# Show that $(X,T)$ is $T_2$

Can someone please help me with this problem?

Assume that $(X, T)$ is a topological space and that $B$ is a basis for the topology $T$. Show that $(X, T)$ is $T_2$ if and only if for all $x$ and $y$ in $X$, there are elements $B_1$ and $B_2$ of $B$, such that $B_1 \cap B_2 = \varnothing$, $x \in B_1$, and $y \in B_2$.

thanks

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Is this a homework? Please add the tag [homework] if this is the case. –  Vadim Apr 23 '12 at 18:08

## 1 Answer

Suppose $x_1 \neq x_2$ and that $(X,T)$ is $T_2$. Then there are disjoint open sets $U_1 \ni x_1$ and $U_2 \ni x_2$. Now by definition of a basis, there are $B_1$ and $B_2$ from the basis such that $x_1 \in B_1 \subset U_1$ and $x_2 \in B_2 \subset U_2$. Then $B_1 \cap B_2 \subset U_1 \cap U_2 = \emptyset$, as desired.

The other direction is a tautology, as elements of the basis are open.

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thank you so much –  user24874 Apr 23 '12 at 18:25
@user24874 Btw, you can accept and upvote answers like this : ) –  Matt N. Apr 23 '12 at 18:28