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

# Comparing two function spaces

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. Furthermore consider these ordinals as topological spaces endowed with the order topology. It is a well known fact that any continuous real-valued function $f$ defined on either $\omega_1$ or $\omega_1+1$ is eventually constant, i.e., there exists some $\alpha<\omega_1$ such that the function $f$ is constant on the ordinals beyond $\alpha$. Now consider the function spaces $C_p(\omega_1)$ and $C_p(\omega_1+1)$. Thus individually, elements of these two function spaces appear identical. Any $f \in C_p(\omega_1)$ matches a function $f^* \in C_p(\omega_1+1)$ where $f^*$ is the result of adding the point $(\omega_1,a)$ to $f$ where $a$ is the eventual constant real value of $f$. This fact may give the impression that the function spaces $C_p(\omega_1)$ and $C_p(\omega_1+1)$ are identical topologically. The goal in this post is to demonstrate that this is not the case. We compare the two function spaces with respect to some convergence properties (countably tightness and Frechet-Urysohn property) as well as normality.

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Tightness

One topological property that is different between $C_p(\omega_1)$ and $C_p(\omega_1+1)$ is that of tightness. The function space $C_p(\omega_1+1)$ is countably tight, while $C_p(\omega_1)$ is not countably tight.

Let $X$ be a space. The tightness of $X$, denoted by $t(X)$, is the least infinite cardinal $\kappa$ such that for any $A \subset X$ and for any $x \in X$ with $x \in \overline{A}$, there exists $B \subset A$ for which $\lvert B \lvert \le \kappa$ and $x \in \overline{B}$. When $t(X)=\omega$, we say that $X$ has countable tightness or is countably tight. When $t(X)>\omega$, we say that $X$ has uncountable tightness or is uncountably tight.

First, we show that the tightness of $C_p(\omega_1)$ is greater than $\omega$. For each $\alpha<\omega_1$, define $f_\alpha: \omega_1 \rightarrow \left\{0,1 \right\}$ such that $f_\alpha(\beta)=0$ for all $\beta \le \alpha$ and $f_\alpha(\beta)=1$ for all $\beta>\alpha$. Let $g \in C_p(\omega_1)$ be the function that is identically zero. Then $g \in \overline{F}$ where $F$ is defined by $F=\left\{f_\alpha: \alpha<\omega_1 \right\}$. It is clear that for any countable $B \subset F$, $g \notin \overline{B}$. Thus $C_p(\omega_1)$ cannot be countably tight.

The space $\omega_1+1$ is a compact space. The fact that $C_p(\omega_1+1)$ is countably tight follows from the following theorem.

Theorem 1
Let $X$ be a completely regular space. Then the function space $C_p(X)$ is countably tight if and only if $X^n$ is Lindelof for each $n=1,2,3,\cdots$.

Theorem 1 is a special case of Theorem I.4.1 on page 33 of [1] (the countable case). One direction of Theorem 1 is proved in this previous post, the direction that will give us the desired result for $C_p(\omega_1+1)$.

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The Frechet-Urysohn property

In fact, $C_p(\omega_1+1)$ has a property that is stronger than countable tightness. The function space $C_p(\omega_1+1)$ is a Frechet-Urysohn space (see this previous post). Of course, $C_p(\omega_1)$ not being countably tight means that it is not a Frechet-Urysohn space.

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Normality

The function space $C_p(\omega_1+1)$ is not normal. If $C_p(\omega_1+1)$ is normal, then $C_p(\omega_1+1)$ would have countable extent. However, there exists an uncountable closed and discrete subset of $C_p(\omega_1+1)$ (see this previous post). On the other hand, $C_p(\omega_1)$ is Lindelof. The fact that $C_p(\omega_1)$ is Lindelof is highly non-trivial and follows from [2]. The author in [2] showed that if $X$ is a space consisting of ordinals such that $X$ is first countable and countably compact, then $C_p(X)$ is Lindelof.

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Embedding one function space into the other

The two function space $C_p(\omega_1+1)$ and $C_p(\omega_1)$ are very different topologically. However, one of them can be embedded into the other one. The space $\omega_1+1$ is the continuous image of $\omega_1$. Let $g: \omega_1 \longrightarrow \omega_1+1$ be a continuous surjection. Define a map $\psi: C_p(\omega_1+1) \longrightarrow C_p(\omega_1)$ by letting $\psi(f)=f \circ g$. It is shown in this previous post that $\psi$ is a homeomorphism. Thus $C_p(\omega_1+1)$ is homeomorphic to the image $\psi(C_p(\omega_1+1))$ in $C_p(\omega_1)$. The map $g$ is also defined in this previous post.

The homeomposhism $\psi$ tells us that the function space $C_p(\omega_1)$, though Lindelof, is not hereditarily normal.

On the other hand, the function space $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$. Note that $C_p(\omega_1+1)$ is countably tight, which is a hereditary property.

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Remark

There is a mapping that is alluded to at the beginning of the post. Each $f \in C_p(\omega_1)$ is associated with $f^* \in C_p(\omega_1+1)$ which is obtained by appending the point $(\omega_1,a)$ to $f$ where $a$ is the eventual constant real value of $f$. It may be tempting to think of the mapping $f \rightarrow f^*$ as a candidate for a homeomorphism between the two function spaces. The discussion in this post shows that this particular map is not a homeomorphism. In fact, no other one-to-one map from one of these function spaces onto the other function space can be a homeomorphism.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Buzyakova, R. Z., In search of Lindelof $C_p$‘s, Comment. Math. Univ. Carolinae, 45 (1), 145-151, 2004.

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

# Cp(omega 1 + 1) is not normal

In this and subsequent posts, we consider $C_p(X)$ where $X$ is a compact space. Recall that $C_p(X)$ is the space of all continuous real-valued functions defined on $X$ and that it is endowed with the pointwise convergence topology. One of the compact spaces we consider is $\omega_1+1$, the first compact uncountable ordinal. There are many interesting results about the function space $C_p(\omega_1+1)$. In this post we show that $C_p(\omega_1+1)$ is not normal. An even more interesting fact about $C_p(\omega_1+1)$ is that $C_p(\omega_1+1)$ does not have any dense normal subspace [1].

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. The set $\omega_1$ is the first uncountable ordinal. Furthermore consider these ordinals as topological spaces endowed with the order topology. As mentioned above, the space $\omega_1+1$ is the first compact uncountable ordinal. In proving that $C_p(\omega_1+1)$ is not normal, a theorem that is due to D. P. Baturov is utilized [2]. This theorem is also proved in this previous post.

For the basic working of function spaces with the pointwise convergence topology, see the post called Working with the function space Cp(X).

The fact that $C_p(\omega_1+1)$ is not normal is established by the following two points.

• If $C_p(\omega_1+1)$ is normal, then $C_p(\omega_1+1)$ has countable extent, i.e. every closed and discrete subspace of $C_p(\omega_1+1)$ is countable.
• There exists an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$.

We discuss each of the bullet points separately.

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The function space $C_p(\omega_1+1)$ is a dense subspace of $\mathbb{R}^{\omega_1}$, the product of $\omega_1$ many copies of $\mathbb{R}$. According to a result of D. P. Baturov [2], any dense normal subspace of the product of $\omega_1$ many separable metric spaces has countable extent (also see Theorem 1a in this previous post). Thus $C_p(\omega_1+1)$ cannot be normal if the second bullet point above is established.

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Now we show that there exists an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$. For each $\alpha$ with $0<\alpha<\omega_1$, define $h_\alpha:\omega_1 + 1 \rightarrow \left\{0,1 \right\}$ by:

$h_\alpha(\gamma) = \begin{cases} 1 & \mbox{if } \gamma \le \alpha \\ 0 & \mbox{if } \alpha<\gamma \le \omega_1 \end{cases}$

Clearly, $h_\alpha \in C_p(\omega_1 +1)$ for each $\alpha$. Let $H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}$. We show that $H$ is a closed and discrete subspace of $C_p(\omega_1 +1)$. The fact that $H$ is closed in $C_p(\omega_1 +1)$ is establish by the following claim.

Claim 1
Let $h \in C_p(\omega_1 +1) \backslash H$. There exists an open subset $U$ of $C_p(\omega_1 +1)$ such that $h \in U$ and $U \cap H=\varnothing$.

First we get some easy cases out of the way. Suppose that there exists some $\alpha<\omega_1$ such that $h(\alpha) \notin \left\{0,1 \right\}$. Then let $U=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in \mathbb{R} \backslash \left\{0,1 \right\} \right\}$. Clearly $h \in U$ and $U \cap H=\varnothing$.

Another easy case: If $h(\alpha)=0$ for all $\alpha \le \omega_1$, then consider the open set $U$ where $U=\left\{f \in C_p(\omega_1 +1): f(0) \in \mathbb{R} \backslash \left\{1 \right\} \right\}$. Clearly $h \in U$ and $U \cap H=\varnothing$.

From now on we can assume that $h(\omega_1+1) \subset \left\{0,1 \right\}$ and that $h$ is not identically the zero function. Suppose Claim 1 is not true. Then $h \in \overline{H}$. Next observe the following:

Observation.
If $h(\beta)=1$ for some $\beta \le \omega_1$, then $h(\alpha)=1$ for all $\alpha \le \beta$.

To see this, if $h(\alpha)=0$, $h(\beta)=1$ and $\alpha<\beta$, then define the open set $V$ by $V=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (-0.1,0.1) \text{ and } f(\beta) \in (0.9,1.1) \right\}$. Note that $h \in V$ and $V \cap H=\varnothing$, contradicting that $h \in \overline{H}$. So the above observation is valid.

Now either $h(\omega_1)=1$ or $h(\omega_1)=0$. We claim that $h(\omega_1)=1$ is not possible. Suppose that $h(\omega_1)=1$. Let $V=\left\{f \in C_p(\omega_1 +1): f(\omega_1) \in (0.9,1.1) \right\}$. Then $h \in V$ and $V \cap H=\varnothing$, contradicting that $h \in \overline{H}$. It must be the case that $h(\omega_1)=0$.

Because of the continuity of $h$ at the point $x=\omega_1$, of all the $\gamma<\omega_1$ for which $h(\gamma)=1$, there is the largest one, say $\beta$. Now $h(\beta)=1$. According to the observation made above, $h(\alpha)=1$ for all $\alpha \le \beta$. This means that $h=h_\beta$. This is a contradiction since $h \notin H$. Thus Claim 1 must be true and the fact that $H$ is closed is established.

Next we show that $H$ is discrete in $C_p(\omega_1 +1)$. Fix $h_\alpha$ where $0<\alpha<\omega_1$. Let $W=\left\{f \in C_p(\omega_1 +1): f(\alpha) \in (0.9,1.1) \text{ and } f(\alpha+1) \in (-0.1,0.1) \right\}$. It is clear that $h_\alpha \in W$. Furthermore, $h_\gamma \notin W$ for all $\alpha < \gamma$ and $h_\gamma \notin W$ for all $\gamma <\alpha$. Thus $W$ is open such that $\left\{h_\alpha \right\}=W \cap H$. This completes the proof that $H$ is discrete.

We have established that $H$ is an uncountable closed and discrete subspace of $C_p(\omega_1 +1)$. This implies that $C_p(\omega_1 +1)$ is not normal.

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Remarks

The set $H=\left\{h_\alpha: 0<\alpha<\omega_1 \right\}$ as defined above is closed and discrete in $C_p(\omega_1 +1)$. However, the set $H$ is not discrete in a larger subspace of the product space. The set $H$ is also a subset of the following $\Sigma$-product:

$\Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}$

Because $\Sigma(\omega_1)$ is the $\Sigma$-product of separable metric spaces, it is normal (see here). By Theorem 1a in this previous post, $\Sigma(\omega_1)$ would have countable extent. Thus the set $H$ cannot be closed and discrete in $\Sigma(\omega_1)$. We can actually see this directly. Let $\alpha<\omega_1$ be a limit ordinal. Define $t:\omega_1 + 1 \rightarrow \left\{0,1 \right\}$ by $t(\beta)=1$ for all $\beta<\alpha$ and $t(\beta)=0$ for all $\beta \ge \alpha$. Clearly $t \notin C_p(\omega_1 +1)$ and $t \in \Sigma(\omega_1)$. Furthermore, $t \in \overline{H}$ (the closure is taken in $\Sigma(\omega_1)$).

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Reference

1. Arhangel’skii, A. V., Normality and Dense Subspaces, Proc. Amer. Math. Soc., 48, no. 2, 283-291, 2001.
2. Baturov, D. P., Normality in dense subspaces of products, Topology Appl., 36, 111-116, 1990.

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

# A useful lemma for proving normality

In this post we discuss a lemma (Lemma 1 below) that is useful for proving normality. In some cases, it is more natural using this lemma to prove that a space is normal than using the definition of normality. The proof of Lemma 1 is not difficult. Yet it simplifies some proofs of normality. One reason is that the derivation of two disjoint open sets that are to separate two disjoint closed sets is done in the lemma, thus simplifying the main proof at hand. The lemma is well known and is widely used in the literature. See Lemma 1.5.15 in [1]. Two advanced examples of applications are [2] and [3]. After proving the lemma, we give three elementary applications of the lemma. One of the applications is a characterization of perfectly normal spaces. This characterization is, in some cases, easier to use, e.g. making it easy to show that perfectly normal implies hereditarily normal.

In this post, we only consider spaces that are regular and $T_1$. A space $X$ is regular if for each open set $U \subset X$ and for each $x \in U$, there exists an open $V \subset X$ with $x \in V \subset \overline{V} \subset U$. A space is $T_1$ if every set with only one point is a closed set.

Lemma 1
A space $Y$ is a normal space if the following condition (Condition 1) is satisfied:

1. For each closed subset $L$ of $Y$, and for each open subset $M$ of $Y$ with $L \subset M$, there exists a sequence $M_1,M_2,M_3,\cdots$ of open subsets of $Y$ such that $L \subset \bigcup_{i=1}^\infty M_i$ and $\overline{M_i} \subset M$ for each $i$.

Proof of Lemma 1
Suppose the space $Y$ satisfies condition 1. Let $H$ and $K$ be disjoint closed subsets of the space $Y$. Consider $H \subset U=Y \backslash K$. Using condition 1, there exists a sequence $U_1,U_2,U_3,\cdots$ of open subsets of the space $Y$ such that $H \subset \bigcup_{i=1}^\infty U_i$ and $\overline{U_i} \cap K=\varnothing$ for each $i$. Consider $K \subset V=Y \backslash H$. Similarly, there exists a sequence $V_1,V_2,V_3,\cdots$ of open subsets of the space $Y$ such that $K \subset \bigcup_{i=1}^\infty V_i$ and $\overline{V_i} \cap H=\varnothing$ for each $i$.

For each positive integer $n$, define the open sets $U_n^*$ and $V_n^*$ as follows:

$U_n^*=U_n \backslash \bigcup_{k=1}^n \overline{V_k}$

$V_n^*=V_n \backslash \bigcup_{k=1}^n \overline{U_k}$

Let $P=\bigcup_{n=1}^\infty U_n^*$ and $Q=\bigcup_{n=1}^\infty V_n^*$. It is clear $P$ and $Q$ are open and that $H \subset P$ and $K \subset Q$. We claim that $P$ and $Q$ are disjoint. Suppose $y \in P \cap Q$. Then $y \in U_n^*$ for some $n$ and $y \in V_m^*$ for some $m$. Assume that $n \le m$. The fact that $y \in U_n^*$ implies $y \in U_n$. The fact that $y \in V_m^*$ implies that $y \notin \overline{U_j}$ for all $j \le m$. In particular, $y \notin U_n$, a contradiction. Thus $P \cap Q=\varnothing$. This completes the proof that the space $Y$ is normal. $\blacksquare$

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Spaces with nice bases

One application is that spaces with a certain type of bases satisfy condition 1 and thus are normal. For example, spaces with bases that are countable and spaces with bases that are $\sigma$-locally finite. Spaces with these bases are metrizable. The proof that these spaces are metrizable will be made easier if they can be shown to be normal first. The Urysohn functions (the functions described in Urysohn’s lemma) can then be used to embed the space in question into some universal space that is known to be metrizable. Using regularity and Lemma 1, it is straightforward to verify the following three propositions.

Proposition 2
Let $X$ be a regular space with a countable base. Then $X$ is normal.

Proposition 3
Let $X$ be a regular space with a $\sigma$-locally finite base. Then $X$ is normal.

Proposition 4
Let $X$ be a regular space with a $\sigma$-discrete base. Then $X$ is normal.

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A characterization of perfectly normal spaces

Another application of Lemma 1 is that it leads naturally to a characterization of perfect normality. Recall that a space $X$ is perfectly normal if $X$ is normal and perfect. A space $X$ is perfect if every closed subset of $X$ is a $G_\delta$ set (i.e. the intersection of countably many open subsets of $X$). Equivalently a space $X$ is perfect if and only if every open subset of $X$ is an $F_\sigma$ set, i.e., the union of countably many closed subsets of $X$. We have the following theorem.

Theorem 5
A space $Y$ is perfectly normal if and only if the following condition holds.

1. For each open subset $M$ of $Y$, there exists a sequence $M_1,M_2,M_3,\cdots$ of open subsets of $Y$ such that $M \subset \bigcup_{i=1}^\infty M_i$ and $\overline{M_i} \subset M$ for each $i$.

Clearly, condition 2 is strongly than condition 1.

Proof of Theorem 5
$\Longrightarrow$
Suppose that the space $Y$ is perfectly normal. Let $M$ be a non-empty open subset of $Y$. Then $M=\bigcup_{n=1}^\infty P_n$ where each $P_n$ is a closed subset of $Y$. Using normality of $Y$, for each $n$, there exists open subset $M_n$ of $Y$ such that $P_n \subset M_n \subset \overline{M_n} \subset M$. Then consition 2 is satisfied.

$\Longleftarrow$
Suppose condition 2 holds, which implies condition 1 of Lemma 1. Then $Y$ is normal. It is clear that condition 2 implies that every open subset of $Y$ is an $F_\sigma$ set. $\blacksquare$

The characterization of perfectly normal spaces in Theorem 5 is hereditary. This means that any subspace of a perfectly normal space is also perfectly normal. In particular, perfectly normal implies hereditarily normal. Thus we have the following theorem.

Theorem 6
Condition 2 in Theorem 5 is hereditary, i.e., if a space satisfies Condition 2, every subspace satisfies Condition 2. Therefore if the space $Y$ is a perfectly normal space, then every subspace of $Y$ is also perfectly normal. In particular, if $Y$ is perfectly normal, then $Y$ is hereditarily normal (i.e. every subspace of $Y$ is normal).

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Normality is hereditary with respect to $F_\sigma$ subsets

Normality is not a hereditary notion. Lemma 1 can used to show that normality is hereditary with respect to $F_\sigma$ subspaces.

Theorem 7
Let $Y$ be a normal space. Then every $F_\sigma$ subspace of $Y$ is normal.

Proof of Theorem 7
Let $H$ be a subspace of $Y$ such that $H=\bigcup_{n=1}^\infty P_n$ where each $P_n$ is closed subset of $Y$. Let $L$ be a closed subset of $H$ and let $M$ be an open subset of $H$ such that $L \subset M$. We need to find $M_1,M_2,M_3,\cdots$, open in $H$, such that $L \subset \bigcup_{i=1}^\infty M_i$ and $\overline{M_i} \subset M$ for all $i$ (closure of $M_i$ is within $H$).

Let $U$ be an open subset of $Y$ such that $M=U \cap H$. For each positive integer $n$, let $H_n=P_n \cap L$. Obviously $H_n$ is closed in $H$. It is also the case that $H_n$ is closed in $Y$. To see this, let $p \in Y$ be a limit point of $H_n$. Then $p$ is a limit point of $P_n$. Hence $p \in P_n$ since $P_n$ is closed in $Y$. We now have $p \in H$. The point $p$ is also a limit point of $L$. Thus $p \in L$ since $L$ is closed in $H$. Now we have $p \in H_n=P_n \cap L$, proving that $H_n$ is closed in $Y$.

Now we have $H_n \subset U$ for all $n$. By Lemma 1, for each $n$, there exists a sequence $U_{n,1},U_{n,2},U_{n,3},\cdots$ of open subsets of $Y$ such that $H_n \subset \bigcup_{j=1}^\infty U_{n,j}$ and $\overline{U_{n,j}} \subset U$ for all $j$. Note that $L=\bigcup_{n=1}^\infty H_n$. Rename $M_{n,j}=U_{n,j} \cap H$ over all $n,j$ by the sequence $M_1,M_2,M_3,\cdots$. Then $L \subset \bigcup_{i=1}^\infty M_i$. It also follows that $\overline{M_i} \subset M$ for all $i$ (closure of $M_i$ is within $H$). This completes the proof that the $F_\sigma$ set $H$ is normal. $\blacksquare$

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Gruenhage, G., Normality in $X^2$ for complete $X$, Trans. Amer. Math. Soc., 340 (2), 563-586, 1993.
3. Nyikos, P., A compact nonmetrizable space $P$ such that $P^2$ is completely normal, Topology Proc., 2, 359-363, 1977.

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$\copyright \ 2014-2015 \text{ by Dan Ma}$ Revised April 14, 2015

# Normal x compact needs not be subnormal

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Reference

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

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

# An example of a normal but not Lindelof Cp(X)

In this post, we discuss an example of a function space $C_p(X)$ that is normal and not Lindelof (as indicated in the title). Interestingly, much more can be said about this function space. In this post, we show that there exists a space $X$ such that

• $C_p(X)$ is collectionwise normal and not paracompact,
• $C_p(X)$ is not Lindelof but contains a dense Lindelof subspace,
• $C_p(X)$ is not first countable but is a Frechet space,
• As a corollary of the previous point, $C_p(X)$ cannot contain a copy of the compact space $\omega_1+1$,
• $C_p(X)$ is homeomorphic to $C_p(X)^\omega$,
• $C_p(X)$ is not hereditarily normal,
• $C_p(X)$ is not metacompact.

A short and quick description of the space $X$ is that $X$ is the one-point Lindelofication of an uncountable discrete space. As shown below, the function space $C_p(X)$ is intimately related to a $\Sigma$-product of copies of real lines. The results listed above are merely an introduction to this wonderful example and are derived by examining the $\Sigma$-products of copies of real lines. Deep results about $\Sigma$-product of real lines abound in the literature. The references listed at the end are a small sample. Example 3.2 in [2] is another interesting illustration of this example.

We now define the domain space $X=L_\tau$. In the discussion that follows, the Greek letter $\tau$ is always an uncountable cardinal number. Let $D_\tau$ be a set with cardinality $\tau$. Let $p$ be a point not in $D_\tau$. Let $L_\tau=D_\tau \cup \left\{p \right\}$. Consider the following topology on $L_\tau$:

• Each point in $D_\tau$ an isolated point, and
• open neighborhoods at the point $p$ are of the form $L_\tau-K$ where $K \subset D_\tau$ is countable.

It is clear that $L_\tau$ is a Lindelof space. The Lindelof space $L_\tau$ is sometimes called the one-point Lindelofication of the discrete space $D_\tau$ since it is a Lindelof space that is obtained by adding one point to a discrete space.

Consider the function space $C_p(L_\tau)$. See this post for general information on the pointwise convergence topology of $C_p(Y)$ for any completely regular space $Y$.

All the facts about $C_p(X)=C_p(L_\tau)$ mentioned at the beginning follow from the fact that $C_p(L_\tau)$ is homeomorphic to the $\Sigma$-product of $\tau$ many copies of the real lines. Specifically, $C_p(L_\tau)$ is homeomorphic to the following subspace of the product space $\mathbb{R}^\tau$.

$\Sigma_{\alpha<\tau}\mathbb{R}=\left\{ x \in \mathbb{R}^\tau: x_\alpha \ne 0 \text{ for at most countably many } \alpha<\tau \right\}$

Thus understanding the function space $C_p(L_\tau)$ is a matter of understanding a $\Sigma$-product of copies of the real lines. First, we establish the homeomorphism and then discuss the properties of $C_p(L_\tau)$ indicated above.

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The Homeomorphism

For each $f \in C_p(L_\tau)$, it is easily seen that there is a countable set $C \subset D_\tau$ such that $f(p)=f(y)$ for all $y \in D_\tau-C$. Let $W_0=\left\{f \in C_p(L_\tau): f(p)=0 \right\}$. Then each $f \in W_0$ has non-zero values only on a countable subset of $D_\tau$. Naturally, $W_0$ and $\Sigma_{\alpha<\tau}\mathbb{R}$ are homeomorphic.

We claim that $C_p(L_\tau)$ is homeomorphic to $W_0 \times \mathbb{R}$. For each $f \in C_p(L_\tau)$, define $h(f)=(f-f(p),f(p))$. Here, $f-f(p)$ is the function $g \in C_p(L_\tau)$ such that $g(x)=f(x)-f(p)$ for all $x \in L_\tau$. Clearly $h(f)$ is well-defined and $h(f) \in W_0 \times \mathbb{R}$. It can be readily verified that $h$ is a one-to-one map from $C_p(L_\tau)$ onto $W_0 \times \mathbb{R}$. It is not difficult to verify that both $h$ and $h^{-1}$ are continuous.

We use the notation $X_1 \cong X_2$ to mean that the spaces $X_1$ and $X_2$ are homeomorphic. Then we have:

$C_p(L_\tau) \ \cong \ W_0 \times \mathbb{R} \ \cong \ (\Sigma_{\alpha<\tau}\mathbb{R}) \times \mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}$

Thus $C_p(L_\tau) \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}$. This completes the proof that $C_p(L_\tau)$ is topologically the $\Sigma$-product of $\tau$ many copies of the real lines.

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Looking at the $\Sigma$-Product

Understanding the function space $C_p(L_\tau)$ is now reduced to the problem of understanding a $\Sigma$-product of copies of the real lines. Most of the facts about $\Sigma$-products that we need have already been proved in previous blog posts.

In this previous post, it is established that the $\Sigma$-product of separable metric spaces is collectionwise normal. Thus $C_p(L_\tau)$ is collectionwise normal. The $\Sigma$-product of spaces, each of which has at least two points, always contains a closed copy of $\omega_1$ with the ordered topology (see the lemma in this previous post). Thus $C_p(L_\tau)$ contains a closed copy of $\omega_1$ and hence can never be paracompact (and thus not Lindelof).

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Consider the following subspace of the $\Sigma$-product $\Sigma_{\alpha<\tau}\mathbb{R}$:

$\sigma_\tau=\left\{ x \in \Sigma_{\alpha<\tau}\mathbb{R}: x_\alpha \ne 0 \text{ for at most finitely many } \alpha<\tau \right\}$

In this previous post, it is shown that $\sigma_\tau$ is a Lindelof space. Though $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is not Lindelof, it has a dense Lindelof subspace, namely $\sigma_\tau$.

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A space $Y$ is first countable if there exists a countable local base at each point $y \in Y$. A space $Y$ is a Frechet space (or is Frechet-Urysohn) if for each $y \in Y$, if $y \in \overline{A}$ where $A \subset Y$, then there exists a sequence $\left\{y_n: n=1,2,3,\cdots \right\}$ of points of $A$ such that the sequence converges to $y$. Clearly, any first countable space is a Frechet space. The converse is not true (see Example 1 in this previous post).

For any uncountable cardinal number $\tau$, the product $\mathbb{R}^\tau$ is not first countable. In fact, any dense subspace of $\mathbb{R}^\tau$ is not first countable. In particular, the $\Sigma$-product $\Sigma_{\alpha<\tau}\mathbb{R}$ is not first countable. In this previous post, it is shown that the $\Sigma$-product of first countable spaces is a Frechet space. Thus $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is a Frechet space.

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As a corollary of the previous point, $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ cannot contain a homeomorphic copy of any space that is not Frechet. In particular, it cannot contain a copy of any compact space that is not Frechet. For example, the compact space $\omega_1+1$ is not embeddable in $C_p(L_\tau)$. The interest in compact subspaces of $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is that any compact space that is topologically embeddable in a $\Sigma$-product of real lines is said to be Corson compact. Thus any Corson compact space is a Frechet space.

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It can be readily verified that

$\Sigma_{\alpha<\tau}\mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \cdots \ \text{(countably many times)}$

Thus $C_p(L_\tau) \cong C_p(L_\tau)^\omega$. In particular, $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$ due to the following observation:

$C_p(L_\tau) \times C_p(L_\tau) \cong C_p(L_\tau)^\omega \times C_p(L_\tau)^\omega \cong C_p(L_\tau)^\omega \cong C_p(L_\tau)$

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As a result of the peculiar fact that $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$, it can be concluded that $C_p(L_\tau)$, though normal, is not hereditarily normal. This follows from an application of Katetov’s theorem. The theorem states that if $Y_1 \times Y_2$ is hereditarily normal, then either $Y_1$ is perfectly normal or every countably infinite subset of $Y_2$ is closed and discrete (see this previous post). The function space $C_p(L _\tau)$ is not perfectly normal since it contains a closed copy of $\omega_1$. On the other hand, there are plenty of countably infinite subsets of $C_p(L _\tau)$ that are not closed and discrete. As a Frechet space, $C_p(L _\tau)$ has many convergent sequences. Each such sequence without the limit is a countably infinite set that is not closed and discrete. As an example, let $\left\{x_1,x_2,x_3,\cdots \right\}$ be an infinite subset of $D_\tau$ and consider the following:

$C=\left\{f_n: n=1,2,3,\cdots \right\}$

where $f_n$ is such that $f_n(x_n)=n$ and $f_n(x)=0$ for each $x \in L_\tau$ with $x \ne x_n$. Note that $C$ is not closed and not discrete since the points in $C$ converge to $g \in \overline{C}$ where $g$ is the zero-function. Thus $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$ is not hereditarily normal.

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It is well known that collectionwise normal metacompact space is paracompact (see Theorem 5.3.3 in [4] where metacompact is referred to as weakly paracompact). Since $C_p(L_\tau)$ is collectionwise normal and not paracompact, $C_p(L_\tau)$ can never be metacompact.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Bella, A., Masami, S., Tight points of Pixley-Roy hyperspaces, Topology Appl., 160, 2061-2068, 2013.
3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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