One point compactification of $\Psi$ spaces of mad families are sequential I am trying to show that if $\mathcal A$ is a infinite mad family and $\Psi(\mathcal A)$ is it's  $\Psi$ space then the one point compactification $\Psi(\mathcal A)\cup \{\infty\}$ is sequential.
The space $\Psi(\mathcal A)$ is the set $\mathcal A \cup \omega$, where every point of $\omega$ is open and the local basis of a point $A \in \mathcal A$ consists of the sets of the form $\{A\}\cup(A \setminus n)$ for $n \in \omega$.
The space $\Psi(\mathcal A)$ is locally compact, Hausdorff, not compact.

I am trying to show that the space $\Psi(\mathcal A)\cup \{\infty\}$ is sequential.
So let $U\subset \Psi(\mathcal A)\cup \{\infty\}$ be sequentially open. We must see that $U$ is open.
Every point of $\omega \cap U$ belongs to the interior of $U$.
If $A \in \mathcal A \cap U$, then if we enumerate $A=\{s_n: n \in \omega\}$. It's easy to see that $s_n\rightarrow A$, and since $U$ is sequentially open, $s_n$ is eventually in $U$. Now it's easy to see that we have constructed a neighborhood of $A$ contained in $U$.
Finnaly, if $\infty \in A$, I don't know what to do. I think I must use the fact that every injective sequence of elements of $\mathcal A$ converges to $\infty$, but I don't know what to do.
 A: Suppose that $\infty\in U$, but $U$ does not contain an open nbhd of $\infty$. Every nbhd of $\infty$ contains all but finitely many points of $\mathscr{A}$, so one way for this to happen would be for $\mathscr{A}\setminus U$ to be infinite. But as you mentioned, in that case there is a sequence in $\mathscr{A}\setminus U$ that converges to $\infty$, contradicting the assumption that $U$ is sequentially open. Thus, $\mathscr{A}\setminus U$ must be finite. This means that $U\setminus\omega$ is a relatively open subset of $\mathscr{A}\cup\{\infty\}$. You’ve already shown that if $A\in U\cap\mathscr{A}$, then $U$ contains a nbhd of $A$. Thus, you’re done if you can prove the following proposition:

Let $U$ be a subset of $\Psi(\mathscr{A})\cup\{\infty\}$ such that $\infty\in U$, $\mathscr{A}\setminus U$ is finite, and $U\setminus A$ is finite for each $A\in\mathscr{A}$; then $U$ is a nbhd of $\infty$.

If $S=\big(\Psi(\mathscr{A})\cup\{\infty\}\big)\setminus U$, this amounts to proving that $S$ is compact in $\Psi(\mathscr{A})$. $S\cap\mathscr{A}$ is finite, so this in turn amounts to showing that if $V$ is any nbhd of $S\cap\mathscr{A}$, then $S\setminus V$ is finite. Suppose that there is a nbhd $V$ of $S\cap\mathscr{A}$ such that $S\setminus V$ is infinite, and use the fact that $\mathscr{A}$ is a mad family to get a contradiction.
