Lemma T.S.1If ScX, A:T->P(X), and M={SnA[t]|t:-T} then u(M) = S n u({A[t]|t:-T}).ProofTake any x. In the following calculations we will use Theorem S.IS.1 (1). x:-u(M) <=> \/(t:-T) (x:-SnA[t]) <=> \/(t:-t) (x:-S and x:-A[t]) <=> x:-S and \/(t:-T) (x:-A[t]) <=> x:-S and x:-u({A[t]|t:-T}) <=> x :- S n u({A[t]|t:-T}). We showed that u(M) = S n u({A[t]|t:-T}).Definition T.S.2 - Relative topologyLet (X,G) be a topological space and let S c X. We define that G|S = {SnU | U:-G}.Theorem T.S.3 - SubspaceIf (X,G) is a topological space and S c X then (S,G|S) is a topological space.ProofWe have that G|S c P(S). Since O :- G, O = S n O :- G|S. Since X :- G, S = S n X :- G|S. We showed that T.1.A0 and T.1.A1 hold for (S,G|S). Take any A,B:-G|S. We have U,V:-G such that A = S n U and B = S n V. Since (X,G) is a topological space, by T.1.A2 we have that U n V :- G. Now, A n B = (S n U) n (S n V) = S n (U n V) :- G|S. We showed that T.1.A2 holds for (S,G|S). In order to show that T.1.A3 holds for (S,G|S) we will introduce the following function s:G|S->G. For each A:-G|S we put s(A)=u({U:-G | A=SnU}). Notice that /\(A:-G|S) s(A):-G and A c s(A). We will show that (*) /\(A:-G|S) (A = S n s(A)). Take any A:-G|S. We have U:-G such that A=SnU. Hence U c s(A). Thus A c S n s(A). Now, take any x :- S n s(A). We have U:-G such that A=SnU and x:-U. Now we have that x:-S and x:-U and A=SnU. So x:-A and we showed that S n s(A) c A. Thus we showed (*). Take any M c G|S. Then by (*) M = {A | A:-M} = {S n s(A) | A:-M}. By Lemma T.S.1, u(M) = S n u({s(A)|A:-M}). We have /\(A:-M) (s(A):-G). Hence by T.1.A3 applied to (X,G) we have that u({s(A)|A:-M}) :- G. Thus u(M) :- G|S. We showed that T.1.A3 holds for (S,G|S). Hence (S,G|S) is a topological space.Definition T.S.4 - Relatively Open SetLet (X,G) be a topological space and let S c X. We define that A is open in S if and only if A:-G|S.Remark T.S.5Compare Definition T.S.4 with Definition T.2 and Remark T.3. Notice that if (X,G) is a topological space and S c X then "A is open" <=> "A is open in X".Definition T.S.6 - Relative InteriorLet (X,G) be a topological space and let A c S c X. We define that int_S_(A) = u({U | UcA and U is open in S}).Definition T.S.7 - Relatively Closed SetLet (X,G) be a topological space and let S c X. We define that A is closed in S if and only if A c S and S\A :- G|S.Remark T.S.8Compare Definition T.S.7 with Definition T.10 and Remark T.11. Notice that if (X,G) is a topological space and S c X then "A is closed" <=> "A is closed in X".Definition T.S.9 - Relative ClosureLet (X,G) be a topological space and let A c S c X. We define that clo_S_(A) = n({F | AcF and F is closed in S}).Theorem T.S.9.1If (X,G) is a topological space, x:-X and A c S c X then x:-clo_S_(A) <=> /\(F) (AcF and F is closed in S => x:-F).ProofApply Theorem T.15.1 to the topological space (S,G|S).Theorem T.S.10If (X,G) is a topological space and A,B c S c X then (1) int_S_(A) c A (2) int_S_(A) is open in S (3) A is open in S <=> A = int_S_(A) (4) A c B => int_S_(A) c int_S_(B) (5) int_S_(A n B) = int_S_(A) n int_S_(B) (6) O and S are closed in S (7) A,B are closed in S => A u B is closed in S (8) M c {E | E is closed in S} and M!=0 => n(M) is closed in S (9) A c clo_S_(A) (10) clo_S_(A) is closed in S (11) A is closed in S <=> A = clo_S_(A) (12) S \ clo_S_(A) = int_S_(S\A) (13) S \ int_S_(A) = clo_S_(S\A) (14) A c B => clo_S_(A) c clo_S_(B) (15) clo_S_(A u B) = clo_S_(A) u clo_S_(B).ProofApply the theorems about topological spaces to the topological space (S,G|S).Theorem T.S.11If (X,G) is a topological space and E c S c X then E is open in S <=> \/(U) (U is open in X and E = S n U).ProofThis follows immediately from the definition: E is open in S <=> E:-G|S <=> E:-{SnU | U:-G} <=> <=> \/(U) (E = S n U and U is open in X).Theorem T.S.12If (X,G) is a topological space and E c S c X then E is closed in S <=> \/(F) (E = S n F and F is closed in X).ProofUse the previous theorem to show (a)<=>(b). E is closed in S <=> (a) S\E is open in S <=> (b) \/(U) (S\E = S n U and U is open in X) <=> \/(U) (E = S n (X\U) and U is open in X) <=> \/(U) (E = S n (X\U) and (X\U) is closed in X) <=> \/(F) (E = S n F and F is closed in X).Theorem T.S.13If (X,G) is a topological space and E c S c X then E is open in X => E is open in S.ProofSuppose that E is open in X. Then E = S n E :- G|S.Theorem T.S.14If (X,G) is a topological space and E c S c X then E is closed in X => E is closed in S.ProofSuppose that E is closed in X. Then E = S n E and by Theorem T.S.12 E is closed in S.Theorem T.S.15If (X,G) is a topological space and E c S c X then S is open in X => [ E is open in X <=> E is open in S ].ProofSuppose that S is open in X. If E is open in X then by Theorem T.S.13 E is open in S. If E is open in S then we have U:-G such that E = S n U, and by T.1.A2 S n U :- G, so E is open in X.Theorem T.S.16If (X,G) is a topological space and E c S c X then S is closed in X => [ E is closed in X <=> E is closed in S ].ProofSuppose that S is closed in X. If E is closed in X then by Theorem T.S.14 E is closed in S. If E is closed in S then by Theorem T.S.12 we have F such that F is closed in X and E = S n F = n({S,F}) and by Theorem T.14 E is closed in X.Theorem T.S.17If (X,G) is a topological space and E c S c X then int(E) c int_S_(E).ProofBy Theorem T.S.13 we have that {U | UcE and U:-G} c {U | UcE and U:-G|S}. Hence u({U | UcE and U:-G}) c u({U | UcE and U:-G|S}). Thus int(E) c int_S_(E).Theorem T.S.18If (X,G) is a topological space and E c S c X then S is open in X => int_S_(E) = int(E).ProofSuppose that S is open in X. Then by Theorem T.S.15 we have that {U | UcE and U:-G|S} = {U | UcE and U:-G}. Hence u({U | UcE and U:-G|S}) = u({U | UcE and U:-G}). Thus int_S_(E) = int(E).Theorem T.S.19If (X,G) is a topological space and E c S c X then clo_S_(E) c clo(E).ProofTake any x:-clo_S_(E). By Theorem T.S.9.1 we have (*) /\(F) (EcF and F is closed in S => x:-F). Take any F such that EcF and F is closed in X. Let M = S n F. By Theorem T.S.12 M is closed in S. Since E c M, by (*) x:-M. Hence x:-F. We showed that /\(F) (EcF and F is closed in X => x:-F). Hence x:-clo(E) by Theorem T.15.1. We showed that clo_S_(E) c clo(E).Theorem T.S.20If (X,G) is a topological space and E c S c X then S is closed in X => clo(E) = clo_S_(E).ProofSuppose that S is closed in X. Take any x:-clo(E). By Theorem T.15.1 we have (*) /\(F) (EcF and F is closed in X => x:-F). Take any F such that EcF and F is closed in S. By Theorem T.S.16 F is closed in X. By (*) x:-F. We showed that /\(F) (EcF and F is closed in S => x:-F). Hence x:-clo_S_(E) by Theorem T.S.9.1. We showed that clo(E) c clo_S_(E). By Theorem T.S.19 clo_S_(E) c clo(E). Hence clo(E) = clo_S_(E).

[ProvenMath]
[Topology]:

[Topological Space]
[Subspace]
[Separation Axioms]