Compact for limit point and continuous function

Question

If $X$ is limit point compact, i.e., every infinite subset of $X$ has a limit point in $X$, and $f:$ $X\mapsto Y$ is continuous function, is $f(X)$ limit point compact?.

Answer 1

It is limit point compact as you said. For an infinite subset $S \subseteq f(X) \to \exists A \subseteq X: S = f(A), A$ is infinite. Then $A$ has a limit point $a \in X$, you can show $f(a)$ is a limit point of $S$ in $f(X)$. There is a sequence of points $a_n$'s of $A$ converging to $a$, since $f$ is continuous, $\{f(a_n)\}$ is a sequence in $f(A)$ that converges to $f(a)$, thus $f(a)$ is a limit point of $S$ in $f(X)$.

Answer 2

Hint : Take an infinite subset $A \in f(X)$. And prove that $f^{-1}(A)$ has a limit point $x$. Then try to prove $f(x)$ is a limit point of $A$. You have to use the continuity of $f$.

Answer 3

The claim is false. Consider the space $X=\coprod_{n\in\Bbb N}(n,n+1)$ with the topology generated by the open intervals $(n,n+1)$ with natural endpoints. This space is limit point compact. Now consider the quotient map $f:X\to\Bbb N$ sending each interval $(n,n+1)$ to the point $\{n\}$.

Maybe it's true for $T_1$ spaces, though. It definitely holds when $f$ is injective.