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Do sets always have open covers exist? I know they are not always finite, but do infinite ones always exist?

I was reading baby rudin and the proofs for non-relative nature for compactness seems to require that. But I couldn't find any explanations on why I can assume that open covers always exist.

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2 Answers 2

Open covers do always exist, and in fact it will always be possible to find a finite one.

If $(X, \tau)$ is a topological space, then by definition, $X$ is open. So if $A \subset X$ is any subset, then $\{X\}$ is a finite open cover of it.

The point is, that in order to be compact, every open cover has to have a finite subcover. But that doesn't stop there being some finite cover for any non-compact set.

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One of the properties of open sets is that each point lies within at least one open set. If $A\subseteq X$, then for each $a\in A$ there is some open set $U$ such that $a\in U$, and in other words $A\cap U\neq\varnothing$.

So taking $\{U\subseteq X\mid U\text{ is open and } U\cap A\neq\varnothing\}$ is an open cover of $A$ by the fact given above.

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It's not very clear, what you want to say. Look at the other answer. It has answered the question correctly. –  Susobhan Jul 20 at 12:19
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What is not clear? One of the properties of a topology is that every point lies within an open set. This shows that if we take all the open sets which intersect with $A$ nontrivially this defines an open cover. Or do you say that I should repeat the arguments that were given by others on purpose, because I usually try to avoid doing that. –  Asaf Karagila Jul 20 at 12:21
    
What is meant by "One of the properties of open sets is that each point lies at least one open set."? You are saying trivial things in a complicated way. The whole space X will be the 'U' you are looking for every point $a\in A$. Your answer is correct. But you have unnecessarily complicated the answer. –  Susobhan Jul 20 at 12:26
    
@Susobhan there is no harm in having different answers of different difficulty levels, as different users may value depth over simplicity –  Mathmo123 Jul 20 at 12:29
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@Mathmo123: There's nothing really to say. You just take all the open sets which intersect with $A$. You can easily manufacture cases where this collection must include $X$ itself and you can easily manufacture any other possible case. So while $\{X\}$ is at one extreme, this is at the other. And truth be told, these are pretty much the only two open covers you can prove to exist (and sometimes they will coincide). –  Asaf Karagila Jul 20 at 12:38

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