As the title says, I'm having trouble understanding a part from the following proof.
Proposition says;
$X$ is a compact metric space if and only if every infinite subset $A\subset X$ have an accumulation point in $X$.
Proof:
For the converse, suppose $X$ is not compact. Let $O$ be an open cover of $X$ of minimal cardinality [such that no finite subcover exists], and let $\{O_\alpha\}_{\alpha<|O|}$ be an enumeration of $O$. Construct a set $A$ as follows:
- let $x_\alpha\in X$ be any element not covered by any of the $O_\beta$ for $\beta<\alpha$.
- let $A=\{x_\alpha:\alpha<|O|\}$. We'll show that $A$ has no complete limit point.Indeed, for any $x\in X$ there is some $\alpha$ such that $O_\alpha$ is a neighbourhood of $x$,but $O_\alpha$ avoids all the $x_\beta$ for $\beta>\alpha$, and so: $$|A\cap O_\alpha|\leq|\{x_\gamma:\gamma\leq\alpha\}|=|\alpha|<|O|=|A|.$$
I'm having trouble understanding the last part of the proof (after the bold part), i understand that we want to show that $A$ has no limit points, but i get lost and don't see were we show that. Thanks in advance for any help.
-I got the proof from here.
Let's suppose $a$ is an accumulation point of $A$.
Then $a$ belongs to some $O_{\alpha}$ since $a\in X$. Thus there is an $\epsilon>0$ such that the ball $B(a,\epsilon)$ centered in $a$ of radius $\epsilon$ is contained in $O_{\alpha}$.
Also, as $a$ is an accumulation point there must be infinitely many elements of $A$ in $B(a,\epsilon)$. But that is impossible, since by the construction of $A$ no $x_{\beta}$ belongs to $A$ if $\alpha < \beta$.
Therefore our supposition was wrong: $A$ has no accumulation points.
The inequalities that follow are rather obscure and can be safely ignored, but to my understanding they are highlighting that the intersection between $A$ and $O_{\alpha}$ is at most finite.