prove that E=union of E_i is compact

**tn11631** · March 28th 2010, 08:23 PM

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

**Drexel28** · March 28th 2010, 08:26 PM

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!

**tn11631** · March 28th 2010, 08:29 PM

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!

**Drexel28** · March 28th 2010, 08:31 PM

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

**Tinyboss** · March 29th 2010, 05:44 AM

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.

**HallsofIvy** · March 29th 2010, 06:21 AM

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

**tn11631** · March 29th 2010, 06:41 AM

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.

**Drexel28** · March 29th 2010, 01:29 PM

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.

**Plato** · March 29th 2010, 02:26 PM

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.

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