# Attempt at showing intersection of two compact sets is compact using open cover definition and why proof doesn't work?

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?

• 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 $\mathbb{R}$. – mi.f.zh Nov 30 '19 at 1:16
• 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. – dc3rd Nov 30 '19 at 1:18
• 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? – mi.f.zh Nov 30 '19 at 1:23
• 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 $S\cap T$. – conditionalMethod Nov 30 '19 at 1:24
• @jcqell, no haven't touched that level of topology yet – dc3rd Nov 30 '19 at 1:26

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?

• 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 $T$)? – dc3rd Nov 30 '19 at 2:19
• @dc3rd it covers all of $\mathbb{R}^n$ and hence each of $S$ and $T$. – user328442 Nov 30 '19 at 2:20
• 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 $x \in T$. – Lee Mosher Nov 30 '19 at 2:24
• Keep your goal in mind, namely to get a finite subcover of $\mathcal U$. – Lee Mosher Nov 30 '19 at 2:25
• @dc3rd well, we would be left with a finite cover of $S \cap T$ using sets from $\mathcal{U}$, as desired. – user328442 Nov 30 '19 at 2:45

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.