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

A useful representation of Cp(X)

Posted on August 4, 2014 by Dan Ma

Let $X$ be a completely regular space. The space $C_p(X)$ is the space of all real-valued continuous functions defined on $X$ endowed with the pointwise convergence topology. In this post, we show that $C_p(X)$ can be represented as the product of a subspace of $C_p(X)$ with the real line $\mathbb{R}$ . We prove the following theorem. See here for an application of this theorem.

Theorem 1
Let $X$ be a completely regular space. Let $x \in X$ . Let $Y$ be defined by:

$Y=\left\{f \in C_p(X): f(x)=0 \right\}$

Then $C_p(X)$ is homeomorphic to $Y \times \mathbb{R}$ .

The above theorem can be found in [1] (see Theorem I.5.4 on p. 37). In [1], the homeomorphism is stated without proof. For the sake of completeness, we provide a detailed proof of Theorem 1.

Proof of Theorem 1
Define $h: C_p(X) \rightarrow Y \times \mathbb{R}$ by $h(f)=(f-f(x),f(x))$ for any $f \in C_p(X)$ . The map $h$ is a homeomorphism.

The map is one-to-one

First, we show that it is a one-to-one map. Let $f,g \in C_p(X)$ where $f \ne g$ . Assume that $f(x) \ne g(x)$ . Then $h(f) \ne h(g)$ . So assume that $f(x)=g(x)$ . Then the functions $f-f(x)$ and $g-g(x)$ are different, which means $h(f) \ne h(g)$ .

The map is onto

Now we show $h$ maps $C_p(X)$ onto $Y \times \mathbb{R}$ . Let $(g,t) \in Y \times \mathbb{R}$ . Let $f=g+t$ . Note that $f(x)=g(x)+t=t$ . Then $f-f(x)=g$ . We have $h(f)=(g,t)$ .

Note. Showing the continuity of $h$ and $h^{-1}$ is a matter of working with the basic open sets in the function space carefully (e.g. making the necessary shifting). Some authors just skip the details and declare them continuous, e.g. [1]. Readers are welcome to work out enough of the details to see the key idea.

The map is continuous

Show that $h$ is continuous. Let $f \in C_p(X)$ . Let $U \times V$ be an open set in $Y \times \mathbb{R}$ such that $h(f) \in U \times V$ and,

$U=\left\{g \in Y: \forall \ i=1,\cdots,n, g(x_i) \in U_i \right\}$

$\forall \ i=1,\cdots,n, \ U_i=(f(x_i)-f(x)-\frac{1}{k},f(x_i)-f(x)+\frac{1}{k})$

$V=(f(x)-\frac{1}{k},f(x)+\frac{1}{k})$

where $x_1,\cdots,x_n$ are arbitrary points in $X$ and $k$ is some large positive integer. Define the following:

$\forall \ i=1,\cdots,n, \ W_i=(f(x_i)-\frac{1}{2k},f(x_i)+\frac{1}{2k})$

$W_{n+1}=(f(x)-\frac{1}{2k},f(x)+\frac{1}{2k})$

$x_{n+1}=x$

Then define the open set $W$ as follows:

$W=\left\{q \in C_p(X): \forall \ i=1,\cdots,n,n+1, q(x_i) \in W_i \right\}$

Clearly $f \in W$ . We need to show $h(W) \subset U \times V$ . Let $q \in W$ . Then $h(q)=(q-q(x),q(x))$ . We need to show that $q-q(x) \in U$ and $q(x) \in V$ . Note that $q(x_{n+1})=q(x) \in W_{n+1}$ . For each $i=1,\cdots,n$ , $q(x_i) \in W_i$ . So we have the following:

$f(x_i)-\frac{1}{2k}<q(x_i)<f(x_i)+\frac{1}{2k}$

$f(x)-\frac{1}{2k}<q(x)<f(x)+\frac{1}{2k}$

Subtracting the above two inequalities, we have the following:

$f(x_i)-f(x)-\frac{1}{k}<q(x_i)-q(x)<f(x_i)-f(x)+\frac{1}{k}$

The above inequality shows that for each $i=1,\cdots,n$ , $q(x_i) -q(x) \in U_i$ . Hence $q-q(x) \in U$ . It is clear that $q(x) \in V$ . This completes the proof that the map $h$ is continuous.

The inverse is continuous

We now show that $h^{-1}$ is continuous. Let $(g,t) \in Y \times \mathbb{R}$ . Note that $h^{-1}(g,t)=g+t$ . Let $M$ be an open set in $C_p(X)$ such that $g+t \in M$ and

$M=\left\{f \in C_p(X): \forall \ i=1,\cdots,n+1, f(x_i) \in M_i \right\}$

$\forall \ i=1,\cdots,n, \ M_i=(g(x_i)+t-\frac{1}{m},g(x_i)+t+\frac{1}{m})$

$x_{n+1}=x$

$M_{n+1}=(t-\frac{1}{m},t+\frac{1}{m})$

where $x_1,\cdots,x_n$ are arbitrary points of $X$ and $m$ is some large positive integer. Now define an open subset $G \times T$ of $Y \times \mathbb{R}$ such that $(g,t) \in G \times T$ and

$G=\left\{q \in Y: \forall \ i=1,\cdots,n+1, q(x_i) \in G_i \right\}$

$\forall \ i=1,\cdots,n, \ G_i=(g(x_i)-\frac{1}{2m},g(x_i)+\frac{1}{2m})$

$T=(t-\frac{1}{2m},t+\frac{1}{2m})$

We need to show that $h^{-1}(G \times T) \subset M$ . Let $(q,a) \in G \times T$ . We then have the following inequalities.

$\forall \ i=1,\cdots,n, \ g(x_i)-\frac{1}{2m}<q(x_i)<g(x_i)+\frac{1}{2m}$

$t-\frac{1}{2m}<a<t+\frac{1}{2m}$

Adding the above two inequalities, we obtain:

$\forall \ i=1,\cdots,n, \ g(x_i)+t-\frac{1}{m}<q(x_i)+a<g(x_i)+t+\frac{1}{m}$

The above implies that $\forall \ i=1,\cdots,n$ , $q(x_i)+a \in M_i$ . It is clear that $q(x_{n+1})+a=q(x)+a=a \in M_{n+1}$ . Thus $q+a \in M$ . This completes the proof that $h^{-1}$ is continuous.

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

A useful embedding for Cp(X)

Posted on August 4, 2014 by Dan Ma

Let $X$ be a Tychonoff space (also called completely regular space). By $C_p(X)$ we mean the space of all continuous real-valued functions defined on $X$ endowed with the pointwise convergence topology. In this post we discuss a scenario in which a function space can be embedded into another function space. We prove the following theorem. An example follows the proof.

Theorem 1
Suppose that the space $Y$ is a continuous image of the space $X$ . Then $C_p(Y)$ can be embedded into $C_p(X)$ .

Proof of Theorem 1
Let $t:X \rightarrow Y$ be a continuous surjection, i.e., $t$ is a continuous function from $X$ onto $Y$ . Define the map $\psi: C_p(Y) \rightarrow C_p(X)$ by $\psi(f)=f \circ t$ for all $f \in C_p(Y)$ . We show that $\psi$ is a homeomorphism from $C_p(Y)$ into $C_p(X)$ .

First we show $\psi$ is a one-to-one map. Let $f,g \in C_p(Y)$ with $f \ne g$ . There exists some $y \in Y$ such that $f(y) \ne g(y)$ . Choose some $x \in X$ such that $t(x)=y$ . Then $f \circ t \ne g \circ t$ since $(f \circ t)(x)=f(t(x))=f(y)$ and $(g \circ t)(x)=g(t(x))=g(y)$ .

Next we show that $\psi$ is continuous. Let $f \in C_p(Y)$ . Let $U$ be open in $C_p(X)$ with $\psi(f) \in U$ such that

$U=\left\{q \in C_p(X): \forall \ i=1,\cdots,n, \ q(x_i) \in U_i \right\}$

where $x_1,\cdots,x_n$ are arbitrary points of $X$ and each $U_i$ is an open interval of the real line $\mathbb{R}$ . Note that for each $i$ , $f(t(x_i)) \in U_i$ . Now consider the open set $V$ defined by:

$V=\left\{r \in C_p(Y): \forall \ i=1,\cdots,n, \ r(t(x_i)) \in U_i \right\}$

Clearly $f \in V$ . It follows that $\psi(V) \subset U$ since for each $r \in V$ , it is clear that $\psi(r)=r \circ t \in U$ .

Now we show that $\psi^{-1}: \psi(C_p(Y)) \rightarrow C_p(Y)$ is continuous. Let $\psi(f)=f \circ t \in \psi(C_p(Y))$ where $f \in C_p(Y)$ . Let $G$ be open with $\psi^{-1}(f \circ t)=f \in G$ such that

$G=\left\{r \in C_p(Y): \forall \ i=1,\cdots,m, \ r(y_i) \in G_i \right\}$

where $y_1,\cdots,y_m$ are arbitrary points of $Y$ and each $G_i$ is an open interval of $\mathbb{R}$ . Choose $x_1,\cdots,x_m \in X$ such that $t(x_i)=y_i$ for each $i$ . We have $f(t(x_i)) \in G_i$ for each $i$ . Define the open set $H$ by:

$H=\left\{q \in \psi(C_p(Y)) \subset C_p(X): \forall \ i=1,\cdots,m, \ q(x_i) \in G_i \right\}$

Clearly $f \circ t \in H$ . Note that $\psi^{-1}(H) \subset G$ . To see this, let $r \circ t \in H$ where $r \in C_p(Y)$ . Now $r(t(x_i))=r(y_i) \in G_i$ for each $i$ . Thus $\psi^{-1}(r \circ t)=r \in G$ . It follows that $\psi^{-1}$ is continuous. The proof of the theorem is now complete. $\blacksquare$

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Example

The proof of Theorem 1 is not difficult. It is a matter of notating carefully the open sets in both function spaces. However, the embedding makes it easy in some cases to understand certain function spaces and in some cases to relate certain 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. As an application of Theorem 1, we show that $C_p(\omega_1+1)$ can be embedded as a subspace of $C_p(\omega_1)$ . Define a continuous surjection $g:\omega_1 \rightarrow \omega_1+1$ as follows:

$g(\gamma) = \begin{cases} \omega_1 & \mbox{if } \ \gamma =0 \\ \gamma-1 & \mbox{if } \ 1 \le \gamma < \omega \\ \gamma & \mbox{if } \ \omega \le \gamma < \omega_1 \end{cases}$

The map $g$ is continuous from $\omega_1$ onto $\omega_1+1$ . By Theorem 1, $C_p(\omega_1+1)$ can be embedded as a subspace of $C_p(\omega_1)$ . On the other hand, $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$ . The function space $C_p(\omega_1+1)$ is a Frechet-Urysohn space, which is a property that is carried over to any subspace. The function $C_p(\omega_1)$ is not Frechet-Urysohn. Thus $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$ . A further comparison of these two function spaces is found in this subsequent post.

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

Cp(omega 1 + 1) is not normal

Posted on August 2, 2014 by Dan Ma

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.

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.

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.

Let $h \in C_p(\omega_1 +1) \backslash H$ . We wish to establish the following claim. Once the claim is established, it follows that $H$ is a closed subset of $C_p(\omega_1 +1)$ .

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

Consider the two mutually exclusive cases. Case 1. There exists some $\alpha<\omega_1$ such that $h(\alpha) \notin \left\{0,1 \right\}$ . Case 2. $h(\omega_1+1) \subset \left\{0,1 \right\}$ .

For Case 1, 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 $U \cap H=\varnothing$ .

Now assume Case 2. Within this case, there are three sub cases. Case 2.1. $h$ is a constant function with value 0. Case 2.2. $h$ is a constant function with value 1. Case 2.3. $h$ is not a constant function.

Case 2.1. 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 $U \cap H=\varnothing$ .

Case 2.2. Suppose $h$ is a constant function with value 1. Then let $U$ be the open set: $U=\left\{f \in C_p(\omega_1 +1): f(\omega_1) \in \mathbb{R} \backslash \left\{0 \right\} \right\}$ . It is clear that no function in $H$ can be in $U$ .

Case 2.3. Suppose $h$ is not a constant function. This case be broken down into two cases. Case 2.3.1. $h(\omega_1)=1$ . Case 2.3.2. $h(\omega_1)=0$ .

Case 2.3.1. Just like in Case 2.2, let $U=\left\{f \in C_p(\omega_1 +1): f(\omega_1) \in \mathbb{R} \backslash \left\{0 \right\} \right\}$ . Then $h \in U$ and $U \cap H=\varnothing$ .

Case 2.3.2. Assume that $h(\omega_1)=0$ . Since $h$ is not a constant function, it must takes on a value of 1 at some point. Let $\alpha<\omega_1$ be the largest such that $h(\alpha)=1$ . This $\alpha$ exists because $h$ is continuous and $h(\omega_1)=0$ . This case can be further broken into 2 cases. Case 2.3.2.1. There exists $\beta<\alpha$ such that $h(\beta)=0$ . Case 2.3.2.2. $h(\beta)=1$ for all $\beta<\alpha$ .

Case 2.3.2.1. Define $U=\left\{f \in C_p(\omega_1 +1): f(\beta) \in (-0.1,0.1) \text{ and } f(\alpha) \in (0.9,1.1) \right\}$ . Note that $h \in U$ and $U \cap H=\varnothing$ .

Case 2.3.2.2. In this case, $h(\beta)=1$ for all $\beta \le \alpha$ and $h(\gamma)=0$ for all $\alpha<\gamma \le \omega_1$ . This means that $h=h_\alpha$ . This is a contradiction since $h \notin H$ .

In all the cases except the last one, Claim 1 is true. The last case is not possible. Thus Claim 1 is established. The set $H$ is a closed subset of $C_p(\omega_1 +1)$ .

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.

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 $\Sigma(\omega_1)$ ).

The function space $C_p(\omega_1)$ , in contrast, is a Lindelof space and hence a normal space. If we restrict the above defined functions $h_\alpha$ to just $\omega_1$ , would the resulting functions form a closed and discrete set in $C_p(\omega_1)$ ? For each $\alpha$ with $0<\alpha<\omega_1$ , let $g_\alpha=h_\alpha \upharpoonright \omega_1$ . Let $G=\left\{g_\alpha: 0<\alpha<\omega_1 \right\}$ .

Is $G$ a closed and discrete subset of $C_p(\omega_1)$ ? It turns out that $G$ is a discrete subspace of $C_p(\omega_1)$ (relatively discrete). However it is not closed in $C_p(\omega_1)$ . Let $g:\omega_1 \rightarrow \{0, 1\}$ that takes on the constant value of 1. It follows that $g \in \overline{G}$ (the closure is in $C_p(\omega_1)$ ).

It seems that the argument above for showing $H$ is closed and discrete in $C_p(\omega_1+1)$ can be repeated for $G$ . Note that the argument for $H$ relies on the fact that the functions $h_\alpha$ takes on a value at the point $\omega_1$ . So the same argument cannot show that $G$ is a closed and discrete set. Thus $G$ is not discrete in $C_p(\omega_1)$ . Because $C_p(\omega_1)$ is Lindelof (hence normal), it has countable extent. It follows that any uncountable discrete subspace of $C_p(\omega_1)$ cannot be closed in $C_p(\omega_1)$ (the set $G$ is a demonstration). Any uncountable closed subset of $C_p(\omega_1)$ cannot be closed.

Reference

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

Revised 9/17/2018

The canonical evaluation map with a function space perspective

Posted on August 1, 2014 by Dan Ma

The evaluation map is a useful tool for embedding a space into a product space and plays an important role in many theorems and problems in topology. See here for a previous discussion. In this post, we present the evaluation map with the perspective that the map can be used for embedding a space into a function space of continuous functions. This post will be useful background for subsequent posts on $C_p(X)$ . In this post, we take a leisurely approach in setting up the scene. Once the map is defined properly, we show what additional conditions will make the evaluation map a homeomorphism. Then a function space perspective is presented as indicated above. After presenting an application, we conclude with some special cases for evaluation maps.

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The general setting

Let $X$ be a set (later we will add a topology). Let $\mathcal{F}$ be a set of real-valued functions defined on $X$ . Another way to view $\mathcal{F}$ is that it is a subspace of the product space $\mathbb{R}^X$ . For each $x \in X$ , consider the map $\pi_x: \mathbb{R}^X \longrightarrow \mathbb{R}$ defined by $\pi_x(f)=f(x)$ for all $f \in \mathbb{R}^X$ . One way to look at $\pi_x$ is that it is the projection map from the product space $\mathbb{R}^X$ to one of the factors. Thus $\pi_x$ is continuous when $\mathbb{R}^X$ has the product topology. In fact, the product topology is the smallest topology that can be defined on $\mathbb{R}^X$ that would make the $\pi_x$ continuous. When we restrict the map $\pi_x$ to the subspace $\mathcal{F}$ , the map $\pi_x: \mathcal{F} \longrightarrow \mathbb{R}$ is still continuous.

One more comment before defining the evaluation map. The set $\mathcal{F}$ of functions is a subspace of the product space $\mathbb{R}^X$ . Therefore the set $\mathcal{F}$ inherits the subspace topology from the product space. It makes sense to consider the function space $C_p(\mathcal{F})$ , the space of all continuous real-valued functions defined on $\mathcal{F}$ endowed with the pointwise convergence topology. Thus we can write $\pi_x \in C_p(\mathcal{F})$ .

We now define the evaluation map. Define the map $E_\mathcal{F}: X \longrightarrow C_p(\mathcal{F})$ by letting $E_\mathcal{F}(x)=\pi_x$ for each $x \in X$ , or more explicitly, by letting, for each $x \in X$ , $E_\mathcal{F}(x)$ be the map such that $E_\mathcal{F}(x)(f)=f(x)$ for all $f \in \mathcal{F}$ .

The map $E_\mathcal{F}$ is called the evaluation map defined by the family $\mathcal{F}$ . When the set $\mathcal{F}$ is understood, we can omit the subscript and denote the evaluation map by $E$ . We say $E_\mathcal{F}$ is the canonical evaluation map.

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What makes the evaluation map works

The goal of the evaluation map is that it be a homeomorphism. For that to happen, we need to make a few more additional assumptions. In defining the evaluation map above, the functions in the family $\mathcal{F}$ are not required to be continuous. In fact, the set $X$ is just a set in the above section. Now we require that $X$ is a topological space (it must be a completely regular space) and that all functions in $\mathcal{F}$ are continuous. Thus we have $\mathcal{F} \subset C_p(X)$ . With this assumption, the evaluation map is then a continuous function. We have the following theorem.

Theorem 1
Let $X$ be a space. Let $\mathcal{F} \subset C_p(X)$ . Then the evaluation map $E_\mathcal{F}: X \longrightarrow C_p(\mathcal{F})$ is always continuous.

Proof of Theorem 1
Let $x \in X$ . Let $U$ be open in $C_p(\mathcal{F})$ with $E_\mathcal{F}(x)=\pi_x \in U$ such that

$U=\left\{q \in C_p(\mathcal{F}): \forall \ i=1,\cdots,n, \ q(f_i) \in U_i \right\}$

where $f_1,\cdots,f_n$ are arbitrary points of $\mathcal{F}$ and each $U_i$ is an open interval of $\mathbb{R}$ . For each $i=1,\cdots,n$ , $\pi_x(f_i)=f_i(x) \in U_i$ . Let $V=\bigcap_{i=1}^n f_i^{-1}(U_i)$ , which is open in $X$ since each $f_i$ is a continuous function. We show that $E_\mathcal{F}(V) \subset U$ . For each $y \in V$ and for each $i=1,\cdots,n$ , $E_\mathcal{F}(y)(f_i)=\pi_y(f_i)=f_i(y) \in U_i$ . This means that for each $y \in V$ , $E_\mathcal{F}(y)=\pi_y \in U$ . The continuity of the evaluation map is established. $\blacksquare$

In order to make the evaluation map a homeomorphism, we consider two more definitions. A family $\mathcal{F} \subset \mathbb{R}^X$ is said to separate points of $X$ if for any $x,y \in X$ with $x \ne y$ , there exists an $f \in \mathcal{F}$ such that $f(x) \ne f(y)$ . A family $\mathcal{F} \subset \mathbb{R}^X$ is said to separate points from closed subsets of $X$ if for each $x \in X$ and for each closed subset $C$ of $X$ with $x \notin C$ , there exists an $f \in \mathcal{F}$ such that $f(x) \notin \overline{f(C)}$ . We have the following theorem.

Theorem 2
Let $X$ be a space. Let $\mathcal{F} \subset C_p(X)$ . Then the following are true about the evaluation map $E_\mathcal{F}: X \longrightarrow C_p(\mathcal{F})$ .

If $\mathcal{F}$ separates the points of $X$ , then the evaluation map $\mathcal{F}$ is a one-to-one.
If $\mathcal{F}$ separates the points from closed subsets of $X$ , then the evaluation map $\mathcal{F}$ is a homeomorphism.

Proof of Theorem 2
To prove the bullet point 1, suppose that $\mathcal{F}$ separates the points of $X$ . Let $x,y \in X$ with $x \ne y$ . Then there is some $f \in \mathcal{F}$ such that $f(x) \ne f(y)$ . It follows that the functions $E_\mathcal{F}(x)=\pi_x$ and $E_\mathcal{F}(y)=\pi_y$ differ at the point $f \in \mathcal{F}$ . This completes the proof for the bullet 1 of Theorem 2.

To prove the bullet point 2, suppose that the family $\mathcal{F}$ separates the points from closed subsets of $X$ . It suffices to show that the evaluation map $E_\mathcal{F}$ is an open map. Let $U \subset X$ be a non-empty open set. We show that $E_\mathcal{F}(U)$ is open in image $E_\mathcal{F}(X)$ . Let $E_\mathcal{F}(x)=\pi_x \in E_\mathcal{F}(U)$ where $x \in U$ . Since $\mathcal{F}$ separates the points from closed subsets of $X$ , there exists an $f \in \mathcal{F}$ such that $f(x) \notin \overline{f(X \backslash U})$ . Let $V=\mathbb{R}-\overline{f(X \backslash U})$ . Consider the following open set.

$W=\left\{q \in E_\mathcal{F}(X): q(f) \in V \right\}$

Clearly $E_\mathcal{F}(x)=\pi_x \in W$ . We show that $W \subset E_\mathcal{F}(U)$ . Choose $q \in W$ . Then $q=E_\mathcal{F}(y)=\pi_y$ for some $y \in X$ . It is also the case that $q(f)=\pi_y(f)=f(y) \in V$ . Thus $f(y) \notin \overline{f(X \backslash U})$ . This means that $y \in U$ and that $q=E_\mathcal{F}(y)=\pi_y \in E_\mathcal{F}(U)$ . This completes the proof for the bullet 2 of Theorem 2. $\blacksquare$

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Embedding every space into a function space

In defining the evaluation map, we start with a space $X$ . Then take a family of continuous maps $\mathcal{F} \subset C_p(X)$ . As long as the family of functions $\mathcal{F}$ separates points from closed sets, we know for sure that the evaluation map is a homeomorphism from $X$ into a subspace of $C_p(\mathcal{F})$ . We now look at some choices for $\mathcal{F}$ . One is that $\mathcal{F} = C_p(X)$ . Then we have the following corollary.

Corollary 3a
Any space $X$ is homeomorphic to a subspace of the function space $C_p(C_p(X))$ .

Because $X$ is a completely regular space, the family $\mathcal{F} = C_p(X)$ clearly separates points from closed sets. Thus Corollary 3a is valid. In fact, the complete regularity of $X$ only requires that we use $\mathcal{F} = C_p(X,I)$ , the set of all continuous functions from $X$ into $I$ where $I=[0,1]$ . We have the following corollary.

Corollary 3b
Any space $X$ is homeomorphic to a subspace of the function space $C_p(C_p(X,I))$ .

We can also let $\mathcal{F} = C_p^0(X)$ , the set of all bounded real-valued continuous functions defined on $X$ . It is clear that $C_p^0(X)$ separates points from closed sets. So we also have:

Corollary 3c
Any space $X$ is homeomorphic to a subspace of the function space $C_p(C_p^0(X))$ .

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One application

We demonstrate one application of Corollary 3a. When $X$ is a separable metric space, $C_p(X)$ has a countable network (see this previous post). It is natural to ask whether every space with a countable network can be embedded in a $C_p(Y)$ for some separable metric space $Y$ ? The answer is yes. We have the following theorem. One direction of the theorem is Theorem III.1.13 in [1].

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

The space $X$ has a countable network.
The space $X$ can be embedded in a $C_p(Y)$ for some separable metric space $Y$ .

Proof of Theorem 4
The direction $2 \longrightarrow 1$ is clear. As shown here, $C_p(Y)$ has a countable network whenever $Y$ has a countable base. Having a countable network carries over to subspaces. The direction $1 \longrightarrow 2$ is the one that uses evaluation map.

$1 \longrightarrow 2$
Suppose that $\mathcal{M}$ is a countable network for $X$ . Then $C_p(X)$ has a countable network, e.g., the set of all $[M,V]$ where $M \in \mathcal{M}$ , $V$ is any open interval with rational endpoints and $[M,V]$ is the set of all $f \in C_p(X)$ such that $f(M) \subset V$ .

Any space with a countable network is the continuous image of a separable metric space. Thus there exists a separable metric space $Y$ such that $C_p(X)$ is the continuous image of $Y$ . Let $g: Y \longrightarrow C_p(X)$ be a continuous surjection. Then $C_p(C_p(X))$ can be embedded into $C_p(Y)$ . The embedding $\rho:C_p(C_p(X)) \longrightarrow C_p(Y)$ is defined by $\rho(f)=f \circ g$ .

By Corollary 3a, $X$ is embedded into $C_p(C_p(X))$ . Then $X$ is embedded into $C_p(Y)$ . $\blacksquare$

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More on the evaluation map

In this section, we consider some special cases. As shown in Theorem 2, what makes the evaluation map a one-to-one map is that the family $\mathcal{F} \subset C_p(X)$ separates points of $X$ (for short, the family is point separating). What makes the evaluation map a homeomorphism is that the family $\mathcal{F}$ separates points from closed subsets of $X$ . In this section, we present one property that implies the property of separating points from closed sets. It is clear that if $\mathcal{F}$ is dense in $C_p(X)$ , then $\mathcal{F}$ separates points of $X$ . In general, the fact that $\mathcal{F}$ is point separating does not mean it separates point from closed sets. We show that whenever $X$ is compact, the fact that $\mathcal{F}$ is dense in $C_p(X)$ does imply that $\mathcal{F}$ separates points from closed sets.

The family $\mathcal{F} \subset C_p(X)$ is said to be a generating set of functions if it determines the topology of $X$ , i.e., the following set is a base for the topology of $X$ .

$\mathcal{B}_{\mathcal{F}}=\left\{f^{-1}(U) \subset X: f \in \mathcal{F} \text{ and } U \text{ is open in } \mathbb{R} \right\}$

Since $X$ is assumed to be a completely regular space, we observe that if $\mathcal{B}_{\mathcal{F}}$ is a base for $X$ , then the family $\mathcal{F}$ separates points from closed subsets of $X$ . The following theorem captures the observations we make.

Theorem 5
Let $X$ be a space. Let $\mathcal{F} \subset C_p(X)$ . Then if $\mathcal{F}$ is a generating set of functions, then $\mathcal{F}$ separates points from closed subsets of $X$ , hence the evaluation map $E_\mathcal{F}: X \longrightarrow C_p(\mathcal{F})$ as defined above is a homeomorphism.

We now show that if $X$ is compact and if $\mathcal{F}$ is dense in $C_p(X)$ , then $\mathcal{F}$ separates points from closed subsets of $X$ , making the evaluation map a homeomorphism.

First one definition. Let $X$ be a space. For any finite $F=\left\{f_1,\cdots,f_n \right\}$ consisting of functions in $C_p(X)$ , define the maximum of $F$ to be the function $f:X \longrightarrow \mathbb{R}$ such that for each $x \in X$ , $f(x)$ is the maximum of the real values in $\left\{f_1(x),\cdots,f_n(x) \right\}$ . In other words, the maximum of $F$ is the pointwise maximum of the functions in $F$ . It is not too difficult to show that the pointwise maximum of finitely many continuous real-valued functions is also continuous. We have the following lemma and corollary.

Lemma 6
Let the space $X$ be compact. Suppose the family $\mathcal{F}$ is dense in $C_p(X)$ such that the pointwise maximum of any finite set of functions in $\mathcal{F}$ is also in $\mathcal{F}$ . Then $\mathcal{F}$ separates points from closed subsets of $X$ .

Proof of Lemma 6
Let $x \in X$ and let $C$ be a closed subset of $X$ such that $x \notin C$ . For each $y \in C$ , consider the open set:

$U_y=\left\{f \in C_p(X): f(x) \in O_1 \text{ and } f(y) \in O_2 \right\}$

where $O_1$ is the open interval $(-0.1,0.1)$ and $O_2$ is the open interval $(2,\infty)$ . For each $y \in C$ , choose $f_y \in \mathcal{F} \cap U_y$ . The set of all $f_y^{-1}(O_2)$ is an open cover of the compact set $C$ . Choose $y_1,y_2,\cdots,y_n \in C$ such that $V_{1},V_{2},\cdots,V_{n}$ cover $C$ where each $V_i=f_{y_i}^{-1}(O_2)$ . Let $g:X \longrightarrow \mathbb{R}$ be the pointwise maximum of $\left\{f_{y_1},\cdots, f_{y_n} \right\}$ . By assumption, $g \in \mathcal{F}$ . It is clear that for all $y \in C$ , $2<g(y)$ . Thus $\overline{g(C)} \subset [2,\infty)$ .

Let $W=\bigcap_{i=1}^n f_{y_i}^{-1}(O_1)$ . It is also clear that $f_{y_i}(x) \in O_1=(-0.1,0.1)$ for all $i$ , implying $g(x) <0.1<1$ . Thus $g(x) \notin \overline{g(C)}$ . Thus completes the proof that $\mathcal{F}$ separates points from closed subsets of $X$ . $\blacksquare$

Corollary 7
Let the space $X$ be compact. If $\mathcal{F}$ is a dense subspace of $C_p(X)$ , then the evaluation map $E_\mathcal{F}: X \longrightarrow C_p(\mathcal{F})$ as defined above is a one-to-one map.

In Corollary 7, even if $\mathcal{F}$ is not closed under taking pointwise maximum of finitely many functions, then throw all pointwise maxima of all finite subsets of $\mathcal{F}$ into $\mathcal{F}$ and then apply Lemma 6. Throwing in all pointwise maxima will not increase the cardinality of $\mathcal{F}$ . For example, suppose that $X$ is compact, $C_p(X)$ is separable and $\mathcal{F}$ is a countable dense subspace of $C_p(X)$ . Even if $\mathcal{F}$ does not contain all the pointwise maxima of finite subspaces, we can then throw in all pointwise maxima and the subspace $\mathcal{F}$ is still countable. Then the compact space $X$ is homeomorphic to a subspace of $C_p(\mathcal{F})$ . Since $C_p(\mathcal{F}) \subset \mathbb{R}^\omega$ , $C_p(\mathcal{F})$ is separable and metrizable. Thus the compact space $X$ is separable and metrizable. The following corollary captures this observation.

Corollary 8
If $X$ is a compact space and the function space $C_p(X)$ is separable, then $X$ is metrizable.

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

Dan Ma's Topology Blog

Expository Articles In General Topology

Monthly Archives: August 2014

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