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Let $f : X \longrightarrow Y$ be a continuous map from a Hausdorff topological space $X$ to a metric space $Y.$ Consider the following two statements $:$

$\text P\ :\ $ $f$ is a closed map and the inverse image $f^{-1} (y) = \left \{ x \in X\ :\ f(x) = y \right \}$ is compact for each $y \in Y.$

$\text Q\ :\ $ For every compact subset $K \subset Y,$ the inverse image $f^{-1} (K)$ is a compact subset of $X.$

Which one of the following is true?

$(\text A)$ $\text Q$ implies $\text P$ but $\text P$ does NOT imply $\text Q.$

$(\text B)$ $\text P$ implies $\text Q$ but $\text Q$ does NOT imply $\text P.$

$(\text C)$ $\text P$ and $\text Q$ are equivalent.

$(\text D)$ Neither $\text P$ implies $\text Q$ nor $\text Q$ implies $\text P.$

My attempt $:$ I can see that $\text Q$ implies second part of $\text P$ i.e. for any $y \in Y,$ the set $f^{-1} (y) = \left \{x \in X\ :\ f(x) = y \right \}$ is compact because any singleton set is obviously compact. To prove that $f$ is a closed map let us take any closed subset $C$ of $X.$ Need to show that $f(C)$ is a closed subset of $Y$ i.e. we need only to prove that $\overline {f(C)} = f(C).$ We already know that $f(C) \subseteq \overline {f(C)}.$ So we need only to prove the other part of the inclusion i.e. $\overline {f(C)} \subseteq f(C).$ So let $y \in \overline {f(C)}.$ So we have a sequence $y_n \in f(C)$ such that $y_n \to y$ as $n \to \infty.$ Let $\{x_n \}$ be a sequence in $C$ such that $f(x_n) = y_n,$ for all $n \in \Bbb N.$ Therefore we get a sequence $\{x_n \}$ in $C$ such that $f(x_n) \to y$ as $n \to \infty.$

Here I have to somehow use $\text Q.$ How do I do that? Well one obvious way is to construct the range set of the sequence $\{f(x_n) \}$ including it's limit. Lets try to do that.

Let $K = \left \{f(x_n)\ :\ n \in \Bbb N \right \} \cup \{y \}.$ Then $K$ is a closed and bounded subset of the metric space $Y$ and hence by Heine-Borel theorem $K$ is compact. So by $\text Q$ we have that $f^{-1} (K)$ is a compact subset of $X.$ It is worth noting that $x_n \in f^{-1} (K),$ for all $n \in \Bbb N.$

At this stage I got stuck and couldn't proceed further. Can anybody please help me in this regard? Also I need some help to prove or disprove $\text P \implies \text Q.$

Thanks.

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math maniac.
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1 Answers1

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$P$ implies $Q$ for any topological spaces $X,Y$ and continuous $f$ between them:

If $K \subseteq Y$ is compact, let $\mathcal{U}$ be an open cover of $f^{-1}[K]$.

For every $y \in K$ cover the set $F_y:= f^{-1}[\{y\}]$ by finitely many $\mathcal{U}_y \subseteq \mathcal{U}$ by compactness of $F_y$, which is part 2 of the assumption P ($\mathcal{U}_y$, could be empty if the fibre $F_y$ is empty, of course).

For each $y \in K$ define $V_y:= Y\setminus f[X\setminus \bigcup \mathcal{U}_y]$ which is open in $Y$ as $f$ is a closed map. One checks that $y \in V_y$ so we have a cover of $K$, which is compact, so finitely many $V_y, y \in K'$ cover $K$ for some finite $K' \subseteq K$. Finally, check that

$$\bigcup \{\mathcal{U}_y: y \in K'\}$$ then is a finite (finite union of finite subfamilies) subcover of $K$, so that $f^{-1}[K]$ is compact.

Note that P implies Q always holds, we did not even use continuity of $f$, nor any assumptions on $X$ and $Y$.

Q implies P does also hold in your circumstances:

One part of P follows indeed trivially from $\{y\}$ being compact in $Y$ (as any finite space). The closedness best uses sequences, as you did:

Suppose $y \in \overline{f[C]}$. In a metric space we then can find $y_n \in f[C]$, so $y_n = f(c_n)$ for some $c_n \in C$, so that $y_n \to y$.

Now define $K= \{y\} \cup \{y_n: n \in \Bbb N\}$ which is compact in any metric space, or any topological space (Heine-Borel not needed): in any open cover of $K$, the cover element that contains $y$ already contains all but finitely many $y_n$, because of convergence, and then finitely many more cover all of $K$ already.. So $f^{-1}[K]$ is compact in $X$ so $C \cap f^{-1}[K]$ is also compact in $X$ (as a closed subset of a compact set). And as $f[C \cap f^{-1}[K]]=K \cap f[C]$ and $f$ is continuous, $K \cap f[C]$ is compact hence closed ($Y$ is Hausdorff) and as all $y_n$ are in it, so is $y$, so $y \in f[C]$ and the latter set is closed. This proofs already essentially works for all compactly generated Hausdorff sapaces $Y$, so $Y$ being metric is a bit of overkill. Locally compact Hausdorff would also have done, e.g.

Henno Brandsma
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  • For $\text Q \implies \text P$ the set $K$ is compact as it is sequentially compact and in any arbitrary metric compactness and sequential compactness are two equivalent concepts. Am I right @Henno Brandsma sir? – math maniac. Jan 31 '20 at 18:03
  • @mathmaniac. Do we don’t use that at all. Compactness by cover only. – Henno Brandsma Jan 31 '20 at 18:14
  • Oh! I see now. If I take an open cover of $K$ then there exists an open set $U$ in that collection which contains $y.$ So $U$ is an open neighbourhood of $y.$ Since $f(x_n) \to y$ as $n \to \infty,$ $U$ contains all but a finitely many terms of the sequence ${f(x_n) }$ which can be covered by finitely many open sets in the collection of open sets in the open cover for $K$ I have taken. Does it make sense now? – math maniac. Jan 31 '20 at 18:23
  • @mathmaniac. Almost. The finite sub cover consists of the one covering $y$ plus the finitely many that are needed for the finitely many that are not covered by that. – Henno Brandsma Jan 31 '20 at 18:31
  • You have said that $C \cap f^{-1} [K]$ is compact being a closed subset of a compact set $f^{-1}[K].$ But how do you know $C \cap f^{-1} [K]$ is closed? Here don't you assume that the domain space is Hausdorff? $f^{-1} [K]$ being a compact subset of a Hausdorff space is closed and since $C$ was taken to be closed it follows that $C \cap f^{-1} [K]$ is indeed closed. – math maniac. Jan 31 '20 at 18:33
  • @mathmaniac. $C$ is closed in $X$ so for any subset $A$, we have that $C \cap A$ is closed in $A$ (sub space topology). And a closed subset of a compact space is always compact. – Henno Brandsma Jan 31 '20 at 18:49
  • Then where do you use Hausdorffness of topological space $X$? – math maniac. Jan 31 '20 at 18:55
  • @mathmaniac. That’s a puzzle for you to find... – Henno Brandsma Jan 31 '20 at 18:55
  • I think that the sequential criterion for continuity is not valid for arbitrary topological space. – math maniac. Jan 31 '20 at 19:18
  • Let us continue this discussion in chat. – math maniac. Jan 31 '20 at 19:25
  • @mathmaniac. I don’t use sequential continuity at all. – Henno Brandsma Jan 31 '20 at 19:47
  • Then how do you get $f(c) \in \overline {f[C']}$ with the help of continuity? Can you help me? I am quite confused at this stage. It may be because uniqueness of the limit. – math maniac. Jan 31 '20 at 19:50
  • @mathmaniac. $c \in f[\overline{C'}] \subseteq \overline{f[C']} = K$ where the middle inclusion is a standard consequence of continuity. – Henno Brandsma Jan 31 '20 at 22:59
  • Then where do you use Hausdorffness of $X$? – math maniac. Feb 01 '20 at 03:19
  • @mathmaniac. Spoiler: it's not needed at all. We just need $Y$ to be Hausdorff and compactly generated. Even metric for $Y$ is overkill. For closed $C \subseteq X$ and all compact $K \subseteq Y$, by assumption $f^{-1}[K] \cap C$ is compact and then continuity of $f$ tells us that $f[f^{-1}[K] \cap C] = f[C] \cap K$ is compact and thus closed. So $f[C]$ is closed. – Henno Brandsma Feb 01 '20 at 07:14
  • Then why do the question gave that condition of Hausdorffness of the topological space $X$? – math maniac. Feb 01 '20 at 07:23
  • @mathmaniac. Because books aren't always "minimal". And the author might have another proof in mind that does use it. Only if we can find an example of a continuous $f: X \to Y$, $Y$ metric and $f$ obeying Q but not P, can we say we need Hausdorff. – Henno Brandsma Feb 01 '20 at 08:00