A property of some Hausdorff topological spaces

Question

Let $X$ be a Hausdorff topological space such that any closed subset of $X$ with empty interior is finite. show that $X$ has an isolated point.

Answer 1

Suppose this is false and take $x \in S$, where $S$ is the subset of non isolated points. Let $y_0 \in S$ and take a neighborhood $U_1$ of $x$ which does not contain $y_0$. Now take $y_1 \in S$, $y_1 \in U_1$ (this is clearly possible because every neighborhood of $x$ contains infinite points of $S$), and $U_2 \subset U_1$ such that $y_1 \notin U_2$. Take $y_2 \in U_2$. By proceeding this way, we can build a set

$$V = x \cup \left\{ \cup_{i=0}^{\infty} {y_i} \right\} $$

which is closed (because it contains all of its limit points, namely, $x$) and has empty interior by construction. By hypothesis, V should be finite, which is an absurd.

Answer 2

It suffices to show that the set of isolated points of $X$ is dense, for in that case the set of non-isolated points of $X$ is a closed set with empty interior and is therefore finite.

Suppose, to get a contradiction, that the set of isolated points is not dense in $X$. Then there is a non-empty open $V_0\subseteq X$ that has no isolated points. Since $X$ is Hausdorff, this implies that every non-empty open subset of $V_0$ is infinite. Pick any $x_0,x_1\in V_0$ such that $x_0\ne x_1$; there are disjoint open sets $U_0,V_1\subseteq V_0$ such that $x_0\in U_0$ and $x_1\in V_1$. Let $x_2\in V_1\setminus\{x_1\}$; there are disjoint open sets $U_1,V_2\subseteq V_1$ such that $x_1\in U_1$ and $x_2\in V_2$. Continue in this fashion: given $x_n\in V_n$, let $x_{n+1}\in V_n\setminus\{x_n\}$, and choose disjoint open sets $U_n,V_{n+1}\subseteq V_n$ such that $x_n\in U_n$ and $x_{n+2}\in V_{n+2}$. In this way we recursively construct distinct points $x_n$ and pairwise disjoint open sets $U_n$ such that $x_n\in U_n$ for each $n\in\Bbb N$.

Let $A=\{x_n:n\in\Bbb N\}$, and let $C=\operatorname{cl}A$; clearly $C$ is an infinite closed set. Let $W=\operatorname{int}C$, and suppose that $W\ne\varnothing$. $A$ is dense in $C$, so $x_n\in W$ for some $n\in\Bbb N$. Let $G=(U_n\cap W)\setminus\{x_n\}$. $U_n\cap W$ is a non-empty open subset of $V$, so it’s infinite, and $G$ is therefore also a non-empty open subset of $V$; let $x\in G$. Then $x\in C$, and $G$ is an open nbhd of $x$, so $G\cap A\ne\varnothing$. However, $G\subseteq U_n\setminus\{x_n\}$, so $G\cap A=\varnothing$. Thus, $W$ must be empty, and $C$ must be an infinite closed set with empty interior, contradicting the hypothesis. This contradiction shows that the set of isolated points of $X$ is dense in $X$, as desired.