Rudin's functional analysis appendix A4 (a)

Question

Quick question about the following theorem:

If $K$ is a closed subset of a complete metric space $X$ then the following three properties are equivalent:

(a) $K$ is compact

(b) Every infinite subset of $K$ ha s a limit point in $K$

(c) $K$ is totally bounded

(a) Is proven by contradiction.

Assume (a). If $E \subset K$ is infinite and no point of $K$ is a limit point of $E$, there's an open cover $\left\{ V_\alpha \right\}$ of $K$ such that each $V_\alpha$ contains at most one point of $E$.

Why is that? I was trying to get the same conclusion by using the definition limit point and compact, but this doesn't seem to lead me anywhere.

Therefore $\left\{ V_\alpha \right\}$ has no finite subcover, a contradiction.

Can you elaborate on this bit as well?

Answer 1

1) By definition of limit point. If no point is a limit point, you can find a neighborhood of each point not containing any other point of $E$. The union of all these neighborhoods cover $K$.

2) If you try to take a finite subcover, no matter how you choose them, you will only find finitely many points of E because every neighborhood you chose only contains at most one point of E. Hence you cannot cover E (and consequently K) with a finite subcover.

Answer 2

For every $x\in E$ you can choose an open set $V_x$ such that $E\cap V_x=\{x\}$.

For every $x\in K-E$ you can choose an open set $V_x$ such that $x\in V_x$ and $E\cap V_x=\varnothing$.

This because every $x\in K$ is not a limit point of $E$.

The collection $\{V_x\mid x\in K\}$ is an open cover of $K$.

Answer 3

Attempt (Credit to drhab).

Assume $E \subset K$ is infinite and no point of $K$ is a limit point.

1) Let $e \in E$. Then , since $e \in K$, $e$ is not a limit point of $E$.

There is an open $U_e$ s.t. $U_e\cap E =$ {$e$}.

2) Let $y \in K-E_y$. then since $y$ is not a limit point of $E$, there is an open $V_y$ s.t. $V_y \cap E = \emptyset$.

3) $\displaystyle{\cup_{e \in E}} U_e \displaystyle{\cup_{y \in K-E}} V_y$ is an open cover of $K$ that does not have a finite subcover(Why?).

Contradiction to $K$ compact.