube $X^3$ is not normal. Where can a non-normal subspace of $X^3$ be found? Is it in $X$, in $X^2$ or in $X^3$? In other words, what is the “dimension” in which the hereditary normality fails for a given compact non-metrizable $X$ (1, 2 or 3)? Katetov’s theorem guarantees that the dimension must be at most 3. Out of curiosity, we gather a few compact non-metrizable spaces. They are discussed below. In this post, we motivate an independence result using these examples.

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Katetov’s theorems

First we state the results of Katetov for reference. These results are proved in this previous post.

Theorem 1
If $X \times Y$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

• The factor $X$ is perfectly normal.
• Every countable and infinite subset of the factor $Y$ is closed.

Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal.

Theorem 3
Let $X$ be a compact space. If $X^3=X \times X \times X$ is hereditarily normal, then $X$ is metrizable.

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Examples of compact non-metrizable spaces

The set-theoretic result presented here is usually motivated by looking at Theorem 3. The question is: Can $X^3$ in Theorem 3 be replaced by $X^2$? We take a different angle of looking at some standard compact non-metric spaces and arrive at the same result. The following is a small listing of compact non-metrizable spaces. Each example in this list is defined in ZFC alone, i.e. no additional axioms are used beyond the generally accepted axioms of set theory.

1. One-point compactification of the Tychonoff plank.
2. One-point compactification of $\psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$.
3. The first compact uncountable ordinal, i.e. $\omega_1+1$.
4. The one-point compactification of an uncountable discrete space.
5. Alexandroff double circle.
6. Double arrow space.
7. Unit square with the lexicographic order.

Since each example in the list is compact and non-metrizable, the cube of each space must not be hereditarily normal according to Theorem 3 above. Where does the hereditary normality fail? For #1 and #2, $X$ is a compactification of a non-normal space and thus not hereditarily normal. So the dimension for the failure of hereditary normality is 1 for #1 and #2.

For #3 through #7, $X$ is hereditarily normal. For #3 through #5, each $X$ has a closed subset that is not a $G_\delta$ set (hence not perfectly normal). In #3 and #4, the non-$G_\delta$-set is a single point. In #5, the the non-$G_\delta$-set is the inner circle. Thus the compact space $X$ in #3 through #5 is not perfectly normal. By Theorem 2, the dimension for the failure of hereditary normality is 2 for #3 through #5.

For #6 and #7, each $X^2$ contains a copy of the Sorgenfrey plane. Thus the dimension for the failure of hereditary normality is also 2 for #6 and #7.

In the small sample of compact non-metrizable spaces just highlighted, the failure of hereditary normality occurs in “dimension” 1 or 2. Naturally, one can ask:

Question. Is there an example of a compact non-metrizable space $X$ such that the failure of hereditary nornmality occurs in “dimension” 3? Specifically, is there a compact non-metrizable $X$ such that $X^2$ is hereditarily normal but $X^3$ is not hereditarily normal?

Such a space $X$ would be an example to show that the condition “$X^3$ is hereditarily normal” in Theorem 3 is necessary. In other words, the hypothesis in Theorem 3 cannot be weakened if the example just described were to exist.

The above list of compact non-metrizable spaces is a small one. They are fairly standard examples for compact non-metrizable spaces. Could there be some esoteric example out there that fits the description? It turns out that there are such examples. In [1], Gruenhage and Nyikos constructed a compact non-metrizable $X$ such that $X^2$ is hereditarily normal. The construction was done using MA + not CH (Martin’s Axiom coupled with the negation of the continuum hypothesis). In that same paper, they also constructed another another example using CH. With the examples from [1], one immediate question was whether the additional set-theoretic axioms of MA + not CH (or CH) was necessary. Could a compact non-metrizable $X$ such that $X^2$ is hereditarily normal be still constructed without using any axioms beyond ZFC, the generally accepted axioms of set theory? For a relatively short period of time, this was an open question.

In 2001, Larson and Todorcevic [3] showed that it is consistent with ZFC that every compact $X$ with hereditarily normal $X^2$ is metrizable. In other words, there is a model of set theory that is consistent with ZFC in which Theorem 3 can be improved to assuming $X^2$ is hereditarily normal. Thus it is impossible to settle the above question without assuming additional axioms beyond those of ZFC. This means that if a compact non-metrizable $X$ is constructed without using any axiom beyond ZFC (such as those in the small list above), the hereditary normality must fail at dimension 1 or 2. Numerous other examples can be added to the above small list. Looking at these ZFC examples can help us appreciate the results in [1] and [3]. These ZFC examples are excellent training ground for general topology.

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Reference

1. Gruenhage G., Nyikos P. J., Normality in $X^2$ for Compact $X$, Trans. Amer. Math. Soc., Vol 340, No 2 (1993), 563-586
2. Katetov M., Complete normality of Cartesian products, Fund. Math., 35 (1948), 271-274
3. Larson P., Todorcevic S., KATETOV’S PROBLEM, Trans. Amer. Math. Soc., Vol 354, No 5 (2001), 1783-1791

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$\copyright \ 2015 \text{ by Dan Ma}$

# When a product space is hereditarily normal

When the spaces $X$ and $Y$ are normal spaces, the product space $X \times Y$ is not necessarily normal. Even if one of the factors is metrizable, there is still no guarantee that the product is normal. So it is possible that the normality of each of the factors $X$ and $Y$ can have no influence on the normality of the product $X \times Y$. The dynamics in the other direction are totally different. When the product $X \times Y$ is hereditarily normal, the two factors $X$ and $Y$ are greatly impacted. In this post, we discuss a theorem of Katetov, which shows that the hereditary normality of the product can impose very strict conditions on the factors, which lead to many interesting results. This theorem also leads to an interesting set-theoretic result, and thus can possibly be a good entry point to the part of topology that deals with consistency and independence results – statements that cannot be proved true or false based on the generally accepted axioms of set theory (ZFC). In this post, we discuss Katetov’s theorem and its consequences. In the next post, we discuss examples that further motivate the set-theoretic angle.

A subset $W$ of a space $X$ is said to be a $G_\delta$-set in $X$ if $W$ is the intersection of countably many open subsets of $X$. A space $X$ is perfectly normal if it is normal and that every closed subset of $X$ is a $G_\delta$-set. Some authors use other statements to characterize perfect normality (here is one such characterization). Perfect normality implies hereditarily normal (see Theorem 6 in this previous post). The implication cannot be reversed. Katetov’s theorem implies that the hereditary normality of the product $X \times Y$ will in many cases make one or both of the factors perfectly normal. Thus the hereditary normality in the product $X \times Y$ is a very strong property.

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Katetov’s theorems

Theorem 1
If $X \times Y$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

• The factor $X$ is perfectly normal.
• Every countable and infinite subset of the factor $Y$ is closed.

Proof of Theorem 1
The strategy we use is to define a subspace of $X \times Y$ that is not normal after assuming that none of the two conditions is true. So assume that $X$ has a closed subspace $W$ that is not a $G_\delta$-set and assume that $T=\left\{t_n: n=1,2,3,\cdots \right\}$ is an infinite subset of $Y$ that is not closed. Let $p \in Y$ be a limit point of $T$ such that $p \notin T$. The candidate for a non-normal subspace of $X \times Y$ is:

$M=X \times Y-W \times \left\{p \right\}$

Note that $M$ is an open subspace of $X \times Y$ since it is the result of subtracting a closed set from $X \times Y$. The following are the two closed sets that demonstrate that $M$ is not normal.

$H=W \times (Y-\left\{p \right\})$

$K=(X-W) \times \left\{p \right\}$

It is clear that $H$ and $K$ are closed subsets of $M$. Let $U$ and $V$ be open subsets of $M$ such that $H \subset U$ and $K \subset V$. We show that $U \cap V \ne \varnothing$. To this end, define $U_j=\left\{x \in X: (x,t_j) \in U \right\}$ for each $j$. It follows that for each $j$, $W \subset U_j$. Furthermore each $U_j$ is an open subspace of $X$. Thus $W \subset \bigcap_j U_j$. Since $W$ is not a $G_\delta$-set in $X$, there must exist $t \in \bigcap_j U_j$ such that $t \notin W$. Then $(t, p) \in K$ and $(t, p) \in V$.

Since $V$ is open in the product $X \times Y$, choose open sets $A \subset X$ and $B \subset Y$ such that $(t,p) \in A \times B$ and $A \times B \subset V$. With $p \in B$, there exists some $j$ such that $t_j \in B$. First, $(t,t_j) \in V$. Since $t \in U_j$, $(t,t_j) \in U$. Thus $U \cap V \ne \varnothing$. This completes the proof that the subspace $M$ is not normal and that $X \times Y$ is not hereditarily normal. $\blacksquare$

Let’s see what happens in Theorem 1 when both factors are compact. If both $X$ and $Y$ are compact and if $X \times Y$ is hereditarily normal, then both $X$ and $Y$ must be perfect normal. Note that in any infinite compact space, not every countably infinite subset is closed. Thus if compact spaces satisfy the conclusion of Theorem 1, they must be perfectly normal. Hence we have the following theorem.

Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal.

Moe interestingly, Theorem 1 leads to a metrization theorem for compact spaces.

Theorem 3
Let $X$ be a compact space. If $X^3=X \times X \times X$ is hereditarily normal, then $X$ is metrizable.

Proof of Theorem 3
Suppose that $X^3$ is hereditarily normal. By Theorem 2, the compact spaces $X^2$ and $X$ are perfectly normal. In particular, the following subset of $X^2$ is a $G_\delta$-set in $X^2$.

$\Delta=\left\{(x,x): x \in X \right\}$

The set $\Delta$ is said to be the diagonal of the space $X$. It is a well known result that any compact space whose diagonal is a $G_\delta$-set in the square is metrizable (discussed here). $\blacksquare$

The results discussed here make it clear that hereditary normality in product spaces is a very strong property. One obvious question is whether Theorem 3 can be improved by assuming only the hereditary normality of $X^2$. This was indeed posted by Katetov himself. This leads to the discussion in the next post.

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Reference

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

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$\copyright \ 2015 \text{ by Dan Ma}$

# Cp(X) where X is a separable metric space

Let $\tau$ be an uncountable cardinal. Let $\prod_{\alpha < \tau} \mathbb{R}=\mathbb{R}^{\tau}$ be the Cartesian product of $\tau$ many copies of the real line. This product space is not normal since it contains $\prod_{\alpha \in \omega_1} \omega=\omega^{\omega_1}$ as a closed subspace. However, there are dense subspaces of $\mathbb{R}^{\tau}$ are normal. For example, the $\Sigma$-product of $\tau$ copies of the real line is normal, i.e., the subspace of $\mathbb{R}^{\tau}$ consisting of points which have at most countably many non-zero coordinates (see this post). In this post, we look for more normal spaces among the subspaces of $\mathbb{R}^{\tau}$ that are function spaces. In particular, we look at spaces of continuous real-valued functions defined on a separable metrizable space, i.e., the function space $C_p(X)$ where $X$ is a separable metrizable space.

For definitions of basic open sets and other background information on the function space $C_p(X)$, see this previous post.

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$C_p(X)$ when $X$ is a separable metric space

In the remainder of the post, $X$ denotes a separable metrizable space. Then, $C_p(X)$ is more than normal. The function space $C_p(X)$ has the following properties:

• normal,
• Lindelof (hence paracompact and collectionwise normal),
• hereditarily Lindelof (hence hereditarily normal),
• hereditarily separable,
• perfectly normal.

All such properties stem from the fact that $C_p(X)$ has a countable network whenever $X$ is a separable metrizable space.

Let $L$ be a topological space. A collection $\mathcal{N}$ of subsets of $L$ is said to be a network for $L$ if for each $x \in L$ and for each open $O \subset L$ with $x \in O$, there exists some $A \in \mathcal{N}$ such that $x \in A \subset O$. A countable network is a network that has only countably many elements. The property of having a countable network is a very strong property, e.g., having all the properties listed above. For a basic discussion of this property, see this previous post and this previous post.

To define a countable network for $C_p(X)$, let $\mathcal{B}$ be a countable base for the domain space $X$. For each $B \subset \mathcal{B}$ and for any open interval $(a,b)$ in the real line with rational endpoints, consider the following set:

$[B,(a,b)]=\left\{f \in C(X): f(B) \subset (a,b) \right\}$

There are only countably many sets of the form $[B,(a,b)]$. Let $\mathcal{N}$ be the collection of sets, each of which is the intersection of finitely many sets of the form $[B,(a,b)]$. Then $\mathcal{N}$ is a network for the function space $C_p(X)$. To see this, let $f \in O$ where $O=\bigcap_{x \in F} [x,O_x]$ is a basic open set in $C_p(X)$ where $F \subset X$ is finite and each $O_x$ is an open interval with rational endpoints. For each point $x \in F$, choose $B_x \in \mathcal{B}$ with $x \in B_x$ such that $f(B_x) \subset O_x$. Clearly $f \in \bigcap_{x \in F} \ [B_x,O_x]$. It follows that $\bigcap_{x \in F} \ [B_x,O_x] \subset O$.

Examples include $C_p(\mathbb{R})$, $C_p([0,1])$ and $C_p(\mathbb{R}^\omega)$. All three can be considered subspaces of the product space $\mathbb{R}^c$ where $c$ is the cardinality of the continuum. This is true for any separable metrizable $X$. Note that any separable metrizable $X$ can be embedded in the product space $\mathbb{R}^\omega$. The product space $\mathbb{R}^\omega$ has cardinality $c$. Thus the cardinality of any separable metrizable space $X$ is at most continuum. So $C_p(X)$ is the subspace of a product space of $\le$ continuum many copies of the real lines, hence can be regarded as a subspace of $\mathbb{R}^c$.

A space $L$ has countable extent if every closed and discrete subset of $L$ is countable. The $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ of the separable metric spaces $\left\{X_\alpha: \alpha \in A \right\}$ is a dense and normal subspace of the product space $\prod_{\alpha \in A} X_\alpha$. The normal space $\Sigma_{\alpha \in A} X_\alpha$ has countable extent (hence collectionwise normal). The examples of $C_p(X)$ discussed here are Lindelof and hence have countable extent. Many, though not all, dense normal subspaces of products of separable metric spaces have countable extent. For a dense normal subspace of a product of separable metric spaces, one interesting problem is to find out whether it has countable extent.

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

# 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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$\copyright \ \ 2012$

# 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 $\mathbb{P}$.