# Proving compactness

1. Nov 16, 2008

### kathrynag

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

If $$E_{1}$$,.....$$E_{n}$$ are compact, prove that E=$$\cup^{n}_{i=1}$$$$E_{i}$$ is compact.

2. Relevant equations

3. The attempt at a solution
A set E is compact iff for every family {$$G_{\alpha}$$}$$_{\alpha\in}A$$ of open sets such that E$$\subsetU_{\alpha\in}A G_{\alpha}$$

Let $$G_{\alpha}$$=$$E_{n}$$.
Let E=(i,n)
If i<x<n, there is a positive integer n such that $$E_{n}$$<x, hence x$$\in$$$$G_{n}$$ and E$$\subset$$$$G_{n}$$.

2. Nov 16, 2008

### VeeEight

There are many definitions of compactness, depending on if you are talking about the real line, a metric space, etc. You may want to review the ones that apply for you. One method that I like to use is that a metric space is compact if and only if it is complete and totally bounded. Since in R, totally bounded and bounded is the same thing, and a closed subset of a complete metric space is complete, we can generalize to say that a subset of R is compact if and only if it is closed and bounded. This is the one I like to use since messing around with open covers can get messy (but if you want to do that then go ahead).

3. Nov 19, 2008

### kathrynag

I'm still a bit confused on the proof.
So, we have $$E_{n}$$ that is closed and bounded.
I'm confused on how to do a proof that shows the subset is compact.

4. Nov 20, 2008

### Vid

Are you using the definition that a set is compact if every open cover contains a finite subcover?

If so, you just have to show that an open cover of the union has a finite subcover. You simply have to construct this finite subcover using the fact that each E_i is compact.

5. Nov 20, 2008

### Dick

I don't even understand your definition of a compact set, iff for any family WHAT? Then it trails into gibberish. Try saying things in words, like "a set E is compact iff for every open cover of E there is a finite subcover". If there is a finite subcover of each of the E_i (since they are compact) how can you make a finite subcover of the union of the E_i?

6. Nov 20, 2008

### HallsofIvy

Staff Emeritus
Katherynag, the fact that this is a union of a finite number of compact sets is crucial. And what Dick said is important: you wrote, "A set E is compact iff for every family {$$G_{\alpha}$$} of open sets such that E$$\subsetU_{\alpha\in}A G_{\alpha}$$". That's the "subject" of the sentence but you have no "verb"! What did you intend to say about those sets?

That makes me worry that you are memorizing definitions without really understanding them. Do you understand why, as I said before, "finiteness" is important here?

7. Nov 20, 2008

### kathrynag

Because of the Heine Borel Theorem. If a set E$$\subset$$R is compact iff E is closed and bounded.
Also for evey family of open sets there is a finite set.