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I'm looking for proof verification/help for the title question. Here is what I have now:

Let $f:X \rightarrow Y$ be continuous with $X$ limit point compact. Let $V$ be an infinite subset of $f(X)$. Then we also have that $f^{-1}(V)$ is also an infinite subset of $X$. Since $X$ is limit point compact there is $x \in X$ such that $x$ is a limit point of $f^{-1}(V)$. Then $x \in \overline{f^{-1}(V)}$ by definition of the closure. Hence since $f$ is continuous, $$ f(x) \in f(\overline{f^{-1}(V)}) \subseteq \overline{f(f^{-1}(V))} \subseteq \overline{V} $$ Then since $f(x) \in \overline{V}$, it is either a limit point of $V$ or an element of $V$. However since is infinite, any neighborhood of $f(x)$ will contain infinitely many points of $V$ in either case. Thus $V$ has a limit point and $f(X)$ is limit point compact.

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  • $\begingroup$ Also if this is wrong, please don't just post the answer. Give me hints/tips on where I went wrong so that I can fix it myself. $\endgroup$
    – Kersen
    May 6, 2015 at 7:56
  • $\begingroup$ You can assume WLOG that $x \notin f^{-1}(V)$, so that $f(x) \notin V$. And then you are done. $\endgroup$
    – Crostul
    May 6, 2015 at 7:59
  • $\begingroup$ I don't believe that assuming $x\notin f^{-1}(V)$ is in fact without loss of generality. And if $f(x)\in V$, then I don't see why any neighborhood of $f(x)$ must contain infinitely many points of $V$. $\endgroup$ May 6, 2015 at 8:11

2 Answers 2

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Your statement does not hold in general.

Here is a counterexample: Take $\mathbb{R}$ with the usual topology and $\{0,1\}$ with the indiscrete topology. Then $X = \mathbb{R} \times \{0,1\}$ with product topology.

$X$ is limit point compact. Take any subset $A\subset X$, then if $(a,0) \in A$, $(a,1)$ is a limit point of $A$ because all neighborhoods of $(a,1)$ also contain $(a,0)$ (by definition of topology of $\{0,1\}$ and product topology). Similarly if $(a,1) \in A$, $(a,0)$ is a limit point of $A$

But the projection map onto $\mathbb{R}$ (just the real line..) is continuous, while the real line is not limit point compact (say $\mathbb{N}$ has no limit point).

For your proof , I don't quite understand "However since is infinite, any neighborhood of $f(x)$ will contain infinitely many points of $V$ in either case". In particular, what if $f(x)$ is an element of $V$?

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  • $\begingroup$ Yeah that's true, I guess I let myself think only about $\mathbb{R}$ instead of a general space. Thank you. $\endgroup$
    – Kersen
    May 6, 2015 at 8:21
  • $\begingroup$ @Kersen I am glad it helps. $\endgroup$
    – John
    May 6, 2015 at 8:24
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Yes. Let A be an infinite subset of$ f(X)$. Then $A$ = {${f(x)|x ∈ B}$} where B ⊆ X is infinite. Since $X$ is limit point compact, B has a limit point $b$. Let $V_b$ be a neighborhood of $f(b)$. Then, since $f$ is continuous, there exists some neighborhood $U_b$ of $ b$ such that $f(U_b) ⊆ Vb$. Since $b$ is a limit point of $B$, there exists some $y ∈ B$ such that $y \neq b$$, y ∈ U_b.$ Thus, $f(y) ∈ f(Ub) ⊆ Vb,$ so, since $f(y) ∈ A$, $V_b$ intersects $A$ at some point other than $f(b)$. Since our choice of neighborhood for $f(b)$ was arbitrary, we conclude that every neighborhood of $f(b)$ intersects $A$ somewhere other than$ f(b)$, meaning $f(b)$is a limit point of $A$. Since our choice of infinite subset$ A $was arbitrary, we conclude that every infinite subset of $f(X)$ has a limit point, so $f(X)$ is limit point compact.

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    $\begingroup$ How do you know $f(b)\neq f(y)$? $\endgroup$
    – Not Euler
    Aug 17, 2020 at 12:37
  • $\begingroup$ John's answer is a counter example. As Not Euler points out you haven't actually proven that $V_b$ intersects $f(X)$ at a point other than $f(b)$. $\endgroup$
    – Henry
    Jul 12, 2022 at 15:59

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