Advanced topology

[Mesfer Alqahtan]

Prove that X is T2 iff for each x belong to X, then the intersection{CL(U): U is a nhd of x }={x}.

 

[MarkFL]

If you post a problem with no work or thoughts, then our helpers don't know where you are stuck, and it may take additional time for you to get meaningful help if they give you a hint or suggestion that you have already applied or know.

This way their valuable time is utilized in the best way.

 

[Amer]

Prove that X is T2 iff for each x belong to X, then the intersection{CL(U): U is a nhd of x }={x}.

Let X be a T2 space, Let x in X fixed
Let y be arbitrary point in X and different from x want to show that there exist an open set containing x such that y is not in it is clusure
Solution: since [tex] x \ne y [/tex] there exist two disjoint open sets U,V such that [tex]x \in U, y \in V[/tex]. which means y is not in U clusure. thats hold for every y in X

Or you can provce it by contradiction
suppose that [tex]\cap\{ \bar{U} : U\; open\; and \; x\in U \} [/tex] contain another point different from x say y , there exist G,H disjoint open sets and x in G , y in H this gives that y is not in Cl(G) but
[tex]\cap\{ \bar{U} : U\; open\; and \; x\in U \}\subseteq \bar{G}[/tex] so y is not in the intersection.

The other direction

Let X be a space such that for every x in X , [tex] \{ \cap \bar{U} : U\; open \; and \; x\in U\} = \{x\} [/tex]

Let x,y in X. y is not in the [tex] \cap \{ \bar{U} : \; U\; open \; and \; x\in U \}[/tex]
which means y is not in the clusure of some open sets containing x say
[tex] y \notin \bar{U}^* [/tex]
note that [tex] x \notin \bar{A} [/tex] there exist and open set containing x and the intersection between A and the open set is phi using this, give us an open set V such that
[tex] y\in V , \; and \; V \cap U^* = \phi [/tex]