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Let $S$ and $T$ both be compact sets in $\mathbb{R}^{n}$. Show that $S \cap T$ is also compact using the open cover definition.

What I thought was going to be a simple decomposition has turned out to be more complicated for me.

Attempt

We want to show that for every open cover $U = \{U_{\alpha}\}$ of $S \cap T$, there exists a finite subcover.

Suppose there exists an open cover $H$ of $S \cap T$ that has no finite subcover. THis would imply that both $S$ and $T$ have an open cover which has no finite subcover. But it is assumed $S$ and $T$ are compact, so a contradiction.

My Issue

I feel my solution has a big hole in its reasoning and I'm not a fan of contradiction either since I'm not truly working with the concepts. I looked at a few of the solutions posted here and I don't fully understand their approach with regards to this question. Particularly this solution: Prove Intersection of Two compact sets is compact using open cover?.

How are we constructing a finite subcover from the compliment?

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  • $\begingroup$ Are you assuming anything on the topologies of $S$ and $T$ themselves? See here - in general, the intersection of two (finite) compact sets is not itself compact. The post you linked takes the two compact sets to be subsets of $\mathbb{R}$. $\endgroup$ – mi.f.zh Nov 30 '19 at 1:16
  • $\begingroup$ There's nothing explicit, but this is a 2nd year analysis course so I'm going to assume we are dealing with things in $\mathbb{R}^{n}$. I'll edit my question to include that. $\endgroup$ – dc3rd Nov 30 '19 at 1:18
  • $\begingroup$ This statement actually works for an intersection of compact subsets (even uncountably many), if we're working in a Hausdorff space. Do you know the definition of what that is/have you covered separability axioms? $\endgroup$ – mi.f.zh Nov 30 '19 at 1:23
  • $\begingroup$ If the two sets are closed, then you can enrich the open cover of $S\cap T$ with the open sets $S^c$ and $T^c$. The new open cover covers $S\cup T$. A finite subcover, which existence follows, will have to consist of a finite subcover of $S\cap T$ together with possibly $S^c$ or $T^c$ or both. But $S^c$ and $T^c$ are both disjoint with $S\cap T$. So the rest of the finite subcover, covers $S\cap T$. $\endgroup$ – conditionalMethod Nov 30 '19 at 1:24
  • $\begingroup$ @jcqell, no haven't touched that level of topology yet $\endgroup$ – dc3rd Nov 30 '19 at 1:26
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You may or may not know what it means for a topological space to be Hausdorff but $\mathbb{R}^n$ has this property. It turns out that compact sets are closed in Hausdorff spaces (and hence $\mathbb{R}^n$) and arbitrary intersections of closed sets are also closed (perhaps you are already aware of these facts for subsets of $\mathbb{R}^n$). So, $S, T$ and $S \cap T$ are closed.

Let $ \mathcal{U} = \{ U_{ \alpha } \}$ be an open cover of $S \cap T$. Then $\mathcal{U} \cup \{ \mathbb{R}^n \setminus (S \cap T) \}$ is an open cover of $S$ as well as for $T$ (why? I leave that to you). So, this cover yields a finite subcover of each $S$ and $T$ (the covers may be distinct but each will cover $S \cap T$). If either finite subcover includes $\mathbb{R}^n \setminus (S \cap T)$ then remove that. What are you left with?

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  • $\begingroup$ This is where i'm stuck...why is $ \mathcal{U} \cup \{ \mathbb{R}^n \setminus (S \cap T) \}$ an open cover of $S$ or $T$? I get that $\mathcal{U}$ covers the intersection, but how does adding that one extra set allow for it to cover all of $S$ (or $T$)? $\endgroup$ – dc3rd Nov 30 '19 at 2:19
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    $\begingroup$ @dc3rd it covers all of $\mathbb{R}^n$ and hence each of $S$ and $T$. $\endgroup$ – user328442 Nov 30 '19 at 2:20
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    $\begingroup$ Consider $x \in S$. If $x \in T$ then $x \in S \cap T$ and so $x$ is in some element of $\mathcal U$. If $x \not\in T$ then $x \in \mathbb R^n \setminus (S \cap T)$. Argue similarly for $x \in T$. $\endgroup$ – Lee Mosher Nov 30 '19 at 2:24
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    $\begingroup$ Keep your goal in mind, namely to get a finite subcover of $\mathcal U$. $\endgroup$ – Lee Mosher Nov 30 '19 at 2:25
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    $\begingroup$ @dc3rd well, we would be left with a finite cover of $S \cap T$ using sets from $\mathcal{U}$, as desired. $\endgroup$ – user328442 Nov 30 '19 at 2:45
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As in the comments, this statement actually holds for all Hausdorff spaces - we are not limited to $\mathbb{R}^n$ - and also for arbitrary families of compact sets. But we can work with $\mathbb{R}^n$ here - the proof here can be easily generalised to Hausdorff spaces.

We let $K_i$ be a family of compact subsets indexed by some arbitrary set $I$. They are closed (you may have this as a characterisation, or by Heine-Borel, but it's also a nice exercise to do, and reasonably straightforward since $\mathbb{R}^n$ is a metric space), and so their intersection $\cap_{i\in I} K_i$ is closed. Then (as another exercise), closed subspaces of compact spaces are themselves compact. Since $\cap_{i\in I} K_i \subseteq K_i$ (for any $i \in I$), the former is compact.

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