The argument in today’s lecture depended at the end on the following fact:
Suppose that is an uncountable subset of Then admits a limit point. In fact, if , then admits a limit point with
The proof should be standard by now but, for completeness, here is one way of seeing this:
Before we begin, recall that we already know that a bounded infinite set has a limit point. So, either is unbounded, or else it has a limit point. But even if we know that , so it certainly is bounded, we are not quite done yet because the limit point could be or
Recall as well that the countable union of countable sets is countable.
First, we may assume is bounded. This is because
so for some we must have is uncountable so, even if is unbounded, we can replace it with the uncountable bounded set
Let or, if we are given that to begin with, let So we can assume . Write so we do not have to distinguish between both cases. For , let
Then , and it follows that for some we must have is uncountable.
Now, we know that any infinite bounded subset of the reals has a limit point, so has a limit point. This point in absolute value has size at most , but this is what we needed.
In fact, one can extend this argument to see that any uncountable set has an uncountable set of limit points, but we did not need this for today’s argument.
Added: You may find interesting this recent question in MathOverflow.