How might I resolve a failure in my own attempt at showing that Alexandroff's extension is compact? Finite intersection property seems to fail

Question

$\newcommand{\F}{\mathcal{F}}\newcommand{\O}{\mathcal{O}}\newcommand{\id}{\operatorname{Id}}$Let $X$ be a nonempty topological space. Let $\omega$ be an arbitrary object, taken to be not in $X$, and let $X^\ast=X\sqcup\{\omega\}$.

The topology on $X^\ast$ is generated by the open sets of $X$ and sets of the form: $$(X\setminus K)\sqcup\{\omega\}:K\subseteq X\text{ is closed and compact}$$

I will call these "type-$2$ open sets" and the former the "type-$1$ open sets".

In order to show $X^\ast$ is compact, it seemed ideal to apply the finite intersection test. Let $\F$ be a family of closed sets in $X^\ast$ with the finite intersection property. I may observe that $\F$ will contain a subcollection $\F_1$ of complements of open sets of the first type and a subcollection $\F_2$ of complements of open sets of the second type. I let $\id:X\to X^\ast$ be the identity embedding.

Examining $\F_1$:

Any set in $\F_1$ is of the form $F=(X\sqcup\{\omega\})\setminus\O=\id(F')\sqcup\{\omega\},\,\O\subseteq X$ open in $X$, $F'\subseteq X$ closed in $X$. We have $\bigcap_{F\in\F_1}F\supseteq\{\omega\}\neq\emptyset$.

Examining $\F_2$:

Any set in $\F_2$ is of the form $F=\id(K)\subseteq X$, $K$ compact. Since all the $\{K\}$ will have the finite intersection property, and they are compact and closed, their intersection is a nonempty compact subset of $X$.

So, the intersections of $\F_1,\F_2$ are nonempty, but the intersection of $\F$ will be the intersections of these intersections, and I note that the intersection of $\F_2$ will not contain $\omega$ and the intersection of $\F_1$ may well only contain $\omega$. Then in general I cannot say that the intersection is nonempty...

What am I doing wrong? I am aware I can find proofs online, I just want to know what went wrong with my proof. It feels very trivial... probably my decomposition $\F=\F_1\sqcup\F_2$ is problematic.

Answer 1

Let $\mathcal F$ be a family of closed sets in $X^∗$ with the finite intersection property. There are two possibilities:

• $\omega$ belongs to each element of $\mathcal F$: then, clearly, the intersection of all elements of $\mathcal F$ is non-empty ($\omega$ belongs to it).
• $\omega\notin F$, for some $F\in\mathcal F$. Then, $X^*\setminus F$ is an open set to which $\omega$ belongs, and therefore $F$ is compact. And $\bigcap_{C\in\mathcal F}C=\bigcap_{C\in\mathcal F}(C\cap F)$, since $F\in\mathcal F$. But $\mathcal F$ has the finite intersection property and $F\in\mathcal F$; therefore $\{C\cap F\mid C\in\mathcal F\}$ has the finite intersection property. But $F$ is compact. So, $\bigcap_{C\in\mathcal F}(C\cap F)\ne\emptyset$.

Answer 2

Just open covers also works: of $\mathcal{U}$ is an open cover of $X^\ast$ there must be some $U_0\in \mathcal{U}$ with $\omega \in U_0$, So this must be a "type 2" set of the form $U_0=(X\setminus K) \sqcup \{\omega\}$ with $K$ compact. This $K$ is covered by $\mathcal{U}$ too and so finitely many form a subcover for $K$, say $U_1, \ldots, U_n \in \mathcal{U}$. Together with $U_0$ we then have a finite sucover of $\mathcal{U}$, QED.