Problem
Let $\langle X,\tau\rangle$ be a Hausdorff topological space. Let $\mathcal{H}$ be a non-empty family of compact subsets of $X$ having the finite intersection property. Then prove or disprove that $\displaystyle\bigcap_{F\in\mathcal{H}} F\ne\emptyset$
So far I had been able to make only the following two observations,
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for all $F\in \mathcal{H}$, $F$ is closed (since $\langle X,\tau\rangle$ is Hausdorff).
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$\displaystyle\bigcap_{F\in\mathcal{H}} F$ is closed and hence compact.
However, I am unable to proceed any further. Can anyone help me?
Best Answer
(*) In general space $Y$ is compact if and only if every family of closed sets that has finite intersection property has a non-empty intersection.
For a proof of that see here.
Choose some $K\in\mathcal H$.
Let $\mathcal H_K$ denote the collection $\{K\cap H\mid H\in\mathcal H\}$ of subsets of $K$ and now focus on compact Hausdorff space $K$.
Prove that $\mathcal H_K$ has finite intersection property and is a collection of sets closed in compact set $K$.
Applying (*) it can be concluded that $\bigcap\mathcal H_K\neq\varnothing$ and consequently $\bigcap\mathcal H\neq\varnothing$.