# Thread: prove that E=union of E_i is compact

1. ## prove that E=union of E_i is compact

If $E_1$..., $E_n$ are compact, prove that E= $\cup$ i=1 to n $E_i$ is compact.

2. Originally Posted by tn11631
If $E_1$..., $E_n$ are compact, prove that E= $\cup$ i=1 to n $E_i$ is compact.
I mean, if you know the Heine-Borel theorem this is easy as pi.

The finite union of closed sets is closed and

$\text{diam }E_1\cup\cdots\cup E_n\leqslant\sum_{j=1}^{n}\text{diam }E_j<\infty$.

Ta-da!

3. Originally Posted by Drexel28
I mean, if you know the Heine-Borel theorem this is easy as pi.

The finite union of closed sets is closed and

$\text{diam }E_1\cup\cdots\cup E_n\leqslant\sum_{j=1}^{n}\text{diam }E_j<\infty$.

Ta-da!
Ah yes! lol I feel so inferior right now for not thinking of these things! lol I'm clearly getting overtired. Thanks again your such a big help!

4. Originally Posted by tn11631
Ah yes! lol I feel so inferior right now for not thinking of these things! lol I'm clearly getting overtired. Thanks again your such a big help!
Don't feel inferior! We all have bad days! :P

5. It's pretty quick from the finite-open-cover definition of compactness, too; any open cover of the union is also an open cover for each compact set, and finite*n=finite.

6. Are you given that these are subsets of the real numbers (with the usual topology)? If not then you cannot use "Heine-Borel".

7. Originally Posted by HallsofIvy
Are you given that these are subsets of the real numbers (with the usual topology)? If not then you cannot use "Heine-Borel".
Um the original problem posted was the only information given.

8. Originally Posted by HallsofIvy
Are you given that these are subsets of the real numbers (with the usual topology)? If not then you cannot use "Heine-Borel".
I don't think the OP is meaning to be that advanced. I think this is a beginning analysis course.

9. Originally Posted by tn11631
If $E_1$..., $E_n$ are compact, prove that E= $\cup$ i=1 to n $E_i$ is compact.
Give a straightforward proof.
Suppose that the collection $\left\{O_{\alpha}\right\}$ is an open cover for $\bigcup\limits_{j = 1}^n {E_j }$.
For each $m,~1\le m\le n$ we have $E_m\subseteq\bigcup\limits_{j = 1}^n {E_j }$
So $E_m$ is covered by a finite subcollection of $\left\{O_{\alpha}\right\}$.
Now the finite union of finite collections is a finite collection.
That proves that $\bigcup\limits_{j = 1}^n {E_j }$ is compact.

BTW: This proof works in general topological spaces.