=== Definition by Open Sets implies Definition by Closed Neighborhoods === Suppose that: :$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ Let $Y, A \subseteq S$. Suppose $A \subseteq Y^\circ$ and $A^- \subseteq Y$.  Then: {{begin-eqn}} {{eqn | l = A^- \cap \relcomp S Y       | o = \subseteq       | r = Y \cap \relcomp S Y       | c = Set Intersection Preserves Subsets: Corollary: from $A^- \subseteq Y$ }} {{eqn | r = \O       | c = Intersection with Complement }} {{eqn | ll = \leadsto       | l = A^- \cap \relcomp S Y       | r = \O       | c = Subset of Empty Set }} {{end-eqn}} We also have: {{begin-eqn}} {{eqn | l = A \cap \paren {\relcomp S Y}^-       | o = \subseteq       | r = Y^\circ \cap \paren {\relcomp S Y}^-       | c = Set Intersection Preserves Subsets/Corollary: from $A \subseteq Y^\circ$ }} {{eqn | r = Y^\circ \cap \relcomp S Y^\circ       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = \O       | c = Intersection with Complement }} {{eqn | ll= \leadsto       | l = A \cap \paren {\relcomp S Y}^-       | r = \O       | c = Subset of Empty Set }} {{end-eqn}} We have by hypothesis:  :$\exists U, V \in \tau: A \subseteq U, \relcomp S Y \subseteq V, U \cap V = \O$ Then: {{begin-eqn}} {{eqn | l = A       | o = \subseteq       | r = U       | c = by hypothesis }} {{eqn | o = \subseteq       | r = \relcomp S V       | c = Empty Intersection iff Subset of Complement: from $U \cap V = \O$ }} {{eqn | o = \subseteq       | r = \relcomp S {\relcomp S Y}       | c = Set Complement inverts Subsets: from $\relcomp S Y \subseteq V$ }} {{eqn | r = Y       | c = Relative Complement of Relative Complement }} {{end-eqn}} So we have demonstrated that there exists a closed neighborhood $\relcomp S V$ of $A$ contained in $Y$.  As $Y$ and $A$ are arbitrary: :$\forall Y, A \subseteq S: \paren {A \subseteq Y^\circ \wedge A^- \subseteq Y} \implies \exists N \subseteq Y: \relcomp S N \in \tau: \exists U \in \tau: A \subseteq U \subseteq N$ {{qed|lemma}} === Definition by Closed Neighborhoods implies Definition by Open Sets === Suppose that: :$\forall Y, A \subseteq S: \paren {A \subseteq Y^\circ \wedge A^- \subseteq Y} \implies \exists N \subseteq Y: \relcomp S N \in \tau: \exists U \in \tau: A \subseteq U \subseteq N$ Let $A, B \subseteq S$ be separated sets. Then $A^- \cap B = A \cap B^- = \O$. Then: {{begin-eqn}} {{eqn | l = A^-       | o = \subseteq       | r = \relcomp S B       | c = Empty Intersection iff Subset of Complement: from $A^- \cap B = \O$ }} {{end-eqn}} We also have: {{begin-eqn}} {{eqn | l = A       | o = \subseteq       | r = \relcomp S {B^-}       | c = Empty Intersection iff Subset of Complement: from $A \cap B^- = \O$ }} {{eqn | r = \paren{\relcomp S B}^\circ       | c = Complement of Interior equals Closure of Complement }} {{end-eqn}} We have by hypothesis that there exists a closed neighborhood $N$ of $A$ contained in $\relcomp S B$. Then: :$\exists U \in \tau: A \subseteq U \subseteq N \subseteq \relcomp S B$ Then we have: {{begin-eqn}} {{eqn | l = B       | r = \relcomp S {\relcomp S B}       | c = Relative Complement of Relative Complement }} {{eqn | o = \subseteq       | r = \relcomp S N       | c = Set Complement inverts Subsets: from $N \subseteq \relcomp S B$ }} {{eqn | o = \in       | r = \tau       | c = $N$ is closed }} {{end-eqn}} Finally: {{begin-eqn}} {{eqn | l = U \cap \relcomp S N       | o = \subseteq       | r = N \cap \relcomp S N       | c = Set Intersection Preserves Subsets: Corollary: from $U \subseteq N$ }} {{eqn | r = \O       | c = Intersection with Complement }} {{eqn | ll= \leadsto       | l = U \cap \relcomp S N       | r = \O       | c = Subset of Empty Set }} {{end-eqn}} Therefore we have: {{begin-eqn}} {{eqn | l = U, \relcomp S N       | o = \in       | r = \tau }} {{eqn | l = A       | o = \subseteq       | r = U }} {{eqn | l = B       | o = \subseteq       | r = \relcomp S N }} {{eqn | l = U \cap \relcomp S N       | r = \O }} {{end-eqn}} As $A$ and $B$ are arbitrary: :$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ {{qed}}