§ Limit point compactness from Munkres

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

§ Compact implies limit compact

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

§ Classical Proof Using Bisection

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