0
$\begingroup$

I am working on the following problem.

Let $X$ be a compact topological space, and let $\{C_i\}_{i \in \mathbb{Z_+}}$ be a collection of nonempty compact closed sets in $X$ satisfying $C_{i+1} \subset C_i$ for all $i \in \mathbb{Z_+}$. Prove $\bigcap_{i=1}^{\infty}C_i \neq \varnothing$.

This is a homework problem ... so I would just like to see if I am going the the correct direction.

We must choose a open Cover say $\{U_{\alpha}\}_{\alpha \in I}$ where the $U_{\alpha}$'s are open in $X$. Then we know that the cover is of the form $U_{\alpha} \cap \bigcap_{i=1}^{\infty}C_i$ and out of that we must construct a finite subcover that maybe will do something for us? But I seem to be unable to get past that point...

Or maybe we should use contradiction?

Thank you

  • 0
    @Riemann-bitcoin, it is enough to assume one of them is compact. The rest (with higher numbers) are now compact as closed subsets of compact sets. – 2017-02-01
  • 0
    @szw1710 I agree – 2017-02-01
  • 2
    Consider the family $V_i = X \setminus C_i$. If the intersection were empty, it would be an open cover of $C_1$. – 2017-02-01

1 Answers 1

1

Take a collection $\{C_i\}$ such that each $C_i$ is compact and nonempty and $C_{i+1}\subset C_i$. Assume that $\bigcap C_i=\emptyset$.

Now define $U_i=X\backslash C_i$. Note that each $U_i$ is open and $U_i\subset U_{i+1}$. Furthermore since $\bigcap C_i=\emptyset$ then $\bigcup U_i=X$. In particular $\{U_i\}$ is a covering of $C_0$. Since $C_0$ is compact, then there exists a finite subfamily of $\{U_i\}$ covering $C_0$. Furthermore since $U_i\subset U_j$ for $i

  • 0
    but what part of this uses that $X$ is hausdorff – 2017-02-01
  • 0
    @Riemann-bitcoin. None. That assumption is not necessary. – 2017-02-01
  • 0
    ok I see @freakish – 2017-02-01