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.


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<y 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<y_2< \cdots < y_n < y_{n+1} < \cdots p \cdots< x_{n+1} <x_n < \cdots < x_2<x_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<t, 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<d and c \ll d cannot both be true. Suppose c<d 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.


Blog posts on Corson compact spaces



  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.

\copyright \ 2014 \text{ by Dan Ma}


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s