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 monolithic and Frechet-Urysohn

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.

1. $C_p(X)$ is a Frechet-Urysohn space.
2. $C_p(X)$ is a k-space.
3. 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

1. 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)

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}

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

Subtracting the above two inequalities, we have the following:

$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}

$t-\frac{1}{2m}

Adding the above two inequalities, we obtain:

$\forall \ i=1,\cdots,n, \ 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

1. 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)

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

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

1. 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

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

1. Arhangelâ€™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.

$\copyright$ 2014-2018 – Dan Ma

Revised 9/17/2018

Every Corson compact space has a dense first countable subspace

In any topological space $X$, a point $x \in X$ is a $G_\delta$ point if the one-point set $\left\{ x \right\}$ is the intersection of countably many open subsets of $X$. It is well known that any compact Hausdorff space is first countable at every $G_\delta$ point, i.e., if a point of a compact space is a $G_\delta$ point, then there is a countable local base at that point. It is also well known that uncountable power of first countable spaces can fail to be first countable at every point. For example, no point of the compact space $[0,1]^{\omega_1}$ can be a $G_\delta$ point. In this post, we show that any Corson compact space has a dense set of $G_\delta$ point. Therefore, any Corson compact space is first countable on a dense set (see Corollary 4 below). However, it is not true that every Corson compact space has a dense metrizable subspace. See Theorem 9.14 in [2] for an example of a first countable Corson compact space with no dense metrizable subspace. A list of other blog posts on Corson compact spaces is given at the end of this post.

The fact that every Corson compact space has a dense first countable subspace is taken as a given in the literature. For one example, see chapter c-16 of [1]. Even though Corollary 4 is a basic fact of Corson compact spaces, the proof involves much more than a direct application of the relevant definitions. The proof given here is intended to be an online resource for any one interested in knowing more about Corson compact spaces.

For any infinite cardinal number $\kappa$, the $\Sigma$-product of $\kappa$ many copies of $\mathbb{R}$ is the following subspace of $\mathbb{R}^\kappa$:

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

A compact space is said to be a Corson compact space if it can be embedded in $\Sigma(\kappa)$ for some infinite cardinal $\kappa$.

For each $x \in \Sigma(\kappa)$, let $S(x)$ denote the support of the point $x$, i.e., $S(x)$ is the set of all $\alpha<\kappa$ such that $x_\alpha \ne 0$.

Proposition 1
Let $Y$ be a Corson compact space. Then $Y$ has a $G_\delta$ point.

Proof of Proposition 1
If $Y$ is finite, then every point is isolated and is thus a $G_\delta$ point. Assume $Y$ is infinite. Let $\kappa$ be an infinite cardinal number such that $Y \subset \Sigma(\kappa)$. For $f,g \in Y$, define $f \le g$ if the following holds:

$\forall \ \alpha \in S(f)$, $f(\alpha)=g(\alpha)$

It is relatively straightforward to verify that the following three properties are satisfied:

• $f \le f$ for all $f \in Y$. (reflexivity)
• For all $f,g \in Y$, if $f \le g$ and $g \le f$, then $f=g$. (antisymmetry)
• For all $f,g,h \in Y$, if $f \le g$ and $g \le h$, then $f \le h$. (transitivity)

Thus $\le$ as defined here is a partial order on the compact space $Y$. Let $C \subset Y$ such that $C$ is a chain with respect to $\le$, i.e., for all $f,g \in C$, $f \le g$ or $g \le f$. We show that $C$ has an upper bound (in $Y$) with respect to the partial order $\le$. We need this for an argument using Zorn’s lemma.

Let $W=\bigcup_{f \in C} S(f)$. For each $\alpha \in W$, choose some $f \in C$ such that $\alpha \in S(f)$ and define $u_\alpha=f_\alpha$. For all $\alpha \notin W$, define $u_\alpha=0$. Because $C$ is a chain, the point $u$ is well-defined. It is also clear that $f \le u$ for all $f \in C$. If $u \in Y$, then $u$ is a desired upper bound of $C$. So assume $u \notin Y$. It follows that $u$ is a limit point of $C$, i.e., every open set containing $u$ contains a point of $C$ different from $u$. Hence $u$ is a limit point of $Y$ too. Since $Y$ is compact, $u \in Y$, a contradiction. Thus it must be that $u \in Y$. Thus every chain in the partially ordered set $(Y,\le)$ has an upper bound. By Zorn’s lemma, there exists at least one maximal element with respect to the partial order $\le$, i.e., there exists $t \in Y$ such that $f \le t$ for all $f \in Y$.

We now show that $t$ is a $G_\delta$ point in $Y$. Let $S(t)=\left\{\alpha_1,\alpha_2,\alpha_3,\cdots \right\}$. For each $p \in \mathbb{R}$ and for each positive integer $n$, let $B_{p,n}$ be the open interval $B_{p,n}=(p-\frac{1}{n},p+\frac{1}{n})$. For each positive integer $n$, define the open set $O_n$ as follows:

$O_n=(B_{t_{\alpha_1},n} \times \cdots \times B_{t_{\alpha_n},n} \times \prod_{\alpha<\kappa,\alpha \notin \left\{ \alpha_1,\cdots,\alpha_n \right\}} \mathbb{R}) \cap Y$

Note that $t \in \bigcap_{n=1}^\infty O_n$. Because $t$ is a maximal element, note that if $g \in Y$ such that $g_\alpha=t_\alpha$ for all $\alpha \in S(t)$, then it must be the case that $g=t$. Thus if $g \in \bigcap_{n=1}^\infty O_n$, then $g_\alpha=t_\alpha$ for all $\alpha \in S(t)$. We have $\left\{t \right\}= \bigcap_{n=1}^\infty O_n$. $\blacksquare$

Lemma 2
Let $Y$ be a compact space such that for every non-empty compact subspace $K$ of $Y$, there exists a $G_\delta$ point in $K$. Then every non-empty open subset of $Y$ contains a $G_\delta$ point.

Proof of Lemma 2
Let $U_1$ be a non-empty open subset of the compact space $Y$. If there exists $y \in U_1$ such that $\left\{y \right\}$ is open in $Y$, then $y$ is a $G_\delta$ point. So assume that every point of $U_1$ is a non-isolated point of $Y$. By regularity, choose an open subset $U_2$ of $Y$ such that $\overline{U_2} \subset U_1$. Continue in the same manner and obtain a decreasing sequence $U_1,U_2,U_3,\cdots$ of open subsets of $Y$ such that $\overline{U_{n+1}} \subset U_n$ for each positive integer $n$. Then $K=\bigcap_{n=1}^\infty \overline{U_n}$ is a non-empty closed subset of $Y$ and thus compact. By assumption, $K$ has a $G_\delta$ point, say $p \in K$.

Then $\left\{p \right\}=\bigcap_{n=1}^\infty W_n$ where each $W_n$ is open in $K$. For each $n$, let $V_n$ be open in $Y$ such that $W_n=V_n \cap K$. For each $n$, let $V_n^*=V_n \cap U_n$, which is open in $Y$. Then $\left\{p \right\}=\bigcap_{n=1}^\infty V_n^*$. This means that $p$ is a $G_\delta$ point in the compact space $Y$. Note that $p \in U_1$, the open set we start with. This completes the proof that every non-empty open subset of $Y$ contains a $G_\delta$ point. $\blacksquare$

Proposition 3
Let $Y$ be a Corson compact space. Then $Y$ has a dense set of $G_\delta$ points.

Proof of Proposition 3
Note that Corson compactness is hereditary with respect to closed sets. Thus every compact subspace of $Y$ is also Corson compact. By Proposition 1, every compact subspace of $Y$ has a $G_\delta$ point. By Lemma 2, $Y$ has a dense set of $G_\delta$ points. $\blacksquare$

Corollary 4
Every Corson compact space has a dense first countable subspace.

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Blog posts on Corson compact spaces

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Reference

1. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
2. Todorcevic, S., Trees and Linearly Ordered Sets, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 235-293, 1984.

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