SEPARATION AXIOMS

Definition T.SA.0 - T_0 Space
(X,G) is a T_0 space if and only if (X,G) is a topological space
such that
/\(x,y:-X, x!=y) \/(A:-G) (x:-A and !(y:-A)) or (y:-A and !(x:-A)).

Definition T.SA.1 - T_1 Space
(X,G) is a T_1 space if and only if (X,G) is a topological space
such that
/\(x,y:-X, x!=y) \/(A:-G) x:-A and !(y:-A).

Definition T.SA.2 - T_2 Space
(X,G) is a T_2 space if and only if (X,G) is a topological space
such that
/\(x,y:-X, x!=y) \/(A,B:-G) AnB=O and x:-A and y:-B.

Definition T.SA.3 - T_3 Space
(X,G) is a T_3 space if and only if (X,G) is a topological space
such that
/\(A c X, A is closed) /\(b:-X\A) \/(U,B:-G) (A c U and b:-B and UnB=O).

Definition T.SA.4 - T_4 Space
(X,G) is a T_4 space if and only if (X,G) is a topological space
such that
/\(A,B) (A,B are closed and AnB=O) => \/(U,V:-G) (AcU and BcV and UnV=O).

Definition T.SA.5 - T_5 Space
(X,G) is a T_5 space if and only if (X,G) is a topological space
such that
/\(A,BcX) clo(A)nB=clo(B)nA=O => \/(U,V:-G) (AcU and BcV and UnV=O).

Definition T.SA.6 - Hausdorff Space
(X,G) is a Hausdorff space if and only if (X,G) is a T_2 space.

Definition T.SA.7 - Regular Space
(X,G) is a regular space if and only if (X,G) is a T_0 space and a T_3 space.

Definition T.SA.8 - Normal Space
(X,G) is a normal space if and only if (X,G) is a T_1 space and a T_4 space.

Theorem T.SA.9
If (X,G) is a topological space then the following conditions are equivalent:
(a) (X,G) is a T_1 space,
(b) /\(x:-X) {x} is closed.

Proof
(a)=>(b)
Take any x:-X.
We will show that X\{x} is open.
Take any a:-X\{x}.
Since (X,G) is T_1, there exists an open set A:-G
such that a:-A and !(x:-A).
Hence A c X\{x} and thus a:-int(X\{x}).
We showed that X\{x} c int(X\{x}), so X\{x} is open.
Thus {x} is closed.

(b)=>(a)
Take any x,y:-X such that x!=y.
Since {y} is closed, X\{y} is open and x:-X\{y}.
Let A=X\{y}.
We showed that
/\(x,y:-X, x!=y) \/(A:-G) x:-A and !(y:-A),
which says that (X,G) is a T_1 space.

Theorem T.SA.10
If (X,G) is a topological space then the following conditions are equivalent:
(a) (X,G) is a T_3 space,
(b) /\(x:-X) /\(U:-G, x:-U) \/(A:-G) x:-A and clo(A) c U.

Proof
(a)=>(b)
Take any x:-X and any U:-G such that x:-U.
Now, X\U is closed and !(x:-X\U).
By T_3, there exist open sets A,B such that
x:-A and X\U c B and AnB=O.
Hence A c X\B c U.
Since B is open, X\B is closed, and clo(A) c X\B c U.
We showed that
/\(x:-X) /\(U:-G, x:-U) \/(A:-G) x:-A and clo(A) c U.

(b)=>(a)
Take any AcX such that A is closed. Take any b:-X\A.
Notice that X\A is open.
By (b), there exists an open set B such that
b:-B and clo(B) c X\A.
Now we have: A c X\clo(B), b:-B, and (X\clo(B)) n B = O.
Let U=X\clo(B). We showed that
/\(A c X, A is closed) /\(b:-X\A) \/(U,B:-G) (A c U and b:-B and UnB=O),
which says that (X,G) is a T_3 space.

Theorem T.SA.11
If (X,G) is a topological space then the following conditions are equivalent:
(a) (X,G) is a T_4 space,
(b) /\(U:-G) /\(A c U, A is closed) \/(E:-G) A c E and clo(E) c U.

Proof
(a)=>(b)
Take any open set U and any closed set A that A c U.
Now A and X\U are both closed and A n (X\U) = O.
By T_4, there exist open sets E,W such that
A c E, X\U c W, and E n W = O.
So E c X\W c U.
Since X\W is closed, clo(E) c X\W c U.
We showed that
/\(U:-G) /\(A c U, A is closed) \/(E:-G) A c E and clo(E) c U.

(b)=>(a)
Take any closed sets A,B such that AnB=O.
Now, A c X\B and X\B is open.
By (b), there exists an open set E such that
A c E and clo(E) c X\B.
Hence B c X\clo(E) and E n (X\clo(E)) = O.
Let V = X\clo(E). We showed that
/\(A,B) (A,B are closed and AnB=O) => \/(E,V:-G) (AcE and BcV and EnV=O),
which says that (X,G) is a T_4 space.

Theorem T.SA.12
If (X,G) is a T_0 space and a T_3 space then (X,G) is a T_2 space.

Proof
Take any x,y:-X such that x!=y.
By T_0, there exists an open set A such that
(1) (x:-A and !(y:-A))
or
(2) (y:-A and !(x:-A)).

Suppose (1).
Then by T_3, there exists an open set E such that x:-E and clo(E) c A.
Since y :- X\A c X\clo(E), we have that
x:-E, y:-X\clo(E), and E n X\clo(E) = O.
Let B=X\clo(E). We showed that
EnB=O and x:-E and y:-B, and E,B are open.

Suppose (2).
Then by T_3, there exists an open set B such that y:-B and clo(B) c A.
Since x :- X\A c X\clo(B), we have that
x:-X\clo(B), y:-B, and B n X\clo(B) = O.
Let E=X\clo(B). We showed that
EnB=O and x:-E and y:-B, and E,B are open.

In any case, we showed that
/\(x,y:-X, x!=y) \/(E,B:-G) EnB=O and x:-E and y:-B,
which says that (X,G) is a T_2 space.

Theorem T.SA.13
If (X,G) is a T_1 space and a T_4 space then (X,G) is a T_3 space.

Proof
Take any x:-X and any open set U such that x:-U.
By T_1 and Theorem T.SA.9, {x} is closed.
So we have two disjoint closed sets: {x} n X\U = 0.
By T_4, there exist open sets A,B such that
{x} c A, X\U c B, and AnB=O.
Hence A c X\B c U.
Since X\B is closed, clo(A) c X\B c U.
We showed that
/\(x:-X) /\(U:-G, x:-U) \/(A:-G) x:-A and clo(A) c U,
which says that (X,G) is a T_3 space.

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