0

"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.

Shweta Aggrawal
  • 5,501
  • 2
  • 15
  • 48
  • Why not post this question to the user who wrote the original post? – lulu Feb 19 '19 at 18:05
  • 1
    In any case, those intervals are clearly disjoint. if $y\in I(x_1,\epsilon_1/2)\cap I(x_2,\epsilon_2/2)$ with $x_1<x_2$ then $|x_2-x_1|<(\epsilon_1+\epsilon_2)/2$ which contradicts the construction of the $\epsilon_i's$. – lulu Feb 19 '19 at 18:11
  • @lulu: StammeringMathematician did indeed leave a question in a comment under my answer. – TonyK Feb 19 '19 at 23:09
  • My answer was a little imprecise, in that I didn't make it clear that $\epsilon$ depends on $x$. I have edited my answer accordingly. And yes, I realise that this doesn't address your concerns directly, but I thought it might help. – TonyK Feb 19 '19 at 23:14
  • 1
    A collection $C$ of non-empty pair-wise disjoint open subsets of $ \Bbb R$ is countable. Because $\Bbb R$ has a countable dense subset $S={s_n:N\in \Bbb N}.$ ( E.g. $S=\Bbb Q$). So for each $c\in C$ choose $f(c)\in c\cap S.$ Then $T={f(c):c\in C}$ is a subset of the countable set $S,$ so $T$ is countable. And $f:C\to T$ is a bijection so $C$ is also countable – DanielWainfleet Feb 20 '19 at 07:51
  • @DanielWainfleet Thanks for the comment. It helped. – Shweta Aggrawal Feb 20 '19 at 13:27

1 Answers1

3

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).

Henno Brandsma
  • 242,131
  • Lulu mentioned in comment "those intervals are clearly disjoint. if $y∈I(x_1,ϵ_1/2)∩I(x_2,ϵ_2/2)$ with $x_1<x_2 $then $|x_2−x_1|<(ϵ_1+ϵ_2)/2$ which contradicts the construction of the $ϵ_i$'s." This argument looks right to me. Just want to double check. Can you please have a look. Thanks for the proof. I like the approach. – Shweta Aggrawal Feb 19 '19 at 23:00
  • Thanks for putting in details. It will really help a person like me who is self learning. I hope I would be able to use the ideas later at some other places. Have a great day! – Shweta Aggrawal Feb 19 '19 at 23:05
  • 1
    @StammeringMathematician If $\varepsilon_1 \le \varepsilon_2$, say, we'd conclude that $|x_2 -x_1| < \varepsilon_2$ and so $x_1 \in (x_2 - \varepsilon_2, x_2+\varepsilon_2)$. The factor two makes a difference (we cannot do that in a general space). – Henno Brandsma Feb 19 '19 at 23:05