# Compact subset of open set?

1. Feb 6, 2014

### Mixer

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

Let $K \subset \mathbb{R^n}$ be compact and $U$ an open subset containing $K$. Verify that there exists $r > 0$ such that $B_r{u} \subset U$ for all $u \in K$.

2. Relevant equations

Every open cover of compact set has finite subcover.

3. The attempt at a solution

I tried to cover my $K$ with open balls therefore there should be finitely many open balls (because $K$ is compact). If I choose $r' = min(r_1,r_2,...,r_n)$ ($r_i$ being the radius of the ball) then every element in $K$ has that required ball-neighbourhood. Because $U$ is open then $B_{r'}{u} \subset U$. Is this correct?

2. Feb 6, 2014

### PeroK

That doesn't prove it. Even if you have a finite subcover of open balls, then that subcover might extend beyond U. You need to find a way to make your subcover be subsets of U.

... also, you need the r to work for every u in K, not just a finite subset.

... maybe using open covers is not the answer? You are working in $$\mathbb{R}^n$$

Last edited: Feb 6, 2014
3. Feb 6, 2014

### Mixer

Erh...

Ok, so my reasoning does not work. How about the fact that since $U$ is open, then every element $x \in U$ has an open ball-neigbourhood $B_r{x} \subset U$. Then every element in $K$ has that neigbourhood also. Take all these neighbourhoods and then this is an open cover of $K$. Because $K$ is compact then it has to have finite subcover which has finitely many of these open balls. Then this subcover has to be subset of $U$ and if I take minimum of the $r$:s, then every element $u \in K$ has that required ball-neighbourhood.

I do not know why $\mathbb{R^n}$ is relevant here.. Am I supposed to use some metric?

4. Feb 6, 2014

### PeroK

Take a ball, radius r, centred on k in U. Let x be in this ball. Then the ball, radius r, centred on x may not be in U.

In general, any finite subcover will give you a finite set of balls, but that isn't easily going to show that a ball of radius r centred on any point in K is in U.

5. Feb 6, 2014

### Mixer

I feel I'm completely lost now.. Of cource since $K$ is compact then it is closed and bounded. Because it is bounded then for every $u \in K$ $\|u\| \leq M$.

I really need some hint now..

6. Feb 6, 2014

7. Feb 6, 2014

### PeroK

Here's an outline of a basic proof:

Assume not, then you get a sequence in K and a sequence not in U that get arbitrarily close.

As K is compact, the sequence in K has a subsequence which converges to k in K.

And, it's easy to show that the corresponding subsequence of points not in U also converges to k.

But, as U is open, this is a contradiction (as k has a neighbourhood in U).

Can you do your own formal proof based on that?

8. Feb 6, 2014

### Mixer

Well, only difficulty I see in the proof is that how to show the corresponding subsequence of points not in U also converges to k. Of cource if points get arbitarily close, so the distance between points is less than r > 0.. ?

9. Feb 6, 2014

### Mixer

I think I got it now.

So let's assume that there exist no such r. Therefore there is sequence $(x_n) \subset \mathbb{R^n}\setminus U$ such that $\|{x_n - k_{n_j}}\| < \frac{\epsilon}{2}$. Now $k_{n_j}$ is a converging subsequence in $K$.

Because $K$ is compact there exists subsequence $k_{n_j}$ which converges to some $k$, so $\|{k_{n_j} - k}\| < \frac{\epsilon}{2}$

Now consider $\|{x_n - k}\| =\|{x_n - k_{n_j} + k_{n_j} - k}\| \leq \|{x_n - k_{n_j}}\| + \|{k_{n_j} - k}\| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

So $(x_n)$ converges too to $k$. This is impossible, since $U$ is open and therefore every element in $U$ has open ball-neigbourhood that is a subset of $U$.

Therefore there is such r.

Q.E.D ?

Last edited: Feb 6, 2014
10. Feb 6, 2014

### Dick

I don't see how those inequalities with a fixed $\epsilon$ are telling you much. Why don't you start by assuming something like $\|{x_{n} - k_{n}}\| < \frac{1}{n}$?

11. Feb 7, 2014

### glb_lub

You could also try out the Lebesgue's number lemma . But I think you will have to prove that lemma for sake of completion. Of course , proof is available online and using the Lemma almost solves the question. http://en.wikipedia.org/wiki/Lebesgue's_number_lemma