## § Limit point compactness from Munkres

Munkres calls "Bolzano Weirstrass" as limit point compactness. He defines a space $X$ to be limit point compact if every infinite subset of $X$ has a limit point.

#### § Compact implies limit compact

We will prove the contrapositive. That is, let $X$ be a compact set. If $A \subseteq X$, does not have any limit point, then $A$ is finite. If $A$ does not have any limit point, then $A$ vacuously contains all of its limit points. Thus, $A$ is closed. Since $A$ is a closed subset of a compact set $X$, $A$ itself is compact. Next, see that for each $a \in A$, we can find an open $U_a$ such that $U_a \cap A = \{ a \}$. If we can't find such a $U_a$, then it means that $a$ is a limit point! (Since all nbhd of $a$ intersect $A$ non-trivially). Clearly, these "isolating" $U_a$ cover $A$. Since $A$ is compact, we have a finite subcover $U_{a_i}$. See that $A = \{ a_i \}$. To show this, since the $U_i$ cover $A$, we have $A \subseteq \cup_i U_{a_i}$. Hence, $A = (\cup_i U_{a_i}) \cap A$, which is equal to $\cup (U_{a_i} \cap A)$ which is $\cup_i a_i$. Hence, $A$ has finitely many points, exactly the $a_i$.

#### § Classical Proof Using Bisection

Let's prove this in $\mathbb R$. Let $C$ be a compact set containing an infinite number of points. We know from Heine Borel that $C$ is closed and bounded. Let the interval containing $C$ be $I[0] \equiv [l, r]$. Bisect the interval into two sub-intervals: $J[0][0] \equiv [l, m]$ and $J[0][1] \equiv [m, r]$ for $m = (l+r)/2$. One of these must contain an infinite number of points (suppose both contain a finite number of points, then $I[0]$ itself must contain a finite number of points, contradiction). We can thus recurse, setting $I[1]$ to be the sub-interval that has an infinite number of points. This gives us a nested sequence of intevals $I[0] \supset I[1] \supset \dots$. The interval $J \equiv \cap_i I[i]$ is closed as it is the intersection of closed intervals. Also, $J$ has length zero since we bisect the interval each time. Hence, $J$ is a single point, ie $J = \{ j \}$. We claim that $j$ is an accumulation point of the original subsequence. Any open set around $O$ will contain some interval $I[o]$