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Let $X$ be a topological space. Prove the equivalence of the following assertions.

a) All the singletons $\{x\}$ are closed sets.

b) If $x$ and $y$ are distinct points of $X$, there is a neighborhood $U$ of $x$ such that $y\notin U$ and a neighborhood $V$ of $y$ such that $x\notin V$.

I know that this is a $T_1$ space, but how do I prove the equivalence?

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    The statement in your title is not correct: a space in which distinct points have disjoint nbhds is a Hausdorff (or $T_2$) space, while a space in which singletons are closed is a $T_1$ space. The statement of the result that you give in the body of the question, however, is correct. – 2017-01-06
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    @BrianM.Scott. Yes, of course, you are right! Do you have a suggestion for a correct title that is not too long? – 2017-01-06
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    I would suggest "$T_{1}$-spaces are characterized by singletons being closed". – 2017-01-06
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    Your (b) is the usual definition of $T_1$, so I’d simply say the equivalence of the $T_1$ property with singletons being closed. – 2017-01-06

2 Answers 2

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A closed set is the complement of an open set. Obviously, the complement of the complement of a set is the set itself, so the complement of a closed set is an open set.

Note: in the sequel, $\{x\}^{c}=\{y\in X\vert y\neq x\}=X-\{x\}=X\setminus\{x\}$ denotes the complement of $\{x\}$ in $X$.

$(a)\implies (b)$ The set $\{x\}$ is closed for all $x\in X$, thus $\{x\}^{c}$ is open for all $x\in X$. Let $x,y\in X$ be distinct points. Then, $x\in\{y\}^{c}$ and $\{y\}^{c}$ is an open set that contains $x$, so that $\{y\}^{c}$ is a neighbourhood that contain $x$ but not $y$. The same reasoning may be applied to $y\in\{x\}^{c}$. We have proved $(b)$.

$(b)\implies (a)$ Let $x\in X$. We want to prove that $\{x\}^{c}$ is open. Take all points $y\in\{x\}^{c}$. For every such point $y$, there exists a neighbourhood $V_{y}$ containing $y$ and such that $x\notin V_{y}$. But we have $\{x\}^{c}=\cup_{y\in\{x\}^{c}}V_{y}$. Any union of open sets is open, hence $\{x\}^{c}$ is open and thus $\{x\}$ is closed.

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$b) \implies a)$: Let $x\in X$ and $y \in X - \{x\}$. Then $x\neq y$ and there exists a nbhd $V$ of $y$ such that $x\notin V$, i.e. $V \subset X - \{x\}$. Thus $X-\{x\}$ is open.

$a)\implies b)$: Let $x\neq y$. We have $X-\{x\}$ is open and $y\in X-\{x\}$, so there exists a nbhd $V$ of $y$ such that $V \subset X-\{x\}$, i.e. $x\notin V$. Similarly for the other nbhd.