Exercise 17.15

Proof. $\text{(⇒)}$ Suppose that a space $X$ satisfies the $T_1$ axiom and consider any two distinct points $x$ and $y$ of $X$. Then the point $\left\{x\right\}$ is closed since it is finite, and hence it also contains all of its limit points by Corollary 17.7. Since the point $y$ is not in $\left\{x\right\}$ (since $y\ne x$), it cannot be a limit point of $\left\{x\right\}$. Hence there is a neighborhood $U$ of $y$ that does not intersect $\left\{x\right\}$. Hence $x\notin U$. An analogous argument involving $\left\{y\right\}$ shows that there is a neighborhood $V$ of $x$ that does not contain $y$. Since $x$ and $y$ were arbitrary points, this shows the desired property.

$\text{(⇐)}$ Now suppose that, for each pair of distinct points in $X$, each point has a neighborhood that does not contain the other point. As in the proof of Theorem 17.8, it suffices to show that every one-point set is closed since any finite set can be expressed as the finite union of such sets, which is also then closed by Theorem 17.1. So let $\left\{x\right\}$ be such a one-point set and consider any $y\notin \left\{x\right\}$ so that clearly $y\ne x$. Then, since $x$ and $y$ are distinct, there is a neighborhood $U$ of $y$ such that $U$ does not contain $x$. Therefore $U$ and $\left\{x\right\}$ are disjoint. This shows that $y$ is not a limit point of $\left\{x\right\}$, which shows that $\left\{x\right\}$ contains all its limit points since $y\notin \left\{x\right\}$ was arbitrary. Hence $\left\{x\right\}$ is closed as desired by Corollary 17.7. □