# 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}).

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).
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[ProvenMath] [Topology]:
[Topological Space] [Subspace] [Separation Axioms]