Let Y be a subset of a topological space X. If any point of Y admits an open neighbourhood U in X such that is closed in U then Y is open in .

Thanks in advance.

Printable View

- June 28th 2009, 07:10 AMBiscaimabout topological spaces
Let Y be a subset of a topological space X. If any point of Y admits an open neighbourhood U in X such that is closed in U then Y is open in .

Thanks in advance. - June 28th 2009, 11:26 PMMoo
Hello,

Okay, maybe I'm just rusty at topology, but what does it mean for a set Y to be open in another set, ? - June 29th 2009, 06:39 AMBiscaim
- June 29th 2009, 10:57 AMMoo
- June 29th 2009, 11:24 AMflyingsquirrel
- June 30th 2009, 12:18 AMEnrique2
It is a nice problem. I'll use cl for closure and \cap for intersection

and \subset for inclusion

a) For each subset U of X open, U \cap cl(Y)\subset cl (U\cap Y).

For proving this, take x in U \cap cl(Y) and let V be an open neighbourhood of x. U\cap V is an open neighbourhood of x, and since x is in the closure of Y, U\cap V \cap Y is nonempty. Thus V cuts U\cap Y therefore we conclude.

b) Let y in Y. Take U an open neighbourhood of y such that U\cap Y is closed. From a) we conclude that U\cap cl(Y)\subset cl(U\cap Y)= U\ cap Y. Hence U\cap cl(Y)=U\ cap Y, and this means that each y in Y

has an open neighbourhood in cl(Y) included in Y, i.e, Y is open in cl(Y) - June 30th 2009, 03:28 AMEnrique2Gap
My argument in b) is not correct- CL(U\cap Y)=U\cap Y is not true

We noly know that U\cap Y is only closed in U! Anyway we have

a) U \cap cl(Y)\subset cl (U\cap Y)\cap U.

Now b) is true because U\cap cl(Y)\subset cl(U\cap Y)\cap U= U\ cap Y