1

Let $X$ be a limit point compact space, and let $f:X \rightarrow Y$ be continuous, then $f(X)$ is limit point compact as well. I have seen a counter example that disproves this statement, but I have also constructed a "proof" that proves this statement. I am wondering if someone can point out the error below

Assume the hypothesis above, and let $A \subset f(X)$ be an infinite subset. Then $f^{-1}(A)$ is an infinite subset in $X$, and thus has a limit point $x \in X$. let $U \subset f(X)$ be a neighborhood of $f(x)$. Then $f^{-1}(U)$ is a neighborhood of $x$ and thus intersects $f^{-1}(A)$. Hence $\exists$ a $x'\in f^{-1}(A)\cap f^{-1}(U)$, which means that $f(x')\in A \cap U$. Therefore $f(x)$ is a limit point of $A$.

Aydin Ozbek
  • 308
  • 2
    You haven't ruled out $f(x') = f(x)$. – Daniel Fischer May 12 '17 at 22:00
  • Daniel Fischer is correct: you need a point in $A\cap U$ distinct from $f(x)$, as otherwise, $f(x)$ might as well be an isolated point. More generally, a good strategy of finding errors in proofs you have counterexamples for is to try and apply the reasoning from the false proof to the counterexample itself. – tomasz May 12 '17 at 22:35

1 Answers1

2

It's well-known that for $T_1$-spaces limit point compactness and countable compactness (every countable open cover has a finite subcover) are equivalent. See this answer, e.g. A variant of the proof for normal compactness, easily shows that countable compactness is preserved by continuous images.

That same $T_1$-ness allows us the strengthen the property of limit point compactness to "for every infinite subset $A$ of $X$ there is a point of $x$ such that for every open neighbourhood $U$ of $x$, $U \cap A$ is infinite. This stronger version is again easily seen to be preserved by continuous maps, using a variant of your faulty "proof" ,which fails on being unable to show that $f(x') \neq f(x)$.

A standard counterexample to your statement and proof: let $X = \mathbb{N} \times \{0,1\}$, where $\mathbb{N}$ has the usual (discrete ) topology, and $\{0,1\}$ the indiscrete (trivial) one, $f$ the projection onto the first coordinate. Note that $X$ is not even $T_0$ as $(n,0)$ and $(n,1)$ are topologically indistinguishable.

But for $T_1$ spaces $X$ and $Y$ we can conclude that if we have a continuous map $f$ from $X$ onto $Y$ and $X$ is limit point compact, then $Y$ is too.

Henno Brandsma
  • 242,131