Compact set in $\mathbb{R}^n$ and the finite intersection property

Asked Oct 24 '20 at 12:21

Active Oct 24 '20 at 12:21

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Show that a set $K ⊂ \mathbb{R}^n$ is compact if and only if every family of closed subsets of K having the finite intersection property has nonempty intersection. I was able to prove one side (That if $K$ is compact then every family of closed subsets of $K$ having the finite intersection property has nonempty intersection), but something doesn't work in the other direction.

Our definition of compact set is that for every open cover $\{F_{i}\}_{i\in I}$ there exists a finite subcover such that $K\subset\bigcup_{i\in J}F_{i}$ where $J\subset I$ is a finite subset of indices.

I tried using De - Morgan, but that fact that $K\subset\bigcup_{i\in J}F_{i}$ and not $K=\bigcup_{i\in J}F_{i}$ got me stuck.

asked Oct 24 '20 at 12:21

Math101

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But $K=\bigcup_{i\in J}(F_i\cap K)$, where each $F_i\cap K$ is open in $K$. – Thorgott Oct 24 '20 at 12:29
@mathcounterexamples.net No because here $K^{c} \neq \emptyset$ – Math101 Oct 24 '20 at 12:30
@Thorgott Im just now starting topology, how can you tell $F_{i}\cap K$ is open? – Math101 Oct 24 '20 at 12:32
That's the definition of the subspace topology on $K$: open sets in $K$ are precisely the intersections of open sets in $\mathbb{R}^n$ with $K$. – Thorgott Oct 24 '20 at 12:35
@GalBenAyun Just use the subspace topology! – mathcounterexamples.net Oct 24 '20 at 12:35
@Thorgott But I still don't understand how you can deduce that $F_{i}$ open $\rightarrow F_{i}\cap K$ without any assumptions about K. – Math101 Oct 24 '20 at 13:29
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It's true by definition for any reasonable definition. If you don't agree, tell me how you define the open/closed subsets of $K$. – Thorgott Oct 24 '20 at 14:11

Compact set in $\mathbb{R}^n$ and the finite intersection property

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