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I feel really ignorant in asking this question but I am really just don't understand how a compact set can be considered closed? I mean by definition of a compact set it means that given an open cover we can find a finite subcover the covers the topological space. I think the word "open cover" is bothering me because if it is an open cover doesn't that mean it consists of open sets in the topology. If that is the case how can we have a closed compact set? I know a topology can be defined with the notion of closed sets rather than open sets but I guess I am just really confused by this terminology. Please any explanation would be helpful to help clear up this confusion. Thank you!

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You are confusing the definitions. A compact set is not an open cover! So why can't a compact set be closed. Actually any compact set is closed in a Hausdorff Topological space and so in metric spaces –  Nameless Nov 18 '12 at 18:27
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First of all, the open and closed are not opposites of each other. –  Hagen von Eitzen Nov 18 '12 at 18:32
    
Thank you for your answer... I guess I am just confused by the definition given in the text (by Munkres) which says "A space X is said to be compact if every open covering A of X contains a finite subcollection that also covers X" which makes it sound like all the elements of the subcover has to be open subsets of X and if that's the case than it makes me think that by construction a compact space is open. –  InsigMath Nov 18 '12 at 18:33
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@InsigMath The elements of the subcover have to be open subsets of $X$, you are correct. They may, however, cover more than just the compact set. Think of a blanket covering a bed. It may hang over the edges and drape onto the floor, covering more than just the bed. See Brian's answer for more on this. –  Austin Mohr Nov 18 '12 at 18:35
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5 Answers

up vote 19 down vote accepted

I think that what you’re missing is that an open cover of a compact set can cover more than just that set. Let $X$ be a topological space, and let $K$ be a compact subset of $X$. A family $\mathscr{U}$ of open subsets of $X$ is an open cover of $K$ if $K\subseteq\bigcup\mathscr{U}$; it’s not required that $K=\bigcup\mathscr{U}$. You’re right that $\bigcup\mathscr{U}$, being a union of open sets, must be open in $X$, but it needn’t be equal to $K$.

For example, suppose that $X=\Bbb R$ and $K=[0,3]$; the family $\{(-1,2),(1,4)\}$ is an open cover of $[0,3]$: it’s a family of open sets, and $[0,3]\subseteq(-1,2)\cup(1,4)=(-1,4)$. And yes, $(-1,4)$ is certainly open in $\Bbb R$, but $[0,3]$ is not.

Note, by the way, that it’s not actually true that a compact subset of an arbitrary topological space is closed. For example, let $\tau$ be the cofinite topology on $\Bbb Z$: the open sets are $\varnothing$ and the sets whose complements in $\Bbb Z$ are finite. It’s a straightforward exercise to show that every subset of $\Bbb Z$ is compact in this topology, but the only closed sets are the finite ones and $\Bbb Z$ itself. Thus, for example, $\Bbb Z^+$ is a compact subset that isn’t closed.

It is true, however, that compact sets in Hausdorff spaces are closed, though a bit of work is required to establish the result.

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Thank you for the example... that makes things a lot more clear! –  InsigMath Nov 18 '12 at 18:35
    
@InsigMath: You’re welcome; I’m glad to hear it. –  Brian M. Scott Nov 18 '12 at 18:37
    
Hey Brian since I have your attention a little bit... can you also explain the notion of a compact subspace. I know subspace is a subset of a topological space which has the subspace topology induced by it but I guess I am having trouble visualizing what a compact subspace really means. Thank you for any help you can give me. –  InsigMath Nov 18 '12 at 18:48
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@InsigMath: ‘$Y$ is a compact subspace of $X$’ simply means that $Y$ is a subspace of $X$, and with the subspace topology $Y$ is a compact space. This is equivalent to saying that $Y$ is a compact subset of $X$ that we happen now to be looking at as a space in its own right with the subspace topology that it inherits from $X$. –  Brian M. Scott Nov 18 '12 at 18:50
    
Oh okay... thank you for your answer! That makes more sense! –  InsigMath Nov 18 '12 at 18:52
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Compact sets need not be closed in a general topological space. For example, consider the set $\{a,b\}$ with the topology $\{\emptyset, \{a\}, \{a,b\}\}$ (this is known as the Sierpinski Two-Point Space). The set $\{a\}$ is compact since it is finite. It is not closed, however, since it is not the complement of an open set.

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Bourbaki calls this merely quasicompact and requires Hausdorff for compactness. Maybe just to get rid of such examples. :) –  Hagen von Eitzen Nov 18 '12 at 18:31
    
Thank you for this example... you and Brian have been very helpful, and everybody else who has commented :) –  InsigMath Nov 18 '12 at 18:41
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For anyone who comes across this question in the future, here is a proof:

Theorem: Compact subsets of metric spaces are closed.

Proof: Let $K$ be a compact subset of a metric space $X$ and to show that $K$ is closed we will show that its complement $K^c$ is open.

Let $p \in K^c$. Now if $q_\alpha \in K$, let $r_\alpha = \frac{1}{2}d(p,q_\alpha)$ and we will denote the neighbourhood of radius $r_\alpha$ around $q_\alpha$ to be $B_{r_\alpha}(q) = \{x \in X \mid d(q,x) < r_\alpha\}$ and the neighbourhood of radius $r_\alpha$ around $p$ to be $B_{r_\alpha}(p) = \{x \in X \mid d(p,x) < r_\alpha\}$. Then the collection of open sets $\{B_{r_\alpha}(q_\alpha)\}_\alpha$ is an open cover of $K$. As $K$ is compact there exists a finite subcover of $\{B_{r_\alpha}(q_\alpha)\}_\alpha$ such that

$$K \subset B_{r_1}(q_1) \cup \cdots \cup B_{r_n}(q_n) = U.$$

I now make the following claim:

Claim: $(B_{r_1}(p) \cap \cdots \cap B_{r_n}(p)) \cap U = \emptyset$.

Proof: Assume that $x \in (B_{r_1}(p) \cap \cdots \cap B_{r_n}(p)) \cap U$. Then we must have that $x\in B_{r_i}(p)$ for $1 \leq i \leq n$ and $x\in U$. As $x\in U$, then there exists an $i (1 \leq i \leq n)$ such that $x\in B_{r_i}(q_i)$ and, without any loss of generality, we assume $x \in B_{r_1}(q_1)$. In particular, we must also have that $x\in B_{r_1}(p)$. Therefore, by the triangle inequality we have,

$$d(p,q_1) \leq d(p,x) + d(x,q_1) < r_1 + r_1 = d(p,q_1).$$

This however is a contradiction. Therefore, $p \in B_{r_1}(p) \cap \cdots \cap B_{r_n}(p) \subset K^c$ which means that $p$ is an interior point of $K^c$. As $p$ was arbitrary, $K^c$ is open and therefore, $K$ is closed as desired. $_\Box$

Hope this may help someone.

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I just want to mention that in fact the idea in @12F8031's answer could be applied to show a stronger result that a compact subset in a Hausdorff space is closed. –  ntk Apr 11 at 9:41
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Every infinite set with complement finite topology is the counterexample. This space is compact, however is not Hausdorff. Let $X=[0,\omega]$ with complement finite topology. Then the space $X\setminus \{\omega\}$ is compact, but it is not closed.

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@12F8031, 
Your proof is good, but I don't see why "In particular, we must also have that x∈Br[1](p)".
We must have that x∈Br[a](p) for some a. Since it's obvious that Br[l](p) and Br[l](q[l])
don't share elements, then we can say a > 1. And by order of radius r[1]<r[2]..etc
hence we have d(p,q[a]) <= d(p,x)+d(x,q[a])
2r[a] <= r[1]+r[a], which is a contradiction since r[1] < r[a],since a>1

BTW Sorry for my bad notation, I haven't learned how to use LateX yet.
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