# countable base of nhoods in $\beta X$

If $p\in X$ has a countable base of nhoods of $X$, it has a countable base of nhoods in stone -cech compactification $\beta X$. To show that could you give me a hint please?

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Theorem. Let $X$ be a dense subspace of a regular space $Y$, let $x\in X$, and let $\ms{B}$ be a local base at $x$ in $X$. Then $\ms B\,'=\{\int_Y\cl_YB:B\in\ms B\}$ is a local base at $x$ in $Y$.
Certainly the members of $\ms B\,'$ are open neighborhoods of $x$ in $Y$. Let $U$ be any open nbhd of $x$ in $Y$; to prove the theorem we must show that there is a $B\in\ms B$ such that $\int_Y\cl_YB\subseteq U$. $Y$ is regular, so there is an open set $V$ in $Y$ such that $x\in V\subseteq\cl_YV\subseteq U$. Now $V\cap X$ is an open nbhd of $x$ in $X$, so there is a $B\in\ms B$ such that $x\in B\subseteq V\cap X$.
Can you finish the proof by showing that $x\in\int_Y\cl_YB\subseteq U$?
sorry maybe very easy, but why $x \in int_Y B$. $x\in B$ but whyit also contains by the interior in $Y$. –  ege Nov 8 '12 at 10:11
Then, since X is dense {cl_{Y}} V ={cl_{Y}} V\cup X$, then the other part clear. – ege Nov 8 '12 at 10:14 @ege: More generally, if$G$is open in$X$and$x\in G$, then$x\in\int_Y\cl_YG$. There is an open$U$in$Y$such that$G=U\cap X$.$X$is dense in$Y$, so$\cl_YG=\cl_YU$. Thus,$x\in U\subseteq\int_Y\cl_YU=\int_Y\cl_YG$. – Brian M. Scott Nov 8 '12 at 17:52 I agree with you, but I have to show that$x\ in B$, the other part is clear by density.$B$is not open in$Y$?? – ege Nov 9 '12 at 5:34 @ege: There’s nothing to show:$B$is an element of$\mathscr{B}$, which is a local base at$x$, so of course$x\in B\$. –  Brian M. Scott Nov 9 '12 at 14:46