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#### Mesfer Alqahtan

##### New member
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

Staff member
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#### Amer

##### Active member
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 $$x \ne y$$ there exist two disjoint open sets U,V such that $$x \in U, y \in V$$. which means y is not in U clusure. thats hold for every y in X

Or you can provce it by contradiction
suppose that $$\cap\{ \bar{U} : U\; open\; and \; x\in U \}$$ 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
$$\cap\{ \bar{U} : U\; open\; and \; x\in U \}\subseteq \bar{G}$$ so y is not in the intersection.

The other direction

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

Let x,y in X. y is not in the $$\cap \{ \bar{U} : \; U\; open \; and \; x\in U \}$$
which means y is not in the clusure of some open sets containing x say
$$y \notin \bar{U}^*$$
note that $$x \notin \bar{A}$$ 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
$$y\in V , \; and \; V \cap U^* = \phi$$

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