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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,

  • for all $F\in \mathcal{H}$, $F$ is closed (since $\langle X,\tau\rangle$ is Hausdorff).

  • $\displaystyle\bigcap_{F\in\mathcal{H}} F$ is closed and hence compact.

However, I am unable to proceed any further. Can anyone help me?

2 Answers2

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(*) 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 $\bigcap\mathcal H\neq\varnothing$.

drhab
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Assuming $X$ is not compact, $\exists \mathcal{O} = \{O_\alpha:\alpha \in I\}$ an open cover which does not have a finite open subcover. Furthermore we can consider the collection of closed sets $\mathcal{H} = \{F_\alpha=X\backslash O_\alpha:\alpha\in I\}$. Since $\mathcal{O}$ does not have a finite open subcover, $X\backslash \bigcup\limits_{\alpha\in J} O_\alpha\neq \emptyset$ for any finite index set $J$. This implies that $\bigcap\limits_{\alpha\in J}X\backslash O_\alpha = \bigcap\limits_{\alpha\in J}F_\alpha\neq \emptyset$ for any finite index set $J$, i.e. we have the finite intersection property for $\mathcal{H}$. Since $\mathcal{O}$ is an open cover we have $\bigcup\limits_{\alpha\in I}O_\alpha = X\implies \bigcap\limits_{\alpha\in I}F_{\alpha} = \emptyset$.

Jürgen Sukumaran
  • 7,859