I need to prove this small fact to understand the proof of a theorem:
Suppose $X, Y$ are countable discrete subespaces of $\beta \omega$, $Y \subset X$, and let $p \in \beta \omega \setminus \omega$ such that $p \in(\bar X^{\beta\omega}\setminus X)\cap(\bar Y^{\beta\omega}\setminus Y)$.
Then there exists an homeomorphism $h: \bar Y^{\beta \omega}\rightarrow \bar X^{\beta \omega}$ such that $h(p)=p$.
Maybe I don't need all these hypothesis to prove what I need to, since I have copied and pasted all the hypothesis of the theorem I am working on.
Notice that since $X, Y \subset \beta \omega$ are countable and discrete, their closures are homeomorph to $\beta \omega$. Maybe this helps.
Since $\operatorname{cl}_{\beta\omega}Y\subseteq\operatorname{cl}_{\beta\omega}X$, you might as well work in the subspace $\operatorname{cl}_{\beta\omega}X$. In that case you can assume without loss of generality that $X=\omega$. Fix $p\in(\operatorname{cl}_{\beta Y}Y)\setminus Y$; if we view $p$ as an ultrafilter on $\omega$, then $Y\in p$, and there is an $A\in p$ such that $A\subseteq Y$ and $|Y\setminus A|=\omega$. Let $h:Y\to\omega$ be any bijection such that $h(n)=n$ for all $n\in A$.
For any $S\subseteq\omega$ we have $h^{-1}[S]\in p$ iff $A\cap h^{-1}[S]\in p$, and $A\cap h^{-1}[S]=h^{-1}[A\cap S]=A\cap S$, so $h^{-1}[S]\in p$ iff $A\cap S\in p$. Finally, $A\in p$, so $A\cap S\in p$ iff $S\in p$. Thus, $h^{-1}[S]\in p$ iff $S\in p$, and $\hat h(p)=p$, where $\hat h$ is the extension of $h$ to $\operatorname{cl}_{\beta\omega}Y$.