Comparing two function spaces

Posted on August 5, 2014 by Dan Ma

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

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

Posted on August 4, 2014 by Dan Ma

This is another post that discusses what $C_p(X)$ is like when $X$ is a compact space. In this post, we discuss the example $C_p(\omega_1+1)$ where $\omega_1+1$ is the first compact uncountable ordinal. Note that $\omega_1+1$ is the successor to $\omega_1$ , which is the first (or least) uncountable ordinal. The function space $C_p(\omega_1+1)$ is monolithic and is a Frechet-Urysohn space. Interestingly, the first property is possessed by $C_p(X)$ for all compact spaces $X$ . The second property is possessed by all compact scattered spaces. After we discuss $C_p(\omega_1+1)$ , we discuss briefly the general results for $C_p(X)$ .

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Initial discussion

The function space $C_p(\omega_1+1)$ is a dense subspace of the product space $\mathbb{R}^{\omega_1}$ . In fact, $C_p(\omega_1+1)$ is homeomorphic to a subspace of the following subspace of $\mathbb{R}^{\omega_1}$ :

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

The subspace $\Sigma(\omega_1)$ is the $\Sigma$ -product of $\omega_1$ many copies of the real line $\mathbb{R}$ . The $\Sigma$ -product of separable metric spaces is monolithic (see here). The $\Sigma$ -product of first countable spaces is Frechet-Urysohn (see here). Thus $\Sigma(\omega_1)$ has both of these properties. Since the properties of monolithicity and being Frechet-Urysohn are carried over to subspaces, the function space $C_p(\omega_1+1)$ has both of these properties. The key to the discussion is then to show that $C_p(\omega_1+1)$ is homeopmophic to a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ .

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Connection to $\Sigma$ -product

We show that the function space $C_p(\omega_1+1)$ is homeomorphic to a subspace of the $\Sigma$ -product of $\omega_1$ many copies of the real lines. Let $Y_0$ be the following subspace of $C_p(\omega_1+1)$ :

$Y_0=\left\{f \in C_p(\omega_1+1): f(\omega_1)=0 \right\}$

Every function in $Y_0$ has non-zero values at only countably points of $\omega_1+1$ . Thus $Y_0$ can be regarded as a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ .

By Theorem 1 in this previous post, $C_p(\omega_1+1) \cong Y_0 \times \mathbb{R}$ , i.e, the function space $C_p(\omega_1+1)$ is homeomorphic to the product space $Y_0 \times \mathbb{R}$ . On the other hand, the product $Y_0 \times \mathbb{R}$ can also be regarded as a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ . Basically adding one additional factor of the real line to $Y_0$ still results in a subspace of the $\Sigma$ -product. Thus we have:

$C_p(\omega_1+1) \cong Y_0 \times \mathbb{R} \subset \Sigma(\omega_1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Thus $C_p(\omega_1+1)$ possesses all the hereditary properties of $\Sigma(\omega_1)$ . Another observation we can make is that $\Sigma(\omega_1)$ is not hereditarily normal. The function space $C_p(\omega_1+1)$ is not normal (see here). The $\Sigma$ -product $\Sigma(\omega_1)$ is normal (see here). Thus $\Sigma(\omega_1)$ is not hereditarily normal.

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A closer look at $C_p(\omega_1+1)$

In fact $C_p(\omega_1+1)$ has a stronger property that being monolithic. It is strongly monolithic. We use homeomorphic relation in (1) above to get some insight. Let $h$ be a homeomorphism from $C_p(\omega_1+1)$ onto $Y_0 \times \mathbb{R}$ . For each $\alpha<\omega_1$ , let $H_\alpha$ be defined as follows:

$H_\alpha=\left\{f \in C_p(\omega_1+1): f(\gamma)=0 \ \forall \ \alpha<\gamma<\omega_1 \right\}$

Clearly $H_\alpha \subset Y_0$ . Furthermore $H_\alpha$ can be considered as a subspace of $\mathbb{R}^\omega$ and is thus metrizable. Let $A$ be a countable subset of $C_p(\omega_1+1)$ . Then $h(A) \subset H_\alpha \times \mathbb{R}$ for some $\alpha<\omega_1$ . The set $H_\alpha \times \mathbb{R}$ is metrizable. The set $H_\alpha \times \mathbb{R}$ is also a closed subset of $Y_0 \times \mathbb{R}$ . Then $\overline{A}$ is contained in $H_\alpha \times \mathbb{R}$ and is therefore metrizable. We have shown that the closure of every countable subspace of $C_p(\omega_1+1)$ is metrizable. In other words, every separable subspace of $C_p(\omega_1+1)$ is metrizable. This property follows from the fact that $C_p(\omega_1+1)$ is strongly monolithic.

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Monolithicity and Frechet-Urysohn property

As indicated at the beginning, the $\Sigma$ -product $\Sigma(\omega_1)$ is monolithic (in fact strongly monolithic; see here) and is a Frechet-Urysohn space (see here). Thus the function space $C_p(\omega_1+1)$ is both strongly monolithic and Frechet-Urysohn.

Let $\tau$ be an infinite cardinal. A space $X$ is $\tau$ -monolithic if for any $A \subset X$ with $\lvert A \lvert \le \tau$ , we have $nw(\overline{A}) \le \tau$ . A space $X$ is monolithic if it is $\tau$ -monolithic for all infinite cardinal $\tau$ . It is straightforward to show that $X$ is monolithic if and only of for every subspace $Y$ of $X$ , the density of $Y$ equals to the network weight of $Y$ , i.e., $d(Y)=nw(Y)$ . A longer discussion of the definition of monolithicity is found here.

A space $X$ is strongly $\tau$ -monolithic if for any $A \subset X$ with $\lvert A \lvert \le \tau$ , we have $w(\overline{A}) \le \tau$ . A space $X$ is strongly monolithic if it is strongly $\tau$ -monolithic for all infinite cardinal $\tau$ . It is straightforward to show that $X$ is strongly monolithic if and only if for every subspace $Y$ of $X$ , the density of $Y$ equals to the weight of $Y$ , i.e., $d(Y)=w(Y)$ .

In any monolithic space, the density and the network weight coincide for any subspace, and in particular, any subspace that is separable has a countable network. As a result, any separable monolithic space has a countable network. Thus any separable space with no countable network is not monolithic, e.g., the Sorgenfrey line. On the other hand, any space that has a countable network is monolithic.

In any strongly monolithic space, the density and the weight coincide for any subspace, and in particular any separable subspace is metrizable. Thus being separable is an indicator of metrizability among the subspaces of a strongly monolithic space. As a result, any separable strongly monolithic space is metrizable. Any separable space that is not metrizable is not strongly monolithic. Thus any non-metrizable space that has a countable network is an example of a monolithic space that is not strongly monolithic, e.g., the function space $C_p([0,1])$ . It is clear that all metrizable spaces are strongly monolithic.

The function space $C_p(\omega_1+1)$ is not separable. Since it is strongly monolithic, every separable subspace of $C_p(\omega_1+1)$ is metrizable. We can see this by knowing that $C_p(\omega_1+1)$ is a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ , or by using the homeomorphism $h$ as in the previous section.

For any compact space $X$ , $C_p(X)$ is countably tight (see this previous post). In the case of the compact uncountable ordinal $\omega_1+1$ , $C_p(\omega_1+1)$ has the stronger property of being Frechet-Urysohn. A space $Y$ is said to be a Frechet-Urysohn space (also called a Frechet space) if for each $y \in Y$ and for each $M \subset Y$ , if $y \in \overline{M}$ , then there exists a sequence $\left\{y_n \in M: n=1,2,3,\cdots \right\}$ such that the sequence converges to $y$ . As we shall see below, $C_p(X)$ is rarely Frechet-Urysohn.

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General discussion

For any compact space $X$ , $C_p(X)$ is monolithic but does not have to be strongly monolithic. The monolithicity of $C_p(X)$ follows from the following theorem, which is Theorem II.6.8 in [1].

Theorem 1
Then the function space $C_p(X)$ is monolithic if and only if $X$ is a stable space.

See chapter 3 section 6 of [1] for a discussion of stable spaces. We give the definition here. A space $X$ is stable if for any continuous image $Y$ of $X$ , the weak weight of $Y$ , denoted by $ww(Y)$ , coincides with the network weight of $Y$ , denoted by $nw(Y)$ . In [1], $ww(Y)$ is notated by $iw(Y)$ . The cardinal function $ww(Y)$ is the minimum cardinality of all $w(T)$ , the weight of $T$ , for which there exists a continuous bijection from $Y$ onto $T$ .

All compact spaces are stable. Let $X$ be compact. For any continuous image $Y$ of $X$ , $Y$ is also compact and $ww(Y)=w(Y)$ , since any continuous bijection from $Y$ onto any space $T$ is a homeomorphism. Note that $ww(Y) \le nw(Y) \le w(Y)$ always holds. Thus $ww(Y)=w(Y)$ implies that $ww(Y)=nw(Y)$ . Thus we have:

Corollary 2
Let $X$ be a compact space. Then the function space $C_p(X)$ is monolithic.

However, the strong monolithicity of $C_p(\omega_1+1)$ does not hold in general for $C_p(X)$ for compact $X$ . As indicated above, $C_p([0,1])$ is monolithic but not strongly monolithic. The following theorem is Theorem II.7.9 in [1] and characterizes the strong monolithicity of $C_p(X)$ .

Theorem 3
Let $X$ be a space. Then $C_p(X)$ is strongly monolithic if and only if $X$ is simple.

A space $X$ is $\tau$ -simple if whenever $Y$ is a continuous image of $X$ , if the weight of $Y$ $\le \tau$ , then the cardinality of $Y$ $\le \tau$ . A space $X$ is simple if it is $\tau$ -simple for all infinite cardinal numbers $\tau$ . Interestingly, any separable metric space that is uncountable is not $\omega$ -simple. Thus $[0,1]$ is not $\omega$ -simple and $C_p([0,1])$ is not strongly monolithic, according to Theorem 3.

For compact spaces $X$ , $C_p(X)$ is rarely a Frechet-Urysohn space as evidenced by the following theorem, which is Theorem III.1.2 in [1].

Theorem 4
Let $X$ be a compact space. Then the following conditions are equivalent.

$C_p(X)$ is a Frechet-Urysohn space.
$C_p(X)$ is a k-space.
The compact space $X$ is a scattered space.

A space $X$ is a scattered space if for every non-empty subspace $Y$ of $X$ , there exists an isolated point of $Y$ (relative to the topology of $Y$ ). Any space of ordinals is scattered since every non-empty subset has a least element. Thus $\omega_1+1$ is a scattered space. On the other hand, the unit interval $[0,1]$ with the Euclidean topology is not scattered. According to this theorem, $C_p([0,1])$ cannot be a Frechet-Urysohn space.

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Reference

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

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

Cp(X) is countably tight when X is compact

Posted on August 3, 2014 by Dan Ma

Let $X$ be a completely regular space (also called Tychonoff space). If $X$ is a compact space, what can we say about the function space $C_p(X)$ , the space of all continuous real-valued functions with the pointwise convergence topology? When $X$ is an uncountable space, $C_p(X)$ is not first countable at every point. This follows from the fact that $C_p(X)$ is a dense subspace of the product space $\mathbb{R}^X$ and that no dense subspace of $\mathbb{R}^X$ can be first countable when $X$ is uncountable. However, when $X$ is compact, $C_p(X)$ does have a convergence property, namely $C_p(X)$ is countably tight.

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Tightness

Let $X$ be a completely regular 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 $Y$ has countable tightness or is countably tight. When $t(X)>\omega$ , we say that $X$ has uncountable tightness or is uncountably tight. Clearly any first countable space is countably tight. There are other convergence properties in between first countability and countable tightness, e.g., the Frechet-Urysohn property. The notion of countable tightness and tightness in general is discussed in further details here.

The fact that $C_p(X)$ is countably tight for any compact $X$ 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 the countable case of Theorem I.4.1 on page 33 of [1]. We prove one direction of Theorem 1, the direction that will give us the desired result for $C_p(X)$ where $X$ is compact.

Proof of Theorem 1
The direction $\Longleftarrow$
Suppose that $X^n$ is Lindelof for each positive integer. Let $f \in C_p(X)$ and $f \in \overline{H}$ where $H \subset C_p(X)$ . For each positive integer $n$ , we define an open cover $\mathcal{U}_n$ of $X^n$ .

Let $n$ be a positive integer. Let $t=(x_1,\cdots,x_n) \in X^n$ . Since $f \in \overline{H}$ , there is an $h_t \in H$ such that $\lvert h_t(x_j)-f(x_j) \lvert <\frac{1}{n}$ for all $j=1,\cdots,n$ . Because both $h_t$ and $f$ are continuous, for each $j=1,\cdots,n$ , there is an open set $W(x_j) \subset X$ with $x_j \in W(x_j)$ such that $\lvert h_t(y)-f(y) \lvert < \frac{1}{n}$ for all $y \in W(x_j)$ . Let the open set $U_t$ be defined by $U_t=W(x_1) \times W(x_2) \times \cdots \times W(x_n)$ . Let $\mathcal{U}_n=\left\{U_t: t=(x_1,\cdots,x_n) \in X^n \right\}$ .

For each $n$ , choose $\mathcal{V}_n \subset \mathcal{U}_n$ be countable such that $\mathcal{V}_n$ is a cover of $X^n$ . Let $K_n=\left\{h_t: t \in X^n \text{ such that } U_t \in \mathcal{V}_n \right\}$ . Let $K=\bigcup_{n=1}^\infty K_n$ . Note that $K$ is countable and $K \subset H$ .

We now show that $f \in \overline{K}$ . Choose an arbitrary positive integer $n$ . Choose arbitrary points $y_1,y_2,\cdots,y_n \in X$ . Consider the open set $U$ defined by

$U=\left\{g \in C_p(X): \forall \ j=1,\cdots,n, \lvert g(y_j)-f(y_j) \lvert <\frac{1}{n} \right\}$

We wish to show that $U \cap K \ne \varnothing$ . Choose $U_t \in \mathcal{V}_n$ such that $(y_1,\cdots,y_n) \in U_t$ where $t=(x_1,\cdots,x_n) \in X^n$ . Consider the function $h_t$ that goes with $t$ . It is clear from the way $h_t$ is chosen that $\lvert h_t(y_j)-f(x_j) \lvert<\frac{1}{n}$ for all $j=1,\cdots,n$ . Thus $h_t \in K_n \cap U$ , leading to the conclusion that $f \in \overline{K}$ . The proof that $C_p(X)$ is countably tight is completed.

The direction $\Longrightarrow$
See Theorem I.4.1 of [1].

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Remarks

As shown above, countably tightness is one convergence property of $C_p(X)$ that is guaranteed when $X$ is compact. In general, it is difficult for $C_p(X)$ to have stronger convergence properties such as the Frechet-Urysohn property. It is well known $C_p(\omega_1+1)$ is Frechet-Urysohn. According to Theorem II.1.2 in [1], for any compact space $X$ , $C_p(X)$ is a Frechet-Urysohn space if and only if the compact space $X$ is a scattered space.

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Reference

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

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$\copyright \ 2014 - 2015 \text{ by Dan Ma}$

Sigma-products of first countable spaces

Posted on April 8, 2014 by Dan Ma

A product space is never first countable if there are uncountably many factors. For example, $\prod_{\alpha < \omega_1}\mathbb{R}=\mathbb{R}^{\omega_1}$ is not first countable. In fact any dense subspace of $\mathbb{R}^{\omega_1}$ is not first countable. In particular, the subspace of $\mathbb{R}^{\omega_1}$ consisting of points which have at most countably many non-zero coordinates is not first countable. This subspace is called the $\Sigma$ -product of $\omega_1$ many copies of the real line $\mathbb{R}$ and is denoted by $\Sigma_{\alpha<\omega_1} \mathbb{R}$ . However, this $\Sigma$ -product is a Frechet space (or a Frechet-Urysohn space). In this post, we show that the $\Sigma$ -product of first countable spaces is a Frechet space.

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$ . Fix a point $a \in X$ . Consider the following subspace of $X$ :

$\Sigma_{\alpha \in A} X_\alpha(a)=\left\{x \in X: x_\alpha \ne a_\alpha \text{ for at most countably many } \alpha \in A \right\}$

The above subspace of $X$ is called the $\Sigma$ -product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ about the base point $a$ . When the base point is understood, we simply say the $\Sigma$ -product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ and use the notation $\Sigma_{\alpha \in A} X_\alpha$ to denote the space.

For each $y \in \Sigma_{\alpha \in A} X_\alpha$ , define $S(y)$ to be the set of all $\alpha \in A$ such that $y_\alpha \ne a_\alpha$ , i.e., the support of the point $y$ . Another notion of support is that of standard basic open sets in the product topology. A standard basic open set is a set $O=\prod_{\alpha \in A} O_\alpha$ where each $O_\alpha$ is an open subset of $X_\alpha$ . The support of $O$ , denoted by $supp(O)$ is the finite set of all $\alpha \in A$ such that $O_\alpha \ne X_\alpha$ .

A space $Y$ is said to be first countable if there exists a countable local base at each point in $Y$ . A space $Y$ is said to be a Frechet space if for each $y \in Y$ and for each $M \subset Y$ , if $y \in \overline{M}$ , then there exists a sequence $\left\{y_n: n=1,2,3,\cdots \right\}$ of points of $M$ such that the sequence converges to $y$ . Frechet spaces also go by the name of Frechet-Urysohn spaces. Clearly, any first countable space is Frechet. The converse is not true (see Example 1 in this post). We prove the following theorem.

Theorem 1

$X_\alpha$

$\Sigma$

$\Sigma_{\alpha \in A} X_\alpha$

Proof of Theorem 1
Let $\Sigma=\Sigma_{\alpha \in A} X_\alpha$ . Let $M \subset \Sigma$ and let $x \in \overline{M}$ . We proceed to define a sequence of points $t_n \in M$ such that the sequence $t_n$ converges to $x$ . For each $\alpha \in A$ , choose a countable local base $\left\{B_{\alpha,j}: j=1,2,3,\cdots \right\}$ at the point $x_\alpha \in X_\alpha$ . Assume that $B_{\alpha,1} \supset B_{\alpha,2} \supset B_{\alpha,3} \supset \cdots$ . Then enumerate the countable set $S(x)$ by $S(x)=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\cdots \right\}$ . Let $C_1=\left\{\beta_{1,1} \right\}$ . The following set $O_1$ is an open subset of $\Sigma$ .

$O_1=\biggl(\prod_{\alpha \in C_1} B_{\alpha,1} \times \prod_{\alpha \in A-C_1} X_\alpha \biggr) \cap \Sigma$

Note that $O_1$ is an open set containing $x$ . Choose $t_2 \in O_1 \cap M$ . Enumerate the support $S(t_2)$ by $S(t_2)=\left\{\beta_{2,1},\beta_{2,2},\beta_{2,3},\cdots \right\}$ . Form the finite set $C_2$ by picking the first two points of $S(x)$ and the first two points of $S(t_2)$ , i.e., $C_2=\left\{\beta_{1,1},\beta_{1,2},\beta_{2,1},\beta_{2,2} \right\}$ . Then form the following open subset of $\Sigma$ .

$O_2=\biggl(\prod_{\alpha \in C_2} B_{\alpha,2} \times \prod_{\alpha \in A-C_2} X_\alpha \biggr) \cap \Sigma$

Choose $t_3 \in O_2 \cap M$ . Enumerate the support $S(t_3)$ by $S(t_3)=\left\{\beta_{3,1},\beta_{3,2},\beta_{3,3},\cdots \right\}$ . Then let $C_3=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\ \beta_{2,1},\beta_{2,2},\beta_{2,3},\ \beta_{3,1},\beta_{3,2},\beta_{3,3} \right\}$ , i.e., picking the first three points of $S(x)$ , the first three points of $S(t_2)$ and the first three points of $S(t_3)$ . Now, form the following open subset of $\Sigma$ .

$O_3=\biggl(\prod_{\alpha \in C_3} B_{\alpha,3} \times \prod_{\alpha \in A-C_3} X_\alpha \biggr) \cap \Sigma$

Choose $t_4 \in O_2 \cap M$ . Let this inductive process continue and we would obtain a sequence $t_2,t_3,t_4,\cdots$ of points of $M$ . We claim that the sequence converges to $x$ . Before we prove the claim, let’s make a few observations about the inductive process of defining $t_2,t_3,t_4,\cdots$ . Let $C=\bigcup_{j=1}^\infty C_j$ .

Each $C_j$ is the support of the open set $O_j$ .
The sequence of open sets $O_j$ is decreasing, i.e., $O_1 \supset O_2 \supset O_3 \supset \cdots$ . Thus for each integer $j$ , we have $t_k \in O_j$ for all $k \ge j$ .
The support of the point $x$ is contained in $C$ , i.e., $S(x) \subset C$ .
The support of the each $t_j$ is contained in $C$ , i.e., $S(t_j) \subset C$ .
In fact, $C=S(x) \cup S(t_2) \cup S(t_3) \cup \cdots$ .
The previous three bullet points are clear since the inductive process is designed to use up all the points of these supports in defining the open sets $O_j$ .
Consequently, for each $j$ , $x_\alpha=(t_j)_\alpha=a_\alpha$ for each $\alpha \in A-C$ . In other words, $x$ and each $t_j$ agree (and agree with the base point $a$ ) on the coordinates outside of the countable set $C$ .

Let $U=\prod_{\alpha \in A} U_\alpha$ be a standard open set in the product space $X=\prod_{\alpha \in A} X_\alpha$ such that $x \in U$ . Let $U^*=U \cap \Sigma$ . We show that for some $n$ , $t_j \in U^*$ for all $j \ge n$ .

Let $F=supp(U)$ be the support of $U$ . Let $F_1=F \cap C$ and $F_2=F \cap (A-C)$ . Consider the following open set:

$U^{**}=\biggl(\prod_{\alpha \in C} U_\alpha \times \prod_{\alpha \in A-C} X_\alpha \biggr) \cap \Sigma$

Note that $supp(U^{**})=F_1$ . For each $\alpha \in F_1$ , choose $B_{\alpha,k(\alpha)} \subset U_\alpha$ . Let $m$ be the maximum of all $k(\alpha)$ where $\alpha \in F_1$ . Then $B_{\alpha,m} \subset U_\alpha$ for each $\alpha \in F_1$ . Choose a positive integer $p$ such that:

$F_1 \subset W=\left\{\beta_{i,j}: i \le p \text{ and } j \le p \right\}$

Let $n=\text{max}(m,p)$ . It follows that there exists some $n$ such that $O_n \subset U^{**}$ . Then $t_j \in U^{**}$ for all $j \ge n$ . It is also the case that $t_j \in U^{*}$ for all $j \ge n$ . This is because $x=t_j$ on the coordinates not in $C$ . $\blacksquare$

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Dan Ma's Topology Blog

Expository Articles In General Topology

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Comparing two function spaces

Cp(omega 1 + 1) is monolithic and Frechet-Urysohn

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Sigma-products of first countable spaces