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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.

user165335
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2 Answers2

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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.

Mathmo123
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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.

Asaf Karagila
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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. – Babai Jul 20 '14 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 '14 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. – Babai Jul 20 '14 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 '14 at 12:29
  • @Susobhan: I don't know what you want to hear? "Oh yes, dear mighty lord of mathematics. You are correct, we should only post the simplest and clearest answers!!!!11" or "You are right. I should have not posted anything at all. I shall go and commit seppuku in the garden to repent my shame." or maybe "I don't care what you have to say"? I have no intention of giving you any of those replies. I posted an answer which I felt was different from the one by Mathmo123 (which I upvoted before typing my own, by the way). I do think it's a pity that you downvote answers which you admit are correct. – Asaf Karagila Jul 20 '14 at 12:31
  • Mathematics is to simplify complicated things. Not the other way. Simpler arguments are always appreciated "whenever possible". – Babai Jul 20 '14 at 12:32
  • @Susobhan: I also see no issue with the fact that this is not as clear and as simple. This answer adds knowledge to the page, which was not there when I first started typing the answer. Again, it's a shame that you downvote answers which make effort to contribute something positive, after admitting that they are correct (even though this is not that complicated, and if you think this is then you are in for a treat when you study mathematics more seriously). But do whatever you want. – Asaf Karagila Jul 20 '14 at 12:33
  • I often feel that for questions such as this, the optimum is for there to be one basic "simple" answer, and another that adds more depth for more advanced users or for the OP to revisit at a later stage. Although... @AsafKaragila, I'm not sure I fully understand this open cover. It will always have a finite sub cover (since it contains ${X}$) and as such is uninteresting from a compactness point of view. I was hoping you could elaborate on your reasons for this choice of an open cover? – Mathmo123 Jul 20 '14 at 12:35
  • If a 'downvote' makes you so depressed, I will upvote it any way. Sorry to hurt your emotions. But I am not able to upvote it now, provided some editing is being done. No harm feelings. Lets be a happy mathematics family. Cheers. :) – Babai Jul 20 '14 at 12:36
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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 '14 at 12:38
  • @Susobhan: I am not hurt, I am simply confused as to what purpose you think a downvote to an answer you admit is correct is useful for. I am also very tired of these comments, so I will stop replying to them. Have a quiet day. – Asaf Karagila Jul 20 '14 at 12:39
  • @Asaf Karagila, OP here. I think I get what you mean. To make it more concrete can I say that $U$ is the neighbourhood of $a$, that is all $q$ such that $d(a,q)<r$? If so, should I set $r$ to be something? Because I am under the impression that it doesn't matter, but my teachers seem always to have preferred a fixed number, e.g. $r = 1$, in proofs. – is it normal Jul 22 '14 at 01:21
  • @Cai: If you can choose a single neighborhood for each point, sure. And in metric spaces it is indeed always possible to do that nicely. In some cases you might have to invoke the axiom of choice (meaning there is no uniform way to choose the neighborhoods), and you can take my solution here: don't choose, take everything. – Asaf Karagila Jul 22 '14 at 11:10