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