0
$\begingroup$

A topological space X is called compact if each of its open covers has a finite subcover.

http://en.wikipedia.org/wiki/Compact_space

The finite subcovers are also open covers of $X$ so there are finite subcovers for finite subcovers and so on. When we get to the smallest integer $J$ for which the subcover $\{ U_j|j\in J\}$ is still a cover of $X$, there is no another subcover for this cover and therefore not all of the open covers of $X$ have subcovers. So the $X$ is not compact - a contradiction. What is wrong with my reasoning?

$\endgroup$
4
$\begingroup$

Any cover is trivially a subcover of itself. If your cover is finite you are done. If your cover is not finite but has a finite subcover you are done.

$\endgroup$
5
$\begingroup$

A subcover doesn't have to be a proper subcover

$\endgroup$
5
$\begingroup$

An open cover is a subcover of itself, because every set is a subset of itself.

$\endgroup$
2
$\begingroup$

your reasoning is wrong! :D The inclusion need not to be proper: if the cover is finite you can trivially choose it as a finite subcover.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.