homeomorphism between limit point compact set and 1st countable hausdorff space

Question

X is a limit point compact space and Y is a 1st countable hausdorff space. Then show that bijective, continuous map f:X->Y is a homeomorphosm All we need to show is f is an open or closed map. I've teied to take closed subset A of X and show f(A) is closed, but i had no idea about using limit point compactness. How should I try?

Answer 1

I have found this problem to be a tricky one (perhaps by design), where you need to use the various specific assumptions in very particular ways. Here is my solution.

The goal is to show that $f(A)$ is a closed subset of $Y$. To this end, suppose that $y$ lies in the closure of $f(A)$. The proof will be complete once it has been shown that $y\in f(A)$.

Since $y$ is in the closure of $f(A)$ and $Y$ is first countable, there exists a sequence $(y_n)_{n\in\mathbb N}$ in $f(A)$ that converges to $y$. For each $n\in\mathbb N$, $y_n\in f(A)$ implies the existence of some $x_n\in A$ with $y_n=f(x_n)$.

Define $B\equiv\{x_n\mid n\in\mathbb N\}$. The case in which $B$ is finite is relatively easy. In this case, at least one member of the sequence $(x_n)_{n\in\mathbb N}$ is infinitely repeated, so there exists a constant subsequence $(x_{n_k})_{k\in\mathbb N}$ with $x_{n_k}=\widetilde x$ for every $k\in\mathbb N$, where $(n_k)_{k\in\mathbb N}$ is a strictly increasing sequence of indices. Obviously, $(x_{n_k})_{k\in\mathbb N}$ converges to $\widetilde x$, so $(y_{n_k})_{k\in\mathbb N}=(f(x_{n_k}))_{k\in\mathbb N}$ converges to $f(\widetilde x)$ by continuity. But $(y_{n_k})_{k\in\mathbb N}$ converges also to $y$, and limits are unique in a Hausdorff space. Therefore, $y=f(\widetilde x)\in f(A)$.

If $B$ is an infinite set, then it has a limit point by the very definition of limit-point compactness. Let such a limit point be denoted as $\overline x$. Since $B\subseteq A$, $\overline x$ is a limit point also of $A$, and since $A$ is closed (so that it contains its limit points), it follows that $\overline x\in A$. Let $\overline y\equiv f(\overline x)$ and define $C\equiv\{y_n\mid n\in\mathbb N\}$. I leave it to you as an exercise to show that $\overline y$ is a limit point of $C$. (You will need the assumptions that $f$ is continuous and bijective.) To obtain a contradiction, suppose that $y\neq\overline y$. Because of the Hausdorff property, there exist disjoint open subsets $U$ and $V$ of $Y$ such that $y\in U$ and $\overline y\in V$. Remember also that $\overline y$ is a limit point of $C$. Since $Y$ is Hausdorff, it is also a $T_1$ space, so $V$ will contain infinitely many elements of $C$ (see Theorem 17.9 in Munkres, 2000, p. 99). But this is impossible, since by convergence to $y$ the members of the sequence $(y_n)_{n\in\mathbb N}$ will eventually be in $U$. This contradiction reveals that $y=\overline y=f(\overline x)\in f(A)$.