Proposition 1

Let $X$ be a topological space. Then the following statements are equivalent:

(H) Any two distinct points of $X$ have disjoint neighbourhoods. (Hi) The intersection of the closed neighbourhoods of any point of $X$ consists of that point alone. (Hii) The diagonal of the product space $X \times X$ is a closed set. (Hiii) For every set $I$, the diagonal of the product space $Y = X^I$ is closed in $Y$. (Hiv) No filter on $X$ has more than one limit point. (Hv) If a filter $\mathfrak{F}$ on $X$ converges to $x$, then $x$ is the only cluster point of $\mathfrak{F}$.

We shall prove the implications

and

(H) $\Longrightarrow$ (Hi): If $x \neq y$ there is an open neighbourhood $U$ of $x$ and an open neighbourhood $V$ of $y$ such that $U \cap V = \varnothing$; hence $y \notin \overline{U}$.

(Hi) $\Longrightarrow$ (Hv): Let $y \neq x$; then there is a closed neighbourhood $V$ of $x$ such that $y \notin V$, and by hypothesis there exists $M \in \mathfrak{F}$ such that $M \subset V$; thus $M \cap C_V = \varnothing$. But $C_V$ is a neighbourhood of $y$; hence $y$ is not a cluster point of $\mathfrak{F}$.

(Hv) $\Longrightarrow$ (Hiv): Clear, since every limit point of a filter is also a cluster point.

(Hiv) $\Longrightarrow$ (H): Suppose $x \neq y$ and that every neighbourhood $V$ of $x$ meets every neighbourhood $W$ of $y$. Then the sets $V \cap W$ form a base of a filter which has both $x$ and $y$ as limit points, which is contrary to hypothesis.

(H) $\Longrightarrow$ (Hiii): Let $(x) = (x_i)$ be a point of $X^I$ which does not belong to the diagonal $\Delta$. Then there are at least two indices $\lambda, \mu$ such that $x_\lambda \neq x_\mu$. Let $V_\lambda$ (resp. $V_\mu$) be a neighbourhood of $x_\lambda$ (resp. $x_\mu$) in $X$, such that $V_\lambda \cap V_\mu = \varnothing$; then the set $W = V_\lambda \times V_\mu \times \prod_{i \neq \lambda, \mu} X_i$ (where $X_i = X$ if $i \neq \lambda, \mu$) is a neighbourhood of $x$ in $X^I$ (§ 4, no. 1) which does not meet $\Delta$. Hence $\Delta$ is closed in $X^I$.

(Hiii) $\Longrightarrow$ (Hii): Obvious.

(Hii) $\Longrightarrow$ (H): If $x \neq y$ then $(x, y) \in X \times X$ is not in the diagonal $\Delta$, hence (§ 4, no. 1) there is a neighbourhood $V$ of $x$ and a neighbourhood $W$ of $y$ in $X$ such that $(V \times W) \cap \Delta = \varnothing$, which means that