§ Limit point compactness from Munkres
Munkres calls "Bolzano Weirstrass" as limit point compactness. He defines a
space to be limit point compact if every infinite subset of has a
§ Compact implies limit compact
We will prove the contrapositive. That is, let be a compact set.
If , does not have any limit point, then is finite.
If does not have any limit point, then vacuously contains all of its limit
points. Thus, is closed. Since is a closed subset of a compact set , itself
is compact. Next, see that for each , we can find an open such that .
If we can't find such a , then it means that is a limit point!
(Since all nbhd of intersect non-trivially). Clearly, these "isolating" cover .
Since is compact, we have a finite subcover . See that .
To show this, since the cover , we have . Hence,
, which is equal to which is .
Hence, has finitely many points, exactly the .
§ Classical Proof Using Bisection
Let's prove this in .
Let be a compact set containing an infinite number of points. We know from
Heine Borel that is closed and bounded. Let the interval containing be .
Bisect the interval into two sub-intervals: and for .
One of these must contain an infinite number of points (suppose both contain a finite number of points, then
itself must contain a finite number of points, contradiction). We can thus recurse, setting to be the sub-interval
that has an infinite number of points. This gives us a nested sequence of intevals .
The interval is closed as it is the intersection of closed intervals. Also, has length zero
since we bisect the interval each time. Hence, is a single point, ie . We claim that is an accumulation
point of the original subsequence. Any open set around will contain some interval