=== Definition 1 implies Definition 2 === Let $T = \struct {S, \tau}$ be a topological space for which: :$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$  Let us take any arbitrary $x, y \in S: x \ne y$. Let $\mathcal C_x$ be the set of all closed neighborhoods of $x$: :$\mathcal C_x = \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$ where $\relcomp S H$ is the complement of $H$ in $S$. We need to demonstrate that the only element in the intersection of $\mathcal C_x$ is $x$: :$\bigcap \mathcal C_x = \set x$ and to do that we show that if $y \ne x$ then $y \notin \bigcap \mathcal C_x$. Let $C = \bigcap C_x$. Clearly $x \in C$ and so $\set x \subseteq C_x$. We have that $\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ by hypothesis. As $x \in U$ it follows that $x \notin V$ and so $x \in \relcomp S V$. Thus $x \in U \subseteq \relcomp S V$. That is, $\relcomp S V$ is a closed neighborhood of $x$ and so $\relcomp S V \in \mathcal C_x$. As $y \in V$ it follows that $y \notin \relcomp S V$. So $\relcomp S V$ is a closed neighborhood of $x$ which does not contain $y$. So $y \notin \bigcap C_x$. As $y$ is arbitrary: :$\forall y \in S, y \ne x: \exists H: \relcomp S H \in \tau: y \notin H$ and so $C_x \subseteq \set x$. That is: :$\displaystyle \forall x \in S: \set x = \bigcap \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$ or, each point is the intersection of all its closed neighborhoods. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $T = \struct {S, \tau}$ be a topological space for which each point is the intersection of all its closed neighborhoods. Let $x, y \in S: x \ne y$. Let $\mathcal C_x$ be the set of all closed neighborhoods of $x$: :$\mathcal C_x = \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$ where $\relcomp S H$ is the complement of $H$ in $S$. This arises from the definition of a closed set as the complement in $S$ of an open set. We have that: :$\displaystyle \set x = \bigcap \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$ Then as $y \notin \set x$ it is not the case that $\forall H \in C_x: y \in H$. So for some $H \in C_x$ it must be the case that $y \in \relcomp S H = V$. But $V = \relcomp S H \in \tau$, that is, $V$ is open in $T$. Also, as $U \subseteq H$, it must follow that $U \cap V = \O$. So: :$\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ As $x$ and $y$ are arbitrary, it follows that: :$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$  {{qed|lemma}} === Definition 1 implies Definition 3 === Let $T = \struct {S, \tau}$ be a topological space for which: :$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$  Let $x, y \in S: x \ne y$ be arbitrary. From Set is Open iff Neighborhood of all its Points, $U$ and $V$ are neighborhoods of $x$ and $y$. Thus as $x$ and $y$ are arbitrary:   :$\forall x, y \in S, x \ne y: \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x \cap N_y = \O$ {{qed|lemma}} === Definition 3 implies Definition 1 === Let $T = \struct {S, \tau}$ be a topological space for which: :$\forall x, y \in S, x \ne y: \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x \cap N_y = \O$ Let $x, y \in S: x \ne y$ be arbitrary. We have that: : $\exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x \cap N_y = \O$ {{AimForCont}} $\exists z \in S: z \in U \cap V$. Then $z \in U, z \in V$ {{begin-eqn}} {{eqn | l = z       | o = \in       | r = U       | c = {{Defof|Set Intersection}} }} {{eqn | lo= \land       | l = z       | o = \in       | r = V       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = z       | o = \in       | r = N_x       | c = {{Defof|Subset}} }} {{eqn | lo= \land       | l = z       | o = \in       | r = N_y       | c = {{Defof|Subset}} }} {{eqn | ll= \leadsto       | l = z       | o = \in       | r = N_x \cap N_y       | c = {{Defof|Set Intersection}} }} {{end-eqn}} From this contradiction it follows that $U \cap V = \O$. As $x$ and $y$ are arbitrary, it follows that: :$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$  {{qed}}