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Let $f : X\to Y$ be a closed map of topological spaces, such that the inverse image of each point in $Y$ is a compact subset of $X$. Is it true that the pre-image of a compact set $K$ is compact?

The answer is yes but I’m not sure how to show it. I just know that the inverse image of a singleton, say $y,$ is contained in the union of finitely many sets which we can call $U_y$. That’s it. Why is the statement of the question true?

I saw this Why is the inverse image of a compact set under a special sort of function compact?

But it’s too old to revive. Can someone post an easy to follow solution. I’ve just read about the finite intersection property but to understand it I had to come read it here: http://mathonline.wikidot.com/finite-intersection-property-criterion-for-compactness-in-a

Stahl
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Jama
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    Why isn't Gregory's proof from what you saw useful? It expands on Brian's answer. I know his proof looks cluttered, but if you can sift through it, looks fine to me. Do you want a proof that is more elaborate, more spaced out (good presentation) and therefore easier to follow? Or do you just want to discuss some aspects of that answer in this thread? If that is the case, mention those in the question, otherwise it will look like a duplicate. But +1, I think you've made some effort. – Sarvesh Ravichandran Iyer Aug 12 '20 at 04:24
  • Thanks for responding Teresa. I think I want a proof that is more elaborate and more spaced out. I now understand his proof of the F.I.P after reading the link from above and it’s not bad at all. Brian’s second assertion is also not clear. – Jama Aug 12 '20 at 05:00
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    Then I am writing an answer doing what you say, but it will take me some time. – Sarvesh Ravichandran Iyer Aug 12 '20 at 05:03
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    @TeresaLisbon I already did such a proof below. – Henno Brandsma Aug 12 '20 at 07:55
  • @HennoBrandsma Thank you, I was doing something else and was planning to come to this later. – Sarvesh Ravichandran Iyer Aug 12 '20 at 07:56

1 Answers1

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Suppose $C \subseteq Y$ is compact.

Let $\mathcal{U}$ be an open cover for $f^{-1}[C]$ by open sets of $X$, so that

$$f^{-1}[C] \subseteq \bigcup \mathcal{U}$$

For each $y \in C$ we know that $f^{-1}[\{y\}]$ is compact by assumption and as $f^{-1}[\{y\}]\subseteq f^{-1}[C]$, this compact set is also covered by $\mathcal{U}$. So for each $y \in C$ we have a finite subset $\mathcal{U}_y$ of $\mathcal{U}$ such that:

$$f^{-1}[\{y\}] \subseteq U_y := \bigcup \mathcal{U}_y \tag{1}$$

Next we define $$O_y = Y\setminus f[X\setminus U_y]\tag{2}$$

and note that $$\forall y \in C: y \in O_y\tag{3}$$

and $$\forall y \in C: f^{-1}[O_y] \subseteq U_y\tag{4}$$

Proof of (3): suppose that $y \notin O_y$. This means by def. $(2)$ that $y \in f[X\setminus U_y]$, so there is some $x \in X\setminus U_y$ with $f(x)=y$. But then this $x \in f^{-1}[\{y\}]$ but $x \notin U_y$, contradicting $(1)$. This shows $(3)$.

Proof of (4): let $x \in f^{-1}[O_y]$ be arbitary. If $x \notin U_y$ we'd have that $f(x) \in f[X\setminus U_y]$ so $f(x) \notin O_y$ contradicting that $x \in f^{-1}[O_y]$. So $x \in U_y$ and the inclusion $(4)$ has been shown.

So $\{O_y: y \in C\}$ is an open cover of $C$ and, as $C$ is compact, there are finitely many $y_1, y_2, \ldots y_n \in C$ such that $C \subseteq \bigcup_{i=1}^n O_{y_i}$

Then (using $(4)$ as well): $$f^{-1}[C] \subseteq f^{-1}[\bigcup_{i=1}^n O_{y_i}] = \bigcup_{i=1}^n f^{-1}[O_{y_i}] \subseteq \bigcup_{i=1}^n \bigcup \mathcal{U}_{y_i}\tag{5}$$

and $(5)$ shows that $\bigcup_{i=1}^n \mathcal{U}_{y_i}$ is a finite subcover of $\mathcal{U}$ (as it is a finite union of finite sets), as we had to find. So $f^{-1}[C]$ is compact.

Note that continuity of $f$ is never used, just onto-ness and closedness of $f$ and its compact fibres.

isometry
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Henno Brandsma
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