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?
countable base of nhoods in $\beta X$
1 Answers
$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}\newcommand{\ms}{\mathscr}$You don’t have to deal with $\beta X$, which can be rather complicated; this is just a special case of a more general result that has nothing to do with $\beta X$.
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 $x\in\int_Y\cl_YB\subseteq U$?
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0@ege: You’re welcome. By the way, $\{X\cap\int_Y\cl_YB:B\in\mathscr{B}\}$ is also a local base at $x$, since $Y$ is regular, and if you use it instead of $\mathscr{B}$, then you automatically get $x$ to be in the $Y$-interior of each member of the local base. But you don’t need this for the argument. – 2012-11-09