open subsets of boundary [closed]

Question

Let $\bar{X}$ be a Hausdorff and compact topological space. Suppose that $X$ is an open and dense subset of $\bar{X}$. Let $\nu X=\bar{X}\setminus X$ and assume that $U\subseteq \nu X$ is an open subset of $\nu X$ with respect to the subspace topology. Is there any open subset of $W\subseteq \bar{X}$ such that $W\bigcap \nu X=U$ and $\bar{W}\bigcap \nu X=\bar{U}$?

Answer 1

Not in general. Let $\overline X=(\omega_1+1)\times(\omega+1)$ where the ordinals are given their usual order topology, $X=\omega_1\times\omega$, and $U=\omega_1\times\{\omega\}$. If $W\subseteq\overline X$ is any open set that includes $U$, then for every $\alpha<\omega_1$, there is $n_\alpha<\omega$ such that $\{\alpha\}\times[n_\alpha,\omega]\subseteq W$. Since $\mathrm{cf}(\omega_1)>\omega$, there is $n<\omega$ such that $\{\alpha:n=n_\alpha\}$ is cofinal in $\omega_1$. Thus, $\overline W\supseteq\{\omega_1\}\times[n,\omega]$. On the other hand, $\overline U=(\omega_1+1)\times\{\omega\}$, so $\langle\omega_1,n\rangle$ is an element of $\overline W\cap\nu X$ outside $\overline U$.

On the positive side, the property holds whenever $\overline X$ is metrizable, or more generally, a completely normal space (not necessarily compact or Hausdorff): if we put $V=\nu X\smallsetminus\overline U$, then $\overline U\cap V=U\cap\overline V=\varnothing$, hence using complete normality, there are disjoint open sets $W\supseteq U$ and $Z\supseteq V$. By shrinking $W$ if necessary, we may assume $W\cap\nu X=U$. Moreover, $\overline U\subseteq\overline W\cap\nu X$ as $\nu X$ is closed, and $\overline W\cap\nu X\subseteq\overline U$ as $\overline W$ is disjoint from $Z$.