0
$\begingroup$

A space $X$ is called Corson compact if there is a set $I$ such that $X$ is homeomorphic to a compact subspace of $\{x \in \mathbb{R}^I:\{i\in I: x(i) \neq 0\} \text{ is countable }\}$, $\mathbb{R}^I$ being eqquipped with the product topology.

I am looking for an example of Corson compact. Is true that $\mathbb{N}$ with the co-finite topology is Corson compact?

1 Answers 1

0

All Corson compact spaces are completely regular (as a subspace of the space $\mathbb{R}^I$), so the cofinite topology is not.

As to examples: it's classical that all compact metric spaces are Corson compact by the Tychonoff embedding theorem, so $[0,1]$ will do, e.g.

for much more info see this blog post, e.g.

  • 0
    Thanks for your explanation. Since every compact metric space is Corson compact, now, I wonder that is there compact Hausdorff but not Corson compact? – 2017-01-26
  • 0
    @flourence Yes, see https://dantopology.wordpress.com/2014/06/03/an-example-of-a-non-metrizable-corson-compact-space/, which has a link to an example with explanations. – 2017-01-26
  • 0
    @flourence $\beta\omega$ is another example... more advanced though. – 2017-01-26