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Is the the intersection of a finite number of compact sets is compact? If not please give a counter example to demonstrate this is not true.

I said that this is true because the intersection of finite number of compact sets are closed. Which therefore means that it will be bounded because the intersection is contained by every set. I am not sure if this is correct.

Thank you for the help

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    $\begingroup$ What is your definition of compactness? $\endgroup$
    – Phira
    Nov 5, 2012 at 16:41

4 Answers 4

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For Hausdorff spaces your statement is true, since compact sets in a Hausdorff space must be closed and a closed subset of a compact set is compact. In fact, in this case, the intersection of any family of compact sets is compact (by the same argument). However, in general it is false.

Take $\mathbb{N}$ with the discrete topology and add in two more points $x_1$ and $x_2$. Declare that the only open sets containing $x_i$ to be $\{x_i\}\cup \mathbb{N}$ and $\{x_1 , x_2\}\cup \mathbb{N}$. (If you can't see it immediately, check this gives a topology on $\{x_1 , x_2\}\cup \mathbb{N}$).

Now $\{x_i\}\cup \mathbb{N}$ is compact for $i=1,2$, since any open cover must contain $\{x_i\}\cup \mathbb{N}$ (it is the only open set containing $x_i$). However, their intersection, $\mathbb{N}$, is infinite and discrete, so definitely not compact.

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    $\begingroup$ How is {xi}∪N compact? What is the finite subcover of the only open cover -- itself? You can't cover N with a finite subcover right? $\endgroup$
    – verticese
    Apr 7, 2016 at 15:56
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    $\begingroup$ As written, that sentence was wrong. There is more than one open cover. However every open cover must contain $\{x_i\}\cup \mathbb{N}$, and hence has a finite subcover consisting of a single set. I've edited the final paragraph to remove this inaccuracy. $\endgroup$
    – user45861
    Apr 7, 2016 at 21:30
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    $\begingroup$ I think you are confused about what finite cover means. A cover is a collection of sets. It is finite if this collection contains only finitely many sets. It does not mean that the sets in the cover are finite (indeed, as in this case the sets themselves can be infinite). $\endgroup$
    – user45861
    Apr 14, 2016 at 13:52
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    $\begingroup$ The example is quite nice, but the final phrase is wrong. The correct phrase is: «... since any open cover must contain either {xi}∪N or {x1,x2}∪N (they are the only open sets containing x_i ). » Etc. $\endgroup$
    – am70
    Feb 28, 2020 at 16:46
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    $\begingroup$ Whether or not the final phrase is "wrong" depends on your precise definition of compactness for a subset. One definition is that a subset is compact if and only if it is compact as a space with the subspace topology. With this definition the final sentence is fine, as $\{x_1, x_2 \} \cup \mathbb{N}$ is not a set in the relevant topology. This definition is clearly equivalent to the definition where one considers covers of the subspace by open sets in the ambient space. $\endgroup$
    – Gooch
    Dec 10, 2021 at 4:21
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I would like to supply another answer to this interesting question. I learnt the following example in a masters thesis of Cecil Eugene Denney (Kansas State University, 1966). Here is a link to the thesis. What follows below is a modified version of Example 3 in page 27 of the thesis.

Example. Let $X=\{a, b\}$ be a topological space whose only open sets are $\emptyset$ and $X$ itself (so $X$ has the indiscrete topology). Let $Y=\mathbb{R}$ be the set of all real numbers with the standard topology. We will consider the topological space $X\times Y$. Consider the following two subsets of $X\times Y$: $$ C = \{(a, y): 1\leq y < 3\} \cup \{(b, y): 3\leq y \leq 4\} $$ and $$ D = \{(a, y): 2<y \leq 4\} \cup \{(b, y): 1\leq y \leq 2\}. $$ We claim that $C$ and $D$ are both compact. Indeed, in given any open cover $\mathcal{U}$ for $C$, each open set must be of the form $X\times U$ where $U$ is an open subset of $Y$ (this is where we use the fact that the only open subsets of $X$ are $\emptyset$ and $X$). Using the projection $\pi_Y: X\times Y \to Y$, we note that the collection of all $U$ such that $X\times U\in\mathcal{U}$ forms an open cover of $\pi_Y(C) = [1, 4]$ which is compact. Thus, there are finitely many $U_i$ which cover $[1, 4]$, and consequently finitely many $X\times U_i$ which cover $C$. This shows $C$ is compact, and the verification for compactness of $D$ is identical.

However, $C\cap D = \{(a, y): 2 < y < 3\}$ is homeomorphic, via the projection $\pi_Y$, to the open interval $(2, 3)$ in $Y=\mathbb{R}$, and it is well-known that $(2, 3)$ is not compact.

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The intersection of two compact subsets is not, in general compact. A possible example is $\mathbb R$ with the lower semicontinuity topology, i.e. the topology generated by sets of the form $(a, +\infty)$.

A subset $A\subseteq\mathbb R$ is compact in this topology if it has a minimum. Now, consider $A=[0, 1)\cup(3, +\infty)$ and $B=[2, +\infty)$. Both $A$ and $B$ are compact, but their intersection $A\cap B=(3, +\infty)$ is not.

Notice that the lower semicontinuity topology on $\mathbb R$ is not Hausdorff. As others have pointed out, the intersection of two compact subspaces of a Hausdorff topological space is still compact.

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In a $T_2$ space (Hausdorff), compact subsets are closed. Let be a $T_2$ space and let $F$ be a family of compact subsets of $S$. To avoid trivial cases let $F$ have at least one non-empty member $f$.

Now $G=\cap F$ is closed and $G^c=S\backslash G$ is open. If $V$ is a family of open sets with $\cup V\supset G,$ then $V'=V\cup \{G^c\}$ is an open cover of $f.$ (Indeed, $\cup V'=S.$)

Since $f$ is compact, there exists a finite $H\subset V'$ with $\cup H\supset f.$ Then $H'=H\backslash \{G^c\}$ is a finite subset of $V$, and $\cup H'\supset G.$

So $G$ is compact.

Note that we needed each member of $F$ to be closed, so that $G$ is closed, so that $G^c$ is open, so that $V'$ is an open family. As an earlier answer shows, if $S$ is not $T_2$ then this can fail.

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