# If A and B are compact, show AUB is compact.

1. Aug 2, 2013

### Zondrina

1. The problem statement, all variables and given/known data

If $A$ and $B$ are compact sets in a metric space $(M, d)$, show that $AUB$ is compact.

2. Relevant equations

A theorem and two corollaries :

$M$ is compact $⇔$ every sequence in $M$ has a sub sequence that converges to a point in $M$.

Let $A$ be a subset of a metric space $M$. If $A$ is compact, then $A$ is closed in $M$.

If $M$ is compact and $A$ is closed, then $A$ is compact.

Heine-Borel theorem : A subset $K$ of $ℝ^n$ is compact $⇔$ $K$ is closed and bounded.

I'm also told that a compact space is the best of all possible worlds :).

3. The attempt at a solution

I'm told by the Heine-Borel theorem that a subset $K$ of $ℝ^n$ is compact $⇔$ $K$ is closed and bounded.

I'm thinking I should use this in particular to prove this because I'm told that $A$ and $B$ are compact. I believe this means I can bound the sets like so :

$min(A) ≤ A ≤ max(A)$
and
$min(B) ≤ B ≤ max(B)$.

The rest of this seems a bit too straightforward? Am I over thinking this or overlooking something?

2. Aug 2, 2013

### WannabeNewton

Take a sequence $(x_i)$ in $A\cup B$. Infinitely many points of the sequence will be contained in either $A$ or $B$. The rest should be straightforward.

Also, Heine-Borel only applies in $\mathbb{R}^{n}$, as you yourself have noted, so it will not work for arbitrary metric spaces. Additionally, $\text{min}(A)$ and $\text{max}(A)$ make no sense in arbitrary metric spaces. A bounded subset of a metric space is one which can be contained in an open ball.

3. Aug 2, 2013

### pasmith

Definition: $A$ is compact if and only if every open cover of $A$ has a finite subcover.

Hint: Every open cover of $A \cup B$ is an open cover of $A$, and also an open cover of $B$.

4. Aug 2, 2013

### WannabeNewton

Personally knowing that Zondrina is working through Carothers' text on real analysis, the open cover definition of compactness is only introduced later on in the chapter, whereas the above problem shows up at the very start of the chapter.

5. Aug 2, 2013

### Zondrina

It also applies to $ℝ$.

It was generalized to $ℝ^n$ right after they stated it for $ℝ$.

6. Aug 2, 2013

### WannabeNewton

$\mathbb{R}$ is just $\mathbb{R}^{n}$ with $n = 1$. Anyways, did you finish the proof (using either the method suggested by pasmith or myself)?

7. Aug 2, 2013

### Zondrina

I think this is correct so far :

Let $x_i$ be an open cover for $A \cup B$. Then we know $x_i$ is also an open cover for $A$ or $B$.

Now I'm given that $A$ and $B$ are compact, so by another one of the hints I've been given I know that there exist finite sub covers for both $A$ and $B$ respectively.

I'm thinking from here I wound define these finite sub covers for $A$ and $B$ and then the union of these two finite sub covers is an open cover for $A \cup B$. Would that not be sufficient to say that $A \cup B$ is compact?

8. Aug 2, 2013

### WannabeNewton

If you're going with open covers then that's the idea yea. But it's insanely trivial if you just use sequences. If $(x_i)$ is a sequence in $A\cup B$ then infinitely many points of the sequence belong to either $A$ or $B$. This defines a subsequence in $A$ or $B$ which itself has a subsequence that converges to a point in $A$ or $B$ hence $(x_i)$ has a convergent subsequence in $A\cup B$. It doesn't matter in the end since sequential compactness and open cover compactness are equivalent for metric spaces (but not so in more general spaces wherein the open cover compactness is the general definition).

9. Aug 2, 2013

### Zondrina

I see what you did there. I love that you can take a sequence which may not necessarily converge, but has a convergent sub sequence and do math. Thanks for clarifying that though.

10. Aug 2, 2013

### WannabeNewton

Anytime. But yes your open cover method is equally fine. Cheers.