An example of a non-metrizable Corson compact space

A compact space is said to be a Corson compact space if it can be embedded in a $\Sigma$-product of real lines. All compact metric spaces are Corson. In this post, we present a non-metrizable example of a Corson compact space found in the literature, found in [1]. A listing of other blog posts on Corson compact spaces is given at the end of this post.

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 $\kappa$. When $\kappa=\omega$, $\Sigma(\omega)$ is simply $\mathbb{R}^\omega$, the product of countably many copies of the real lines. Any compact metrizable space can be embedded in $\mathbb{R}^\omega$; see Theorem 4.2.10 in [2]. Thus any compact metrizable space is Corson compact. One easily described non-metrizable Corson compact space is the one-point compactification of an uncountable discrete space.

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Describing the example

We define a Corson compact space $C$ that is not metrizable. To define the example, let $W$ be a set of real numbers such that the cardinality of $W$ is $\omega_1$. Let $\ll$ be a well ordering on the set $W$ such that for each $y \in W$, the initial segment $S_y=\left\{x \in W: x \ll y \right\}$ is countable. In other words, the well-ordered set $(W,\ll)$ is of type $\omega_1$. Let $<$ be the usual order on the real numbers. Consider the following collection of subsets of $W$:

$\mathcal{H}=\left\{H \subset W: \text{ the order } < \text{ coincides with the well order } \ll \text{ on the set } H \right\}$

Note that $H \in \mathcal{H}$ if and only if this condition holds: if $x,y \in W$ with $x \ne y$, then

$x if and only if $x \ll y$.

Thus by default, $\varnothing \in \mathcal{H}$ and $\left\{ x \right\} \in \mathcal{H}$ for any $x \in W$. More importantly, every $H \in \mathcal{H}$ is a countable set. To see this, suppose that $J \subset W$ is uncountable. Then $J$ has a two-sided limit point, say $p$ (see this previous post). It does not matter whether $p \in J$. What matters is that for any open interval $(a,b)$ with $p \in (a,b)$, both $(a,p)$ and $(p,b)$ contain points of $J$. Then there is a sequence $\left\{x_n: n=1,2,3,\cdots \right\}$ of points of $J$ converging to $p$ and there is a sequence $\left\{y_n: n=1,2,3,\cdots \right\}$ of points of $J$ converging to $p$ such that

$y_1

Suppose that $(J,<)$ is a well-ordered set. Let $t \in J$ such that $t$ is the least number that is an upper bound of $Y=\left\{y_n: n=1,2,3,\cdots \right\}$. It must be the case that $p \le t$. If $p=t$, then $p$ is an element of $J$ that has no immediate successor, a contradiction. If $p, then $t$ is not the least upper bound of $Y$. It follows that $(J,<)$ cannot be a well-ordered set. Then the orders $<$ and $\ll$ do not agree on the set $J$. Thus no uncountable set can be in the family $\mathcal{H}$.

Consider the compact product space $X=\left\{0,1 \right\}^{W}$. For each $f \in X$, let $S(f)$ be the support of the point $f$, i.e., the set of all $x \in W$ such that $f(x) \ne 0$. Consider the following subspace of $X$:

$C=\left\{f \in X: S(f) \in \mathcal{H} \right\}$

Note that $z \in C$ where $z(x)=0$ for all $x \in W$ and that for each $t \in W$, $f_t \in C$ where $f_t(t)=1$ and $f_t(x)=0$ for all $x \in W-\left\{t \right\}$. Since each $H \in \mathcal{H}$ is countable, $C$ is a subspace of the $\Sigma$-product $\Sigma(\omega_1)$. Since $C$ is a subspace of the compact space $X=\left\{0,1 \right\}^{W}$, in order to show that $C$ is compact, we only need to show that $C$ is closed in $X$. To this end, let $g \in X-C$. Then $(S(g),<)$ and $(S(g),\ll)$ do not coincide. Then there exist $c,d \in S(g)$ such that $c and $c \ll d$ cannot both be true. Suppose $c and $c \not \ll d$. Consider the following open subset of $X$:

$U=\left\{f \in X: f(c)=f(d)=1 \right\}$

It is clear that $U \cap C=\varnothing$. Thus $C$ is a closed subspace of $X=\left\{0,1 \right\}^{W}$. Since $C$ is a compact space and is a subspace of a $\Sigma$-product of real lines, $C$ is a Corson compact space.

If a compact space is metrizable then it is separable. Thus if we can show that $C$ is not separable, then $C$ is not metrizable. We show that no countable subspace of $C$ can be dense in $C$. Let $D$ be a countable subspace of $C$. Let $S=\bigcup_{f \in D} S(f)$, which is countable. Choose $x \in W-S$. Then the open set $\left\{g \in C: g(x)=1 \right\}$ contains no point of $D$. This completes the proof that $C$ is a non-metrizable Corson compact space.

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Blog posts on Corson compact spaces

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Reference

1. Alster, K., Pol, R., On function spaces of compact subspaces of $\Sigma$-products of the real line, Fund. Math., 107, 35-46 1980.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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