(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

For this set is the inverse image of the diagonal of $Y \times Y$ under the mapping $x \mapsto (f(x), g(x))$, which is continuous (§ 4, no. 1, Proposition 1). The result therefore follows from (Hii) and § 2, no. 1, Theorem 1.

In other words, a continuous map of $X$ into $Y$ (Hausdorff) is uniquely determined by its values at all points of a dense subset of $X$.

For this graph is the set of all $(x, y) \in X \times Y$ such that $f(x) = y$, and the two mappings $(x, y) \to y$ and $(x, y) \to f(x)$ are continuous.

The proof is by induction on $n$: the case $n = 2$ is just the axiom (H). Let then $W_i$ ($1 \leq i \leq n - 1$) be a neighbourhood of $x_i$ such that the $W_i$ are mutually disjoint. On the other hand, for $1 \leq i \leq n - 1$ there is a neighbourhood $T_i$ of $x_i$ and a neighbourhood $U_i$ of $x_n$ which do not intersect. If we take $V_i$ to be $W_i \cap T_i$ for $1 \leq i \leq n - 1$ and

the conditions of the proposition are satisfied.

For every subset consisting of a single point is closed by reason of axiom (H\textsuperscript{i}).

Let $V'$ and $W'$ be disjoint neighbourhoods of $f(x)$ and $f(y)$ respectively in $X'$; then $\overline{f}^{-1}(V')$ and $\overline{f}^{-1}(W')$ are disjoint neighbourhoods of $x$ and $y$ respectively in $X$.

Let $x \in X$ and let $V$ be a closed neighbourhood of $x$ in $X$ such that the subspace $V$ is Hausdorff. Then the closed neighbourhoods of $x$ in $V$ have $\{x\}$ as their intersection (axiom (H\textsuperscript{'})); but they are also closed neighbourhoods of $x$ in $X$ (§ 3, no. 1) and therefore $X$ satisfies (H\textsuperscript{i}).

There exist non-Hausdorff spaces in which every point has a Hausdorff neighbourhood (Exercise 7).

Let $X = \prod_{i \in I} X_i$ be a product of topological spaces. Then if $x, y$ are two distinct points of $X$, we have $\operatorname{pr}_i x \neq \operatorname{pr}_i y$ for some index $i$, and Proposition 5 of no. 1 shows that $X$ is Hausdorff if the $X_i$ are. Conversely, if $X$ is Hausdorff and the $X_i$ are non-empty, then each $X_i$ is homeomorphic to a subspace of $X$ (§ 4, no. 2, Proposition 4) and is therefore Hausdorff.

The condition is sufficient by reason of Proposition 5 of no. 1. Conversely, suppose $X$ is Hausdorff; let $Y = \prod_{i \in I} Y_i$ and let $f = (f_i)_{i \in I}$ be the mapping $x \to (f_i(x))$. By Proposition 7 above $Y$ is Hausdorff, and by Proposition 3 of § 4, no. 1, $\mathfrak{T}$ is the inverse image under $f$ of the topology of $Y$. If $f(x) = f(y)$ for two distinct points $x, y$ of $X$ it is clear that every open set (in the topology $\mathfrak{T}$) which contains $x$ also contains $y$, contrary to the hypothesis that $X$ is Hausdorff.

The first assertion follows from the fact that $X$ is a subspace of the Hausdorff space $\prod_\alpha X_\alpha$ (Proposition 7). To show that $X$ is closed in the product space, let $F_{\alpha \beta}$ ($\alpha \leq \beta$) be the subset of $\prod_\alpha X_\alpha$ consisting of points $x$ for which $\operatorname{pr}_\alpha x = f_{\alpha \beta}(\operatorname{pr}_\beta x)$; the $F_{\alpha \beta}$ are closed in $\prod_\alpha X_\alpha$ (no. 1, Proposition 2), hence so is their intersection $X$.

Evidently, any sum of Hausdorff spaces (§ 2, no. 4, Example 3) is a Hausdorff space.

Let $\varphi : X \to X/R$ be the canonical mapping; then $C$ is the inverse image under $\varphi \times \varphi : X \times X \to (X/R) \times (X/R)$ of the diagonal $\Delta$ of $(X/R) \times (X/R)$. The first part of the proposition therefore follows from the continuity of $\varphi \times \varphi$ [Axiom (Hii) and Theorem 1 of § 2, no. 1]. If $R$ is open then $(X/R) \times (X/R)$ can be identified with the quotient space $(X \times X)/(R \times R)$ ($§ 5$, no. 3, corollary to Proposition 8), and $\Delta$ is then identified with the canonical image in $(X \times X)/(R \times R)$ of the set $C$, which is saturated with respect to $R \times R$. Hence $\Delta$ is closed in $X \times X$ and therefore $X$ is Hausdorff.

If $R$ is not open, there are examples where $C$ is closed but $R$ is not Hausdorff (Exercises 10 and 28).

To show that $X/R$ is Hausdorff we can also apply Proposition 5 of no. 1: $M$ and $N$ being two distinct equivalence classes of $R$ it is sufficient that there should be a continuous mapping $f$ of an open subset $A$ of $X$, saturated with respect to $R$ and containing $M$ and $N$, into a Hausdorff space $X'$, such that 1) $f$ is constant on each equivalence class mod $R$ contained in $A$, 2) $f$ takes distinct values on $M$ and $N$. For since $A/R_A$ can be identified with an open subset of $X/R$ ($§ 3$, no. 6, Proposition 10, Corollary 1), we can apply Proposition 5 of no. 1 to the mapping $g : A/R_A \to X'$ induced by $f$, since $g$ is continuous ($§ 3$, no. 4, Proposition 6).

In particular:

For ($§ 3$, no. 5) $X/R$ is homeomorphic to the subspace $s(X/R)$ of $X$, which is Hausdorff. Furthermore $s(X/R)$ is the set of all $x \in X$ such that $s(\varphi(x)) = x$, where $\varphi : X \to X/R$ is the canonical mapping; hence the second assertion follows from no. 1, Proposition 2.

(O_{III}) The set of closed neighbourhoods of any point of $X$ is a fundamental system of neighbourhoods of the point.

(O'_{III}) Given any closed subset $F$ of $X$ and any point $x \notin F$ there is a neighbourhood of $x$ and a neighbourhood of $F$ which do not intersect.

(O_{III}) \Longrightarrow (O'_{III}): If $F$ is closed and $x \notin F$, then there is a closed neighbourhood $V$ of $x$ contained in the neighbourhood $C_F$ of $x$; $V$ and $C_V$ are neighbourhoods of $x$ and $F$ respectively, and have no point in common.

(O'_{III}) \Longrightarrow (O_{III}): If $W$ is an open neighbourhood of $x \in X$, then there is a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $C_W$ which are disjoint, and therefore $\overline{U} \subset W$.

Let $A$ be a subspace of a regular space $X$. Since $X$ is Hausdorff, so is $A$ (no. 2); on the other hand, every neighbourhood of a point $x \in A$ with respect to $A$ is of the form $V \cap A$, where $V$ is a neighbourhood of $x$ in $X$. Since $X$ is regular there is a neighbourhood $W$ of $x$ in $X$ which is closed in $X$ and contained in $V$; $W \cap A$ is then a neighbourhood of $x$ in $A$, closed in $A$ and contained in $V \cap A$. Hence the result. Conversely:

$X$ is Hausdorff by Proposition 6 of no. 2. Let $x$ be any point of $X$ and let $V$ be a closed regular neighbourhood of $x$. If $U$ is any neighbourhood of $x$ contained in $V$, then $U$ is a neighbourhood of $x$ relative to $V$; hence by hypothesis there is a neighbourhood $W$ of $x$ in $V$ which is closed in $V$ and contained in $U$. But $W$ is a neighbourhood of $x$ in $X$ since $V$ is a neighbourhood of $x$ in $X$, and $W$ is closed in $X$ since $V$ is closed in $X$.

The uniqueness of $\overline{f}$ follows from the principle of extension of identities (no. 1, Proposition 2, Corollary 1). It is clear that the condition is necessary, for if $\overline{f}$ is continuous on X, then for each $x \in X$ we have

(§ 7, no. 5). Conversely, suppose that the condition is satisfied and define for each $x \in X$; $\overline{f}(x)$ is a well-defined element of Y, since Y is Hausdorff. We have to show that $\overline{f}$ is continuous at each point $x \in X$. Let then V' be a closed neighbourhood of $\overline{f}(x)$ in Y; then by hypothesis there is an open neighbourhood V of x in X such that $f(V \cap A) \subset V'$. Since V is a neighbourhood of each of its points, we have for each $z \in V$, and from this it follows that $\overline{f}(z) \in \overline{f}(V \cap A) \subset V'$, since V' is closed. The result now follows from the fact that the closed neighbourhoods of $f(x)$ form a fundamental system of neighbourhoods of $f(x)$ in Y.

The mapping $\overline{f}$ is said to be obtained by extending f by continuity to X.

In the statement of Theorem 1 the hypothesis that Y is regular cannot be weakened without imposing additional restrictions on X, A or f (Exercise 19).

b) $\lim_{x_2, \mathfrak{F}_2} f(x_1, x_2) = g(x_1)$ exists for all $x_1 \in X_1$.

Then $\lim_{x_1, \mathfrak{F}_1} g(x_1)$ exists and is equal to $\lim_{\mathfrak{F}_1 \times \mathfrak{F}_2} f$.

Let $X'_1 = X_1 \cup \{\omega_1\}$ (resp. $X'_2 = X_2 \cup \{\omega_2\}$) be the topological space associated with the filter $\mathfrak{F}_1$ (resp. $\mathfrak{F}_2$) (§ 6, no. 5, Example). In the product space $X' = X'_1 \times X'_2$ let $X''$ be the union of the subspaces $X_1 \times X'_2$ and $\{(\omega_1, \omega_2)\}$. $X$ is clearly a dense subspace of $X''$, and the hypotheses imply that $f(y_1, y_2)$ tends to a limit when $(y_1, y_2)$ tend to any point $(x_1, x_2)$ of $X''$ whilst remaining in $X$. The existence of the extension of $f$ by continuity to $X''$ then follows from Theorem 1. Since also $(\omega_1, \omega_2)$ lies in the closure of $X_1 \times \{\omega_2\}$ relative to $X''$, the result follows immediately (§ 7, no. 5).

Let $(a, b)$ be a point of $X \times X$ in the closure of $C$, and let $V$ (resp. $W$) be a closed neighbourhood of $a$ (resp. a neighbourhood of $b$) in $X$; then there is a point $(x, y) \in C \cap (V \times W)$. Since $x \in V$, $y$ belongs to the saturation $S$ of $V$ with respect to $R$; hence each neighbourhood $W$ of $b$ meets $S$. By hypothesis $S$ is closed, and therefore $b \in S$. Now let $B$ be the saturation of $\{b\}$ with respect to $R$, then each closed neighbourhood $V$ of $a$ meets $B$; since by hypothesis $B$ is closed and $X$ is regular, it follows that $a \in B$ and therefore that $(a, b) \in C$. This completes the proof.

This follows from Proposition 14 and Proposition 8 of no. 3.

Let $M$ and $N$ be two distinct equivalence classes in $X$. If each of them consists of a single point in the complement of $F$, then there exist two disjoint open neighbourhoods of $M$ and $N$ in the Hausdorff subspace $CF$; these are neighbourhoods of $M$ and $N$ in $X$ and are saturated with respect to $R$. If $M = F$ (so that $F \neq \emptyset$) and $N = \{ b \}$ where $b \notin F$, then since $X$ is regular there is an open neighbourhood of $b$ and an open neighbourhood of $F$ which do not intersect; these neighbourhoods are saturated with respect to $R$, and the proposition is proved.

Note that the quotient space $X/R$ is not necessarily regular (Chapter IX, § 4, Exercise 14).