# 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 G|S = {SnU | U:-G}.

Theorem T.S.3 - Subspace
If (X,G) is a topological space and S c X
then (S,G|S) is a topological space.

Proof
We 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 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:-G|S.

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 :- G|S.

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,G|S).

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,G|S).

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:-G|S <=> 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 :- G|S.

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:-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.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:-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.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]