Understanding Regular and T3 Spaces in Topology

Regular and T3-space Definition: A topological space (X,?) is said to be regular iff for every closed set F and every point p F, there exist open sets G and H such that p G,F H and G H= The regular space which is also T1-space is called a T3-space Ex: Let X={a,b,c}, and Let ?={ ,{a},{b,c},X}, show that (X,?) is regular space but not a T3-space Sol: The closed subsets are X, {b,c},{a}, . Consider the closed set {b,c}, and the point a not belong to it. Then {b,c} and {a} are open sets such that {b,c} {b,c} and a {a} and {a} {b,c}= Similarly consider the closed set {a} and a point b not belonging to it , then{b,c} and {a} are open sets such that b {b,c} and {a} {a} and {a} {b,c}= . Again for the closed set {a} and the point c , there exist open set {a} and {b,c} such that {a} {b,c}= . It follows that X is regular. Since there dose not exist an open set containing the point b and not containing the point c, the space is not T1and consequently it is neither T2nor T3.

Ex: Show that (R,U) is a T3-space . Sol: Let F be a U-closed subset and let x R, such that x F. Since F is closed set and x F, it follows that x is not an accumulation point of F . hence there exists an >0 such that (x- , x+ ) F= ..(1) Let G=(x- \4,x+ |4) and Let H=U{(y- \4,y+ \4); y F}. Then x G and F H. we now show that G H= . If possible, let z G H, the z G and z H, that is z (x- \4,x+ |4) and z (y0- \4,y0+ \4) for some y0 F. Then |x-z|< \4 and |z- y0|< \4. Hence |x- y0|=|(x-z)+(z- y0)| |x-z|+|z- y0|< \4 + \4< . It follows that y0 (x- , x+ ). Since y0 F, this contradicts (1). Hence G H= . Thus there exist open sets G and H such that G H= , so (R,U) is regular space and since (R,U) is T1-space , it follows that (R,U) is T3-space.

Theorem: A topological space X is regular iff for every point x X and every nbd N of x , there exist a nbd M of x such that clM N . Proof :"only if part" let N be any nbd of x then there exist an open set G such that x G N. Since Gcis closed and x Gc, But the space is regular , two disjoint open set L and M such that Gc L and x M. So that M Lcit follows that, clM clLc=Lc--------- (1) Also Gc L Lc G N-----(2) From (1) and (2) we get clM N . The"if part" let the condition hold. Let F be any closed subset of X and x F then x Fc, Since Fcis an open set containing x, so by ypothesis an open set M such that x M and clM Fc F (clM)c. Then (clM)cis an open set containing F also M Mc= , then M (clM)c= . So the space is regular

Ex: Every T3-space is a T2-space Sol: Let (X,?)be a T3-space , and let x,y be any two distinct point. Now by definition of X, the space is T1 and so {x} is a closed set also y {x}. Since X is regular. two open set G and H such that y G {x} H & G H= , but x {x} H, hence the space is T2. Theorem: the property of space being a regular space is a topological property Proof: Let (X,?) be a regular space and let (Y, ) be a homeomorphic image of (X,?) under a homeomorphism f. To show that (Y, ) is also regular space. Let F be a closed subset of Y and let q be a point of Y such that q F. Since f is one-to-one onto function there exist p X such that f(p)=q iff f-1(q)=p. Again since f is continuous f-1(F) is closed in X. Also q F then f-1(q) f-1(F). Thus f-1(F) is a closed set and p is a point of X such that p f-1(F). Since X is regular space there exist open sets G and H such that p G, f-1(F) H and G H= . p G then f( p) f(G) then q f(G) also f-1(F) H then f( f-1(F)) f(H) then F f(H) and G H= then f(G) f(H)=f(G H)=f( )= , Since f is an open function G1=f(G) and H1= f(H) are open set in Y. So (Y, ) is regular. Theorem: The property of a space being a regular space is hereditary.

Theorem: Every compact Hausdorff space is a T3-space Proof : Let (X,?) be compact Hausdorff space To show that (X,?) is a T3-space . Since a Hausdorff space is a T1-space , it suffices to show that (X,?)is a regular , let F be a closed subset of X and let p X such that p F, so p X\F, since (X,?) is a Hausdorff space so for every x F, there must exist two open sets G(x) and H(x), such that p G(x), x H(x) and G(x) H(x) = ..(*) The collection C={H(x); x F } is open cover of F. Since F is a closed subset of a compact space X, so that F is compact (by theorem ). Hence there exists a finite numbers of points x1,x2,..,xnin F such that F U{H(xi),i=1,2,...,n},let H=U{H(xi),i=1,2,...,n} and G = {G(xi), i=1,...,n }. Then p G , since p G(xi) for each xialso G H= , [other wise G (xk) H (xk) for some xk F this contradict(*)] hence the space is regular