topological space?

Question

let (X, T) be a topological space with A closed subset of X, B a connected subset of X s.t. A intersect B non-empty. if there is non-empty W a subset of B with W interset A non-empty. show B intersect boundary of A.

Curt, no W exists in your example

Answer 1

Are you sure you stated that correctly? It doesn't look at all true.

Example: X = the x-y plane, A = the unit disk (closed), B = the interior of A.

You probably are missing additional assumptions, perhaps that B isn't entirely contained in A.

Answer 2

"Gf is a subset of XxY and XxY is a Hausdorff area so Gf is a Hausdorff area. " Y is Hausdorff area, XxY isn't inevitably Hausdorff area. hence Gf isn't inevitably Hausdorff area "enable (x1,y1) be a shrink aspect to Gf. would favor to tutor that (x1,y1) is in Gf. " awesome acceptable "seeing that (x1,y1) is a shrink aspect to Gf, there exists an open set U in Gf such that (x1,y1) is in U and some aspect (x2,y2) is in U also." If, as you're declaring, U is in Gf and (x1,y1) is in U, then (x1,y1) would nicely be in Gf, isn't it? merely so might want to coach the priority. the dazzling acceptable definition of shrink aspect is that " Any open set U in XxY such that (x1,y1) is in U will contain also a aspect in Gf." "Gf is a Hausdorff area so for all (x2,y2),(x3,y3) in Gf there exists open contraptions U,V in Gf such that (x2,y2) is in U and (x3,y3) is in V and the intersection of U and V is the null set" awesome acceptable "(x1,y1) is in Gf utilising actuality that's in some open set U in Gf and the intersection of this set with yet another arbitrary open set in Gf is the null set. " it isn't sensible universal, you likely did no longer use at each and every of the definition of Gf. does no longer this make the data suspect? No offence.