Let $X$ be a locally compact ,Hausdorff space and $\{U_n\}$ be a sequence of dense open sets in $X$ .Show that $\cap_nU_n$ is dense in $X$.

Question

How can I prove that if $X$ be a locally compact , Hausdorff space and $\{U_n\}$ be a sequence of dense open sets in $X$ then $\cap_nU_n$ is dense in $X$. Let, X be a Hausdorff space , X a locally compact dense subspace of Y then show that X is open in Y.

Answer 1

Let $\,U\ne\emptyset\,$ be open in $\,X.\ $ We want to prove that

$$ U\cap\bigcap_{n=1}^\infty U_n\,\ne\emptyset $$

By local compactness, there is compact subspace $\,W\subseteq U\,$ which has non-empty interior $\,W_0\,$ in $\,X.\, $ Then the closure $\,V\,$ of $\,W_0\,$ is a non-empty compact contained in $\,W\subseteq U.\ $ It's elementary that

$$ \forall_{n=1}^\infty\quad V_n\,:=\,V\cap U_n $$

is dense and open in $\,V.$

Since compact spaces have the Baire property, we obtain

$$ \emptyset\,\ne\,\bigcap_{n=1}^\infty V_n \,\subseteq\,U\cap\bigcap_{n=1}^\infty U_n $$

This proves the theorem.  Great!