## 414/514 – Limit points

The argument in today’s lecture depended at the end on the following fact:

Suppose that $A$ is an uncountable subset of ${\mathbb R}.$ Then $A$ admits a limit point. In fact, if $A\subseteq(-r,r)$, then $A$ admits a limit point $x_0$ with ${}|x_0|

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 $A$ is unbounded, or else it has a limit point. But even if we know that $A\subseteq(-r,r)$, so it certainly is bounded, we are not quite done yet because the limit point could be $r$ or $-r.$

Recall as well that the countable union of countable sets is countable.

First, we may assume $A$ is bounded. This is because $\displaystyle A=\bigcup_n A\cap(-n,n)$,

so for some $n$ we must have $A\cap(-n,n)$ is uncountable so, even if $A$ is unbounded, we can replace it with the uncountable bounded set $A\cap(-n,n).$

Let $I_0=(-n,n)$ or, if we are given that $A\subseteq(-r,r)$ to begin with, let $I_0=(-r,r).$ So we can assume $A\subseteq I_0$. Write $I_0=(-a,a)$ so we do not have to distinguish between both cases. For $n\ge1$, let $\displaystyle I_n=\left(-a+\frac1n,a-\frac1n\right).$

Then $A=\bigcup_{n\ge1} A\cap I_n$, and it follows that for some $n\ge1$ we must have $A\cap I_n$ is uncountable.

Now, we know that any infinite bounded subset of the reals has a limit point, so $A\cap I_n$ has a limit point. This point in absolute value has size at most $\displaystyle a-\frac1n, 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.