about topological spaces

Biscaim · #1 $Old$ June 28th, 2009, 09:10 AM

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

Thanks in advance.

Moo · #2 $Old$ June 29th, 2009, 01:26 AM

Hello,

Okay, maybe I'm just rusty at topology, but what does it mean for a set Y to be open in another set, $\bar{Y}$ ?

Biscaim · #3 $Old$ June 29th, 2009, 08:39 AM

Quote:

Originally Posted by Moo $View Post$

Hello,

Okay, maybe I'm just rusty at topology, but what does it mean for a set Y to be open in another set, $\bar{Y}$ ?

The closure of any subset of a topological space is a closed subspace.

Moo · #4 $Old$ June 29th, 2009, 12:57 PM

Quote:

Originally Posted by Biscaim $View Post$

The closure of any subset of a topological space is a closed subspace.

I forgot to stress the word open.
It's correct to say that a subset is closed/open in a topological space.
But as far as I remember, I've never seen that a subset is closed/open in another subset...

flyingsquirrel · #5 $Old$ June 29th, 2009, 01:24 PM

Quote:

Originally Posted by Moo $View Post$

what does it mean for a set Y to be open in another set, $\bar{Y}$ ?

It means that $Y$ is an open subspace of the topological space $\overline{Y}$ , the topology on $\overline{Y}$ being the topology induced by $X$ on $\overline{Y}$ . In other words, $Y$ is open in $\overline{Y}$ if and only if there exists a subset $O$ of $X$ which is open in $X$ and such that $Y=O\cap \overline{Y}$ .

Enrique2 · #6 $Old$ June 30th, 2009, 02:18 AM

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)

Enrique2 · #7 $Old$ June 30th, 2009, 05:28 AM

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