Nets, Cluster points, and closure: Showing a space is compact if every net has a cluster point

Question

I'm trying to understand the following proof that a space $X$ is compact if and only if every net has a cluster point. I have a specific confusion with how cluster points relate to closure which is expanded on after the proof.

A cluster point is defined as

Here is the proof of the implication that if every net has a cluster point then $X$ is compact.

How does closure relate to cluster points?

I do not understand why $x$ must be in $\overline{X \backslash U_\alpha}$. As far as I understand the closure of a space is the space with all its limit points. But a cluster point need not be a limit point? Is there some other reason $x$ must be in the closure?

Answer 1

Proposition: If $(x_\alpha)_{\alpha \in A}$ is a net in $X$ with cluster point $y$, then $y$ is in $\overline{ \{ x_\alpha | \alpha \in A\}}$.

Proof: Suppose $y \not \in \{ x_\alpha | \alpha \in A\}$. We will then show it is a limit point. Let $U$ an open set containing $y$. Then since $y$ is a cluster point there exists some $\alpha \in A$ s.t. $x_\alpha \in U$. So $\{ x_\alpha | \alpha \in A\} \cap U = \{ x_\alpha | \alpha \in A\} \cap U \backslash\{y\} \neq \emptyset$. So $y$ is a limit point.

Answer 2

Munkres uses that $x \in \overline{A}$ iff every open set that contains $x$ intersects $A$. This is Theorem 17.5(a) in his 2nd edition.

He shows (using the definition of cluster point of a net) that for any $\alpha$ , and any open set $U$ containing $x$, $U$ intersects $X\setminus U_\alpha$. That last part says (by the quoted theorem!) that $x \in \overline{X\setminus U_\alpha}$, and as $U_\alpha$ is open, $X\setminus U_\alpha$ is closed, so in fact $\overline{X\setminus U_\alpha} = X\setminus U_\alpha$. Ergo, $x \in X\setminus U_\alpha$, or $x \notin U_\alpha$. But this was for an arbitrary $\alpha$!. So this contradicts that we have a cover in $\{U_\alpha, \alpha \in \Lambda\}$ ($x$ isn't covered), and that contradiction finishes the proof.