

SUBSPACE OF TOPOLOGICAL SPACE
Lemma T.S.1 If ScX, A:T>P(X), and M={SnA[t]t:T} then u(M) = S n u({A[t]t:T}). Proof Take 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 topology Let (X,G) be a topological space and let S c X. We define that GS = {SnU  U:G}. Theorem T.S.3  Subspace If (X,G) is a topological space and S c X then (S,GS) is a topological space. Proof We have that GS c P(S). Since O : G, O = S n O : GS. Since X : G, S = S n X : GS. We showed that T.1.A0 and T.1.A1 hold for (S,GS). Take any A,B:GS. 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) : GS. We showed that T.1.A2 holds for (S,GS). In order to show that T.1.A3 holds for (S,GS) we will introduce the following function s:GS>G. For each A:GS we put s(A)=u({U:G  A=SnU}). Notice that /\(A:GS) s(A):G and A c s(A). We will show that (*) /\(A:GS) (A = S n s(A)). Take any A:GS. 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 GS. 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) : GS. We showed that T.1.A3 holds for (S,GS). Hence (S,GS) is a topological space. Definition T.S.4  Relatively Open Set Let (X,G) be a topological space and let S c X. We define that A is open in S if and only if A:GS. Remark T.S.5 Compare 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 Interior Let (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 Set Let (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 : GS. Remark T.S.8 Compare 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 Closure Let (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.9.1 If (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). Proof Apply Theorem T.15.1 to the topological space (S,GS). Theorem T.S.10 If (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). Proof Apply the theorems about topological spaces to the topological space (S,GS). Theorem T.S.11 If (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). Proof This follows immediately from the definition: E is open in S <=> E:GS <=> E:{SnU  U:G} <=> <=> \/(U) (E = S n U and U is open in X). Theorem T.S.12 If (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). Proof Use 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.13 If (X,G) is a topological space and E c S c X then E is open in X => E is open in S. Proof Suppose that E is open in X. Then E = S n E : GS. Theorem T.S.14 If (X,G) is a topological space and E c S c X then E is closed in X => E is closed in S. Proof Suppose 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.15 If (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 ]. Proof Suppose 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.16 If (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 ]. Proof Suppose 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.17 If (X,G) is a topological space and E c S c X then int(E) c int_S_(E). Proof By Theorem T.S.13 we have that {U  UcE and U:G} c {U  UcE and U:GS}. Hence u({U  UcE and U:G}) c u({U  UcE and U:GS}). Thus int(E) c int_S_(E). Theorem T.S.18 If (X,G) is a topological space and E c S c X then S is open in X => int_S_(E) = int(E). Proof Suppose that S is open in X. Then by Theorem T.S.15 we have that {U  UcE and U:GS} = {U  UcE and U:G}. Hence u({U  UcE and U:GS}) = u({U  UcE and U:G}). Thus int_S_(E) = int(E). Theorem T.S.19 If (X,G) is a topological space and E c S c X then clo_S_(E) c clo(E). Proof Take 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.20 If (X,G) is a topological space and E c S c X then S is closed in X => clo(E) = clo_S_(E). Proof Suppose 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]