# prove that E=union of E_i is compact

• Mar 28th 2010, 08:23 PM
tn11631
prove that E=union of E_i is compact
If $\displaystyle E_1$...,$\displaystyle E_n$ are compact, prove that E= $\displaystyle \cup$ i=1 to n $\displaystyle E_i$ is compact.
• Mar 28th 2010, 08:26 PM
Drexel28
Quote:

Originally Posted by tn11631
If $\displaystyle E_1$...,$\displaystyle E_n$ are compact, prove that E= $\displaystyle \cup$ i=1 to n $\displaystyle E_i$ 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

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

Ta-da!
• Mar 28th 2010, 08:29 PM
tn11631
Quote:

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

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

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!
• Mar 28th 2010, 08:31 PM
Drexel28
Quote:

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
• Mar 29th 2010, 05:44 AM
Tinyboss
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.
• Mar 29th 2010, 06:21 AM
HallsofIvy
Are you given that these are subsets of the real numbers (with the usual topology)? If not then you cannot use "Heine-Borel".
• Mar 29th 2010, 06:41 AM
tn11631
Quote:

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.
• Mar 29th 2010, 01:29 PM
Drexel28
Quote:

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.
• Mar 29th 2010, 02:26 PM
Plato
Quote:

Originally Posted by tn11631
If $\displaystyle E_1$...,$\displaystyle E_n$ are compact, prove that E= $\displaystyle \cup$ i=1 to n $\displaystyle E_i$ is compact.

Give a straightforward proof.
Suppose that the collection $\displaystyle \left\{O_{\alpha}\right\}$ is an open cover for $\displaystyle \bigcup\limits_{j = 1}^n {E_j }$.
For each $\displaystyle m,~1\le m\le n$ we have $\displaystyle E_m\subseteq\bigcup\limits_{j = 1}^n {E_j }$
So $\displaystyle E_m$ is covered by a finite subcollection of $\displaystyle \left\{O_{\alpha}\right\}$.
Now the finite union of finite collections is a finite collection.
That proves that $\displaystyle \bigcup\limits_{j = 1}^n {E_j }$ is compact.

BTW: This proof works in general topological spaces.