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]