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A set can be either open or closed, and there can either be a finite or infinite number of them.

A "compact" set is one where every open cover has finite subcover.

Is there such a thing as a set that is covered by an infinite cover of open subsets, and what would it be called?

How about a set where the open cover has a finite cover of closed subsets?

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I don't think there are names for them, but you sure can build them. As to the last question: Consider the standard topology on the real numbers, every set of the form $[x]$ is closed, and there's an infinite union of closed subsets $\cup_{x\in\mathbb{R}} \left\lbrace x \right\rbrace$, that covers every subset of the reals. – MyUserIsThis Jun 8 '13 at 23:25
A topological space is compact if every cover by open sets has a finite subcover. This is much different than having a finite cover. Every topological space has a finite cover by open and closed sets, namely the space itself. – dc2814 Jun 8 '13 at 23:30
BEGIN QUOTE: A "compact" set is one that is "covered" by a finite number of open subsets. END QUOTE That is WRONG, and this particular error leads to other mistakes. It's not merely a set with a finite open cover; it's a set for which EVERY open cover has a finite subcover. – Michael Hardy Jun 8 '13 at 23:33
Additionally a set can also fail to be open or closed, or be both. – James Jun 9 '13 at 1:22
up vote 2 down vote accepted

Here is something which you may be interested in:

open (closed) cover: A cover $\mathcal U$ of $X$ is called an open cover (or a closed cover) if each member of $\mathcal U$ is open (closed) in $X$.

Note that closed cover is not often appeared in the general topology. We always consider open covers of $X$.

Lindelof: A regular space $X$ is a Linfdelof space if and only if every open cover of $X$ has a countable subcover.

countably compact: A topological space $X$ is called countably compact pace if $X$ is a Hausdorff space and every countable open cover of $X$ has a finite subcover.

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You might be looking for Lindelöf spaces. In those, every open cover contains a countable subcover.

In $T_1$ spaces, if we try to apply the definition for "compact set" and replace "open" with "closed," we run into a problem, since singleton's are closed. For any infinite set $X$,

$$\bigcup_{x \in X} \{ x \}$$

is a "closed cover" of $X$ with no finite subcover.

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It seems that spaces where every closed cover has finite subcover are sometimes called strongly S-closed spaces. Equivalent condition is that the space contains a finite dense subset (see Theorem 3.3 in this paper.) I have found out about this when I was researching for this question. – Martin Sleziak Apr 9 '14 at 10:30

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