Let $X$ be a Hausdorff topological space and let $S$ be a subset of $X$ that has an accumulation point $x_0$ in $X$. Then show that $S$ is infinite.
Try Since $x_0$ is an accumulation point of $S$, every deleted neighborhood of $x_0$ intersects $S$. Consider any neighborhood $V_1$ of $x_0$, then $V_1\setminus \{x_0\}\cap S\neq \emptyset$. Choose $x_1\in V_1\setminus \{x_0\}\cap S$. Due to $X$ is Hausdorff, there exists a neighborhood $V_2$ of $x_0$ that excludes $x_1$. Also, $V_2\setminus \{x_0\}\cap S\neq \emptyset$. Choose $x_2\in V_2\setminus \{x_0\}\cap S$. Thus, we can recursively construct neighborhoods $V_k$ of $x_0$ such that for any distinct $m,n\in \mathbb{N}$, there exists $x_m\in V_m\cap S$, $x_n\in V_n\cap S$ and $x_m\neq x_n$. Therefore, $\{x_k:~k\in \mathbb{N}\}$ is a countably infinite set contained in $S$.
Query We only have used the $T_1$ property of $X$. So is this proof valid for even if $X$ is $T_1$?
Thanks in advance!!
Best Answer
Yes, the claim is true for all $T_1$ spaces. Here's a proof by contrapositive:
Proof. Suppose that $S \subseteq X$ is finite, say $S = \{s_1, \dots, s_n\}$. Let $x \in X$ be arbitrary. Since $X$ is $T_1$, let $U_1, \dots, U_n$ be open neighborhoods of $x$ such that $s_i \notin U_i$ for all $i \in \{1, \dots, n\}$. Then $U := U_1 \cap \dots \cap U_n$ is an open neighborhood of $x$ such that $(U \setminus \{x\}) \cap S \subseteq U \cap S = \varnothing$, so $x$ is not an accumulation point of $S$. Since $x$ was arbitrary, $S$ has no accumulation points.
Your proof also works, but I recommend being more precise about the recursive construction of $x_m$. From what you wrote in your proof, it seems like you might try to define $x_m$ as an arbitrary element of $V_m \setminus \{x_0\}$ where $V_m$ is an arbitrary open neighborhood of $x_0$ which does not contain $x_{m-1}$. This would not work!