=== Definition by Open Sets implies Definition by Closed Neighborhoods === Let $T = \struct {S, \tau}$ be a topological space for which: :$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$  Let $U \in \tau$, and let $x \in U$. Then: :$\relcomp S U$ is closed and: :$x \notin \relcomp S U$ We have by hypothesis: :$\exists A, B \in \tau: \relcomp S U \subseteq A, x \in B: A \cap B = \O$  It follows that: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = B       | c = by hypothesis }} {{eqn | o = \subseteq       | r = \relcomp S A       | c = Empty Intersection iff Subset of Complement: from $A \cap B$ }} {{eqn | o = \subseteq       | r = \relcomp S {\relcomp S U}       | c = Set Complement inverts Subsets: from $\relcomp S U \subseteq A$ }} {{eqn | r = U       | c = Relative Complement of Relative Complement }} {{end-eqn}} That is: :$x \in \relcomp S A \subseteq U$ So we have demonstrated that there exists a closed neighborhood $\relcomp S A$ of $x$ contained in $U$. As $U, x$ are arbitrary: :$\forall U \in \tau: \forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq U$ {{qed|lemma}} === Definition by Closed Neighborhoods implies Definition by Intersection of Closed Neighborhoods === Let $T = \struct {S, \tau}$ be a topological space for which: :$\forall U \in \tau: \forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq U$ where $N_x$ denotes a neighborhood of $x$. Let $H \subseteq S$ such that $\relcomp S H \in \tau$.  Let $C_H$ be the set of all closed neighborhoods of $H$: :$C_H = \set {N_H: \relcomp S {N_H} \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}$ By construction: :$H \subseteq \bigcap C_H$ Let $x \notin H$. Then: :$x \in \relcomp S H \in \tau$ We have by hypothesis: :$\exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq \relcomp S H$ It follows that: {{begin-eqn}} {{eqn | l = H       | r = \relcomp S {\relcomp S H}       | c = Relative Complement of Relative Complement }} {{eqn | o = \subseteq       | r = \relcomp S {N_x}       | c = Set Complement inverts Subsets: from $N_x \subseteq \relcomp S H$ }} {{eqn | o = \subseteq       | r = \relcomp S V       | c = Set Complement inverts Subsets: from $V \subseteq N_x$ }} {{end-eqn}} Therefore by construction of $C_H$: :$\relcomp S V \in C_H$ Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = V       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = x       | o = \notin       | r = \relcomp S V       | c = {{Defof|Relative Complement}} }} {{eqn | ll= \leadsto       | l = x       | o = \notin       | r = \bigcap C_H       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \relcomp S {\bigcap C_H}       | c = {{Defof|Relative Complement}} }} {{end-eqn}} That is: {{begin-eqn}} {{eqn | l = x \in \relcomp S H       | o = \implies       | r = x \in \relcomp S {\bigcap C_H}       | c =  }} {{eqn | ll= \leadsto       | l = \relcomp S H       | o = \subseteq       | r = \relcomp S {\bigcap C_H}       | c = {{Defof|Subset}} }} {{eqn | ll= \leadsto       | l = \bigcap C_H       | o = \subseteq       | r = H       | c = Set Complement inverts Subsets }} {{end-eqn}} We have shown that: :$\bigcap C_H \subseteq H$ and: :$H \subseteq \bigcap C_H$ So by definition of set equality: :$\bigcap C_H = H$ As $H$ is arbitrary: :$\forall H \subseteq S: \relcomp S H \in \tau: H = \bigcap \set {N_H: \relcomp S {N_H} \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}$ {{qed|lemma}} === Definition by Intersection of Closed Neighborhoods implies Definition by Open Sets === Let $T = \struct {S, \tau}$ be a topological space for which: :$\forall H \subseteq S: \relcomp S H \in \tau: H = \bigcap \set {N_H: \relcomp S {N_H} \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}$ Let $F \subseteq S$ and $\relcomp S F \in \tau$. We have by hypothesis: :$F = \bigcap \set {N_F: \relcomp S {N_F} \in \tau, \exists V \in \tau: F \subseteq V \subseteq N_F}$ Pick arbitrary $x \notin F$.  Then: :$\exists N \subseteq S: \relcomp S N \in \tau, \exists V \in \tau: F \subseteq V \subseteq N$ Because $x \notin F \subseteq N$: :$x \in \relcomp S N$ Because $V \subseteq N$, it follows from Empty Intersection iff Subset of Complement that: :$V \cap \relcomp S N = \O$ Therefore we have: {{begin-eqn}} {{eqn | l = \relcomp S N, V       | o = \in       | r = \tau }} {{eqn | l = x       | o = \in       | r = \relcomp S N }} {{eqn | l = F       | o = \subseteq       | r = V }} {{eqn | l = V \cap \relcomp S N       | r = \O }} {{end-eqn}} As $F$ and $x$ are arbitrary: :$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$  {{qed}}