Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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?

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer

A subcover doesn't have to be a proper subcover

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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