# Difficulty in understanding the proof : S must have infinitely many accumulation points. [duplicate]

"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.

## marked as duplicate by lulu, Xander Henderson, Cesareo, Leucippus, stressed outFeb 20 at 0:42

• Why not post this question to the user who wrote the original post? – lulu Feb 19 at 18:05
• 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 at 18:11
• @lulu: StammeringMathematician did indeed leave a question in a comment under my answer. – TonyK Feb 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 at 23:14
• 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 at 7:51

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).
• 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. – StammeringMathematician Feb 19 at 23:00
• @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 at 23:05