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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?

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4 Answers 4

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A subcover doesn't have to be a proper subcover

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

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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.

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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.

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