Let Y be a subset of a topological space X. If any point of Y admits an open neighbourhood U in X such that $\displaystyle U\cap{Y}$ is closed in U then Y is open in $\displaystyle \bar{Y}$.

Thanks in advance.

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- Jun 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 $\displaystyle U\cap{Y}$ is closed in U then Y is open in $\displaystyle \bar{Y}$.

Thanks in advance. - Jun 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, $\displaystyle \bar{Y}$ ? - Jun 29th 2009, 06:39 AMBiscaim
- Jun 29th 2009, 10:57 AMMoo
- Jun 29th 2009, 11:24 AMflyingsquirrel
It means that $\displaystyle Y$ is an open subspace of the topological space $\displaystyle \overline{Y}$, the topology on $\displaystyle \overline{Y}$ being the topology induced by $\displaystyle X$ on $\displaystyle \overline{Y}$. In other words, $\displaystyle Y$ is open in $\displaystyle \overline{Y}$ if and only if there exists a subset $\displaystyle O$ of $\displaystyle X$ which is open in $\displaystyle X$ and such that $\displaystyle Y=O\cap \overline{Y}$.

- Jun 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) - Jun 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