1

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?

D.C. the III
  • 5,619
  • 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 01:16
  • 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. – D.C. the III Nov 30 '19 at 01:18
  • 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 01:23
  • 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 01:24
  • @jcqell, no haven't touched that level of topology yet – D.C. the III Nov 30 '19 at 01:26
  • @conditionalMethod, so to restate what you said. I have my arbitrary open cover of $S \cap T$, at this moment I don't know if it is compact, but I can add onto this open cover the two sets $S^{c}$ and $T^{c}$. Why does the new open cover cover $S \cup T$?......So going along assuming it does, since $S \cup T$ is compact and $S$ and $T$ are compact separately it must follow that $S \cap T$ is also compact?...........Still not sure why the union is though. – D.C. the III Nov 30 '19 at 01:32
  • You don't even need to look at $S\cup T$. The new cover covers $S$, for example, since $S\subset T^c\cup S\cap T$. So, $T^c$ covers $T^c$ and the original cover covers $S\cap T$. – conditionalMethod Nov 30 '19 at 01:34
  • The big hole is in claiming, without proof, that if $H$ is an open cover of $S\cap T$ with no finite subcover then $S$ or $T$ (or both have open cover(s) without finite subcover(s). You $can$ prove this in $\Bbb R^n$ by using the fact that compact subsets of $\Bbb R^n$ must be closed..... There are many examples of a topological space with a pair $S, T$ of non-closed compact subsets such that $S\cap T$ is not compact. – DanielWainfleet Dec 01 '19 at 00:13

2 Answers2

2

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?

user328442
  • 2,705
  • 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 $T$)? – D.C. the III Nov 30 '19 at 02:19
  • 1
    @dc3rd it covers all of $\mathbb{R}^n$ and hence each of $S$ and $T$. – user328442 Nov 30 '19 at 02:20
  • 1
    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 02:24
  • AH...ok that makes sense after pondering on it for a second, but now why do we remove $\mathbb{R}^n \setminus (S \cap T)$ from either of our finite subcovers? – D.C. the III Nov 30 '19 at 02:24
  • 2
    Keep your goal in mind, namely to get a finite subcover of $\mathcal U$. – Lee Mosher Nov 30 '19 at 02:25
  • @LeeMosher, let me see if my reasoning is right. So we arrive at the point where we have a finite subcover for each of $S$ and $T$, now we would remove the set $\mathbb{R}^n \setminus (S \cap T)$ if it is in either finite subcover because this set explicitly includes the points not in $S \cap T$ and as such not desired points we want to include in our set. – D.C. the III Nov 30 '19 at 02:31
  • And to answer @user328442, we would be left with $S \cap T$ once we remove $\mathbb{R}^n \setminus (S \cap T)$, but now we have a finite subcover over it. – D.C. the III Nov 30 '19 at 02:38
  • 1
    @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 02:45
1

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.

mi.f.zh
  • 1,219