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Let $f$ be a continuous closed function from $X$ to $Y$ where $X$ and $Y$ are topological spaces. (Closed means that for any closed set $C$, $f(C)$ is also closed).

Suppose that for any $y$ in $Y$, the inverse image of $y$ is compact.

Show that if $K$ is a compact subset of $Y$, then the inverse image of $K$ is also compact.

I'm having trouble figuring out how to prove this.

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  • $\begingroup$ What've you tried? $\endgroup$ – dfeuer Jun 23 '13 at 7:16
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    $\begingroup$ Hint: use the finite intersection property. $\endgroup$ – dfeuer Jun 23 '13 at 7:17
  • $\begingroup$ Is it possible just to use the definition of "FIP"? I just know the definition of FIP, but theorem does not. $\endgroup$ – hobin Jun 23 '13 at 7:37
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HINT: Say that a family $\mathscr{A}$ of sets has the finite intersection property (FIP) if $\bigcap\mathscr{F}\ne\varnothing$ whenever $\mathscr{F}\subseteq\mathscr{A}$ is finite. First prove the following useful result:

Proposition $1$. A set $K$ in a space $X$ is compact if and only if $\bigcap\mathscr{F}\ne\varnothing$ whenever $\mathscr{F}$ is a family of of closed subsets of $K$ with the finite intersection property.

This is pretty straightforward; use the fact that if $\bigcap\mathscr{F}=\varnothing$, then $\{X\setminus F:F\in\mathscr{F}\}$ is an open cover of $K$.

Then prove this little result:

Proposition $2$. Let $\mathscr{F}$ be a family of subsets of a set $X$ with the FIP. Let $$\mathscr{F}^*=\left\{\bigcap\mathscr{A}:\mathscr{A}\text{ is a finite subset of }\mathscr{F}\right\}\;,$$ the closure of $\mathscr{F}$ under finite intersections; then $\mathscr{F}^*$ has the FIP, and $\bigcap\mathscr{F}=\bigcap\mathscr{F}^*$.

Now let $H=f^{-1}[K]$, and let $\mathscr{F}$ be a family of closed subsets of $H$ with the FIP; you want to show that $\bigcap\mathscr{F}\ne\varnothing$. By Prop. $2$ you can work with $\mathscr{F}^*$ instead: it has the FIP, and it has the same intersection as $\mathscr{F}$. You know that $$\bigcap\{f[F]:F\in\mathscr{F}^*\}\ne\varnothing\;;$$ why?

Let $y\in\bigcap\{f[F]:F\in\mathscr{F}^*\}$, let $C=f^{-1}[\{y\}]$, and let $\mathscr{F}_C^*=\{F\cap C:F\in\mathscr{F}^*\}$. Show that $\mathscr{F}_C^*$ has the FIP and conclude (how?) that $\bigcap\mathscr{F}=\bigcap\mathscr{F}^*\supseteq\bigcap\mathscr{F}_C^*\ne\varnothing$.

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    $\begingroup$ A good way for hobin to check whether (s)he has really understood this answer is that, in addition to answering the "why?" and "how?" that Brian left, it should become clear why Brian introduced and used $\mathcal F^*$ instead of just working with $\mathcal F$. $\endgroup$ – Andreas Blass Jun 24 '13 at 1:37
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    $\begingroup$ @hobin: Andreas makes a very good point, but I’ll give you a bit of a push in the right direction. I originally started this proof working directly with $\mathscr{F}$, and it was only when I got to the point at which I wanted to prove that $\{F\cap C:F\in\mathscr{F}\}$ had the FIP that I realized that I needed to replace $\mathscr{F}$ with $\mathscr{F}^*$. $\endgroup$ – Brian M. Scott Jun 24 '13 at 5:28
  • $\begingroup$ @Brian M. Scott Sorry, but I don't understand "proposition2". What is the concept of closure of collection? $\endgroup$ – hobin Jun 25 '13 at 12:46
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    $\begingroup$ @hobin: $\mathscr{F}^*$ includes not only the members of the family $\mathscr{F}$, but also all intersections of finitely many members of $\mathscr{F}$. Proposition $2$ says that no harm is done if you replace $\mathscr{F}$ by $\mathscr{F}^*$: the family still has the FIP, and it has the same intersection. And it turns out that you need to make this replacement in order to be able to prove the result that you want. In my previous comment I indicated the point in the proof where you need to have $\mathscr{F}^*$ instead of $\mathscr{F}$. $\endgroup$ – Brian M. Scott Jun 25 '13 at 18:27
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I thank Brian Scott for his marvelous proof. It took me quite a while to understand his proof in complete detail. So I'm posting a write-up with hopefully enough details to make it light reading.

Definition: Let $\mathcal F$ be a collection of sets. Then $\mathcal F$ has the finite intersection property (FIP) if whenever $F_1,\dots,F_n\in \mathcal F$, $F_1\cap\cdots\cap F_n\not=\emptyset$.

Claim: Let $X$ be a topological space. Then $X$ is compact $\Longleftrightarrow$ for every collection $\mathcal F$ of closed sets in $X$ with the FIP, $\cap_{F\in\mathcal F}{F}\not=\emptyset$.

Proof: $(\Leftarrow)$ Let $\{U_\alpha\}$ be an open cover of $X$. Then $\mathcal F=\{U_\alpha^c\}$ is a collection of closed sets such that $\cap U_\alpha^c=(\cup U_\alpha)^c=X^c=\emptyset$. Thus $\mathcal F$ cannot have the FIP. So $\exists$ a finite set such that $U_{\alpha_1}\cup\cdots\cup U_{\alpha_n}=\emptyset$. Thus $U_{\alpha_1}\cup\cdots\cup U_{\alpha_n}=X$. Thus $\{U_\alpha\}$ has a finite subcover. Thus $X$ is compact.

$(\Rightarrow)$ Let $\{C_\alpha\}$ be a collection of closed sets with the FIP. Then $\{C_\alpha^c\}$ is a collection of open sets. Since $\{C_\alpha\}$ has the FIP, no finite subset of it has empty intersection. Note that $\cap A_i=\emptyset$ $\Leftrightarrow$ $\cup A_i^c=X$. Thus no finite subset of $\{C_\alpha^c\}$ is an open cover. Since $X$ is compact, it follows that $\{C_\alpha^c\}$ cannot be an open cover. In other words $\cup C_\alpha^c\not=X$. Thus $\cap C_\alpha\not=\emptyset$. $\square$

Now we use this characterization of compactness to prove the theorem. Let $K\in Y$ be compact. Let $\mathcal F$ be a family of closed subsets of $H=f^{-1}(K)$ with the finite intersection property (FIP). We need to show that $\cap_{F\in\mathcal F}F\not=\emptyset$. Let $\mathcal F^*=\{F_1\cap\cdots\cap F_n\mid F_1,\dots,F_n\in\mathcal F\}$. Then $\cap_{F\in\mathcal F}F=\cap_{F\in\mathcal F^*}F$. And $\mathcal F^*$ also has the FIP. Now consider the sets $f(F)$ where $F\in\mathcal F^*$. Since $F\in\mathcal F^*$ are closed subsets of $H$, $\exists$ closed sets $C_F\in X$ such that $C_F\cap H=F$. The collection $\{C_F\}_{F\in\mathcal F^*}$ also has the FIP. It follows that $\{f(C_F)\}_{F\in\mathcal F^*}$ has the FIP (since it's always true that $f(A\cap B)\subseteq f(A)\cap f(B)$). Notice that for any finite subset of $\{C_F\}_{F\in\mathcal F^*}$ we have $H\cap(C_{F_1}\cap\cdots\cap C_{F_n})\not=\emptyset$. It follows that $\{f(C_F)\cap K\}_{F\in\mathcal F^*}$ has the FIP. Since $C_F$ is closed and $f$ is a closed map, $f(C_F)$ is closed in $Y$. Since $K$ is compact and $\{f(C_F)\cap K\}_{F\in\mathcal F^*}$ is a collection of closed sets in $K$, it must be that $\cap_{F\in\mathcal F^*}f(C_F)\cap K\not=\emptyset$. Next we will show that $f(C_F)\cap K=f(F)$. To see this let $x\in f(C_F)\cap K$. Then $x=f(c)$ for $c\in C_F$. And $f(c)=x\in K$, so $c\in f^{-1}(K)=H$. Thus $c\in C_F\cap H$. Thus $c\in F$. Thus $x=f(c)\in f(F)$. So we have shown $f(C_F)\cap K\subseteq f(F)$. Now let $y\in f(F)$. Then $\exists$ $x\in F$ such that $f(x)=y$. Since $F\subseteq f^{-1}(K)$, $y=f(x)\in K$. Also $x\in F\subseteq C_F$. So $y=f(x)\in f(C_F)$. Thus $y\in f(C_F)\cap K$. Thus we can conclude that $f(C_F)\cap K= f(F)$. Thus we have shown that $\{f(F)\}_{F\in\mathcal F^*}$ is a collection of closed sets in $K$ that have the FIP. Thus $\cap_{F\in\mathcal F^*}f(F)\not=\emptyset$. Let $y\in\cap_{F\in\mathcal F^*}f(F)$. Then $C=f^{-1}(y)$ is compact. Now consider $\mathcal{F}_C^*=\{F\cap C:F\in\mathcal{F}^*\}$. Let $F_1,\dots,F_n\in \mathcal{F}^*$. Then by the way $\mathcal F^*$ was defined, $F_1\cap\cdots\cap F_n=F\in \mathcal F^*$. Thus $(F_1\cap C)\cap\cdots\cap(F_n\cap C)=(F_1\cap\cdots\cap F_n)\cap C=F\cap C$. Now $y\in f(F)$ $\Rightarrow$ $\exists$ $x\in F$ s.t.\ $f(x)=y$ $\Rightarrow$ $x\in C$. Thus $F\cap C\not=\emptyset$. Thus $(F_1\cap C)\cap\cdots\cap(F_n\cap C) \not=\emptyset$. Thus $\{F \cap C\mid {F\in\mathcal F^*}$ has the finite intersection property. Since by assumption $C$ is compact, $\cap_{F\in\mathcal F^*} F \cap C\not=\emptyset$. Thus $\cap_{F\in\mathcal F^*} F\not=\emptyset$. Thus $\cap_{F\in\mathcal F} F\not=\emptyset$. Thus $H$ is compact.

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