If ..., are compact, prove that E= i=1 to n is compact.

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- March 28th 2010, 09:23 PMtn11631prove that E=union of E_i is compact
If ..., are compact, prove that E= i=1 to n is compact.

- March 28th 2010, 09:26 PMDrexel28
- March 28th 2010, 09:29 PMtn11631
- March 28th 2010, 09:31 PMDrexel28
- March 29th 2010, 06:44 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.

- March 29th 2010, 07:21 AMHallsofIvy
Are you

**given**that these are subsets of the real numbers (with the usual topology)? If not then you cannot use "Heine-Borel". - March 29th 2010, 07:41 AMtn11631
- March 29th 2010, 02:29 PMDrexel28
- March 29th 2010, 03:26 PMPlato
Give a straightforward proof.

Suppose that the collection is an open cover for .

For each we have

So is covered by a finite subcollection of .

Now the finite union of finite collections is a finite collection.

That proves that is compact.

BTW: This proof works in general topological spaces.