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#### Mesfer Alqahtan

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- Feb 28, 2013

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

- Thread starter Mesfer Alqahtan
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- Thread starter
- #1

- Feb 28, 2013

- 3

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

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

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

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]

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