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 $\bar{Y}$.

2. Hello,

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

3. Originally Posted by Moo
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.

4. Originally Posted by Biscaim
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...

5. Originally Posted by Moo
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 $Y=O\cap \overline{Y}$.

6. 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)

7. ## Gap

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