Definition T.1 - Topological Space(X,G) is a topological space if and only if the following conditions hold. T.1.A0 X, G are sets and G c P(X) T.1.A1 O :- G and X :- G T.1.A2 /\(A,B:-G) AnB :- G T.1.A3 /\(McG) u(M) :- GDefinition T.2 - Open SetLet (X,G) be a topological space. We define that A is an open subset of the topological space (X,G) if and only if A :- G.Remark T.3Whenever the context is clear we will simply write "A is an open set" or "A is open".Definition T.4 - InteriorLet (X,G) be a topological space and let A c X. We define that int(A) = u({U | U c A and U is open}).Theorem T.5If (X,G) is a topological space and A c X then int(A) c A.ProofTake any x:-int(A). We get an open set U such that x:-U and U c A. Hence x:-A. We showed that int(A) c A.Theorem T.6If (X,G) is a topological space and A c X then int(A) is open.ProofLet M = {U | U c A and U is open}. Notice that M c G. By T.1.A3 u(M) :- G. But int(A) = u(M). So int(A) :- G.Theorem T.7If (X,G) is a topological space and A c X then A is open <=> A = int(A).Proof(<=) Since A = int(A), A is open by Theorem T.6. (=>) Take any x:-A. Since A is open, x:-u({E | E c A and E is open}). So x:-int(A). We showed that A c int(A). By Theorem T.5 int(A) c A, so we have that A = int(A).Theorem T.8If (X,G) is a topological space and A,B c X then A c B => int(A) c int(B).ProofBy Theorem T.5 we have int(A) c A c B. Take any x:-int(A). By Theorem T.6 int(A) is open and we have int(A) c B. Hence x:-u({E | E c B and E is open}). So x:-int(B). We showed that int(A) c int(B).Theorem T.9If (X,G) is a topological space and A,B c X then int(A n B) = int(A) n int(B).ProofTake any x:-int(A n B). We get an open set U such that x:-U and U c A n B. Then U c A and U c B. Hence x:-u({E | E c A and E is open}). So x:-int(A). Analogously, x:-int(B). Thus x :- int(A) n int(B). We showed that int(A n B) c int(A) n int(B). Take any x :- int(A) n int(B). We have an open set U such that x:-U and U c A. We have an open set V such that x:-V and V c B. By T.1.A2 U n V is open. Since x :- U n V, and U n V c A n B, we have that x :- int(A n B). We showed that int(A) n int(B) c int(A n B).Definition T.10 - Closed SetLet (X,G) be a topological space. We define that A is a closed subset of the topological space (X,G) if and only if A c X and X\A :- G.Remark T.11Whenever the context is clear we will simply write "A is a closed set" or "A is closed".Theorem T.12If (X,G) is a topological space then O and X are closed.ProofBy T.1.A01 O:-G and X:-G. Hence X\O is closed and X\X is closed. Thus X and O are closed.Theorem T.13If (X,G) is a topological space and A,B are closed then A u B is closed.ProofWe have that X\A :- G and X\B :- G. By T.1.A2 X\A n X\B :- G. Hence X \ (A u B) = X\A n X\B :- G. So A u B is closed.Theorem T.14If (X,G) is a topological space, M c {E : E is closed} and M!=0 then n(M) is closed.ProofNotice that n(M) = n({E | E:-M}). By De Morgan's Law Theorem S.IS.3 we have that X \ n({E | E:-M}) = u({X\E | E:-M}). Now, {X\E | E:-M} = {X\E | E is closed and E:-M} = {X\E : X\E is open and E:-M} = {U : U is open and X\U:-M} c G. Hence by T.1.A3 u({X\E | E:-M}) = u({U : U is open and X\U:-M}) :- G. So X\n(M) = X\n({E | E:-M}) = u({X\E | E:-M}) :- G. Thus n(M) is closed.Definition T.15 - ClosureLet (X,G) be a topological space and let A c X. We define that clo(A) = n({F | A c F and F is closed}).Theorem T.15.1If (X,G) is a topological space, A c X and x:-X then x:-clo(A) <=> /\(F) (AcF and F is closed => x:-F).ProofLet M = {F | AcF and F is closed}. Since X is closed, X:-M, and thus M!=O. By Theorem S.IS.5 we have that n(M) = {x:-X | /\(A:-M) x:-A}. Since n(M) = clo(A) we have that x:-clo(A) <=> /\(A:-M) x:-A. Hence x:-clo(A) <=> /\(F) (F:-M => x:-F). Thus x:-clo(A) <=> /\(F) (AcF and F is closed => x:-F).Theorem T.16If (X,G) is a topological space and A c X then A c clo(A).ProofTake any x:-A. Take any F such that F is closed and A c F. Then x:-F. We showed that /\(F) (F is closed and A c F => x:-F ). Hence by Theorem T.15.1 x:-clo(A). We showed that A c clo(A).Theorem T.17If (X,G) is a topological space and A c X then clo(A) is closed.ProofLet M = {F | A c F and F is closed}. Since X:-M, by Theorem T.14 n(M) is closed. Since clo(A) = n(M), clo(A) is closed.Theorem T.18If (X,G) is a topological space and A c X then A is closed <=> A = clo(A)Proof(<=) Since A = clo(A), A is closed by Theorem T.17. (=>) Take any x:-clo(A). Then /\(E) (E is closed and AcE => x:-E). Since A is closed, x:-A. We showed that clo(A) c A. By Theorem T.16, A c clo(A) so we have that A = clo(A).Theorem T.19If (X,G) is a topological space and A c X then X \ clo(A) = int(X \ A).ProofBy De Morgan's Law Theorem S.IS.2 we have that X\u({U | U c X\A and U is open}) = n({X\U | U c X\A and U is open}). Follow the calculations below. clo(A) = n({F | A c F and F is closed}) = n({F | X\F c X\A and X\F is open}) = n({X\U | U c X\A and U is open}) = X \ u({U | U c X\A and U is open}) = X \ int(X\A). Hence int(X\A) = X \ clo(A).Theorem T.20If (X,G) is a topological space and A c X then X \ int(A) = clo(X\A).ProofBy Theorem T.19 we have X \ clo(X\A) = int(X\(X\A)) = int(A). Hence X \ int(A) = clo(X\A).Theorem T.21If (X,G) is a topological space and A,B c X then A c B => clo(A) c clo(B).ProofSince A c B, we have X\B c X\A. By Theorem T.8 we have that int(X\B) c int(X\A). By Theorem T.19 we have that int(X\B) = X \ clo(B) and int(X\A) = X \ clo(A). Hence X \ clo(B) c X \ clo(A). Thus clo(A) c clo(B).Theorem T.22If (X,G) is a topological space and A,B c X then clo(A u B) = clo(A) u clo(B).ProofApply Theorem T.19 and Theorem T.9 X \ clo(A u B) = int(X\(AuB)) = int(X\A n X\B) = int(X\A) n int(X\B) = X\clo(A) n X\clo(B) = X \ (clo(A) u clo(B)). Hence clo(A u B) = clo(A) u clo(B).

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[Topological Space]
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