

TOPOLOGICAL SPACE
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) : G Definition T.2  Open Set Let (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.3 Whenever the context is clear we will simply write "A is an open set" or "A is open". Definition T.4  Interior Let (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.5 If (X,G) is a topological space and A c X then int(A) c A. Proof Take 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.6 If (X,G) is a topological space and A c X then int(A) is open. Proof Let 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.7 If (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.8 If (X,G) is a topological space and A,B c X then A c B => int(A) c int(B). Proof By 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.9 If (X,G) is a topological space and A,B c X then int(A n B) = int(A) n int(B). Proof Take 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 Set Let (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.11 Whenever the context is clear we will simply write "A is a closed set" or "A is closed". Theorem T.12 If (X,G) is a topological space then O and X are closed. Proof By 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.13 If (X,G) is a topological space and A,B are closed then A u B is closed. Proof We 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.14 If (X,G) is a topological space, M c {E : E is closed} and M!=0 then n(M) is closed. Proof Notice 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  Closure Let (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.15.1 If (X,G) is a topological space, A c X and x:X then x:clo(A) <=> /\(F) (AcF and F is closed => x:F). Proof Let 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.16 If (X,G) is a topological space and A c X then A c clo(A). Proof Take 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.17 If (X,G) is a topological space and A c X then clo(A) is closed. Proof Let 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.18 If (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.19 If (X,G) is a topological space and A c X then X \ clo(A) = int(X \ A). Proof By 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.20 If (X,G) is a topological space and A c X then X \ int(A) = clo(X\A). Proof By Theorem T.19 we have X \ clo(X\A) = int(X\(X\A)) = int(A). Hence X \ int(A) = clo(X\A). Theorem T.21 If (X,G) is a topological space and A,B c X then A c B => clo(A) c clo(B). Proof Since 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.22 If (X,G) is a topological space and A,B c X then clo(A u B) = clo(A) u clo(B). Proof Apply 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).
[ProvenMath]
[Topology]:
[Topological Space]
[Subspace]
[Separation Axioms]