0
$\begingroup$

Suppose that $X$ is a compact Hausdorff space. Let $\{X_i\}_{i \in \mathbb{N}}$ be a sequence of dense open subsets of $X$. Let $E$ be a closed subset of $X$. Can it happen that $\left(\cap_{i \in \mathbb{N}}X_i\right)$ has empty intersection with $E$?

First by Baire Category theorem, $\left(\cap_{i \in \mathbb{N}}X_i\right)$ is dense in $X$. Moreover, since $E$ is closed, it is compact. If the intersection were empty, then $E \subset \cup_{i \in \mathbb{N}}X\setminus X_i$. But, this doesn't yield any contradiction.

A hint would be greatly appreciated.

Thanks for the help!!

$\endgroup$
1
$\begingroup$

Yes, the intersection can be empty. Suppose that $X=[0,1]$, with its usual topology. Take $E=\{0\}$ and $X_n=(0,1]$, for each natural $n$.

$\endgroup$
  • $\begingroup$ You can also take a Cantor set $C$ and $X_n = [0,1] \setminus C$. Then $C$ is closed and uncountable! Since $C$ is nowhere dense, $X_n$ is dense. $\endgroup$ – p4sch Nov 21 '18 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.