"Let S be an uncountable subset of R. Prove that S must have infinitely many accumulation points. Must it have uncountably many?"
This question was already asked on this website before. I have some difficulty in understanding the proof. Here is the accepted answer:
Let $T$ be the set of elements of $S$ that are not accumulation points of $S$. Then for each $x \in T$ there exists $\epsilon > 0$ such that the interval $I(x,\epsilon) = (x-\epsilon,x+\epsilon)$ contains no other points of $S$. The intervals $\{I(x,\epsilon/2) | x\in T\}$ are disjoint, and each contains a rational number; so there can only be a countable number of them.
Hence $T$ is countable, and the set of accumulation points of $S$, which contains $S-T$, is uncountable.
Must an uncountable subset of R have uncountably many accumulation points?
I can not understand why:
The intervals $\{I(x,\epsilon/2) | x\in T\}$ are disjoint.
Second question: Each $I(x,\epsilon)$ contains a rational number due to the denseness of $\mathbb{Q}$ in $\mathbb{R}$ but why does it imply that there is only countable number of them.
Thanks a lot.
I don't see why the intervals should be disjoint. But there is a better proof: $\mathbb{R}$ has a countable base $B_n, n \in \mathbb{N}$, namely all open intervals with rational endpoints (we can enumerate $\mathbb{Q} \times \mathbb{Q}$ using the natural numbers), and for every open set $O$ of the reals and every $x \in O$ we can find some $n$ such that $x \in B_n \subseteq O$.
Now for every $x \in T$ we have thus some $n(x)$ such that $x \in B_{n(x)}$ and $B_{n(x)} \cap S = \{x\}$ (the intersection with $S$ contains at most $x$, but contains $x$ as $T \subseteq S$).
Then the function $f: T \to \mathbb{N}: x \to n(x)$ is injective: suppose for $x,y \in T$ we have $n(x) = n(y)$, set $B=B_{n(x)}=B_{n(y)}$ then $B \cap S = \{x\}$ while also also $B \cap S = \{y\}$, by the defining conditions on $B_{n(x)}$, resp. $B_{n(y)}$. This clearly implies that $x=y$ and so $f$ is injective and a set $T$ mapping injectively into $\mathbb{N}$ is at most countable.
So $S\setminus T$ is uncountable and by definition consists of accumulation ponts of $S$ (rather, limit points).