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

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- Mar 28th 2010, 08:23 PMtn11631prove 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 PMDrexel28
- Mar 28th 2010, 08:29 PMtn11631
- Mar 28th 2010, 08:31 PMDrexel28
- Mar 29th 2010, 05:44 AMTinyboss
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 AMHallsofIvy
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 AMtn11631
- Mar 29th 2010, 01:29 PMDrexel28
- Mar 29th 2010, 02:26 PMPlato
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.