Must an uncountable subset of R have uncountably many accumulation points?

Question

This question is taken from problem 4.1.8 of "Real Analysis and Foundations" by Krantz

The question reads:

"Let S be an uncountable subset of $\mathbb{R}$. Prove that S must have infinitely many accumulation points. Must it have uncountably many?"

The first part of the question took some work but ended up coming out pretty smoothly; however, I'm at a complete loss as to how to go about addressing the second problem. Intuitively, I think the answer is yes.

My first attempt was to try to show that a countable number of accumulation points allowed one to order the elements of S in such a way that they are countable (ie prove the contrapositive), but I did not manage to get much further than that.

My second attempt was to show that $S-{s_{1},s_{2}...}$ where $s_{1},s_{2},...$ are countably many accumulation points of S is uncountable and thus must have an accumulation point, so S has an accumulation point that is not one of the countably infinite set. Hence, the number of accumulation points is uncountable.

My question is, is this logic valid? I would have to prove that an uncountable set minus a countable set is uncountable, which shouldn't be too difficult.

Any hints/points in the right direction/outright answers are greatly appreciated.

Your second proof assumes that the accumulation points of $S$ belong to $S$. This isn't true unless your set is closed. — EuYu, Feb 05 '14 at 22:05
@AsafKaragila This particular one doesn't seem like it, note that this question asks about the cardinality of acc-pts, while the linked one only asks about existence. — AlexR, Feb 05 '14 at 22:45
@AsafKaragila Oh, thanks for the hint; I only looked at the question and the accepted answer. (Close +1) — AlexR, Feb 05 '14 at 22:50
Ah, that's my bad. I even found that thread while searching for the answer to this question but didn't read beyond the accepted answer. Thank you all for the clarification! — Upside, Feb 05 '14 at 23:30

TonyK · Accepted Answer · 2019-02-19T23:11:52.687

8

Let $T$ be the set of elements of $S$ that are not accumulation points of $S$. Then for each $x \in T$ there exists $\epsilon_x > 0$ such that the interval $I(x,\epsilon_x) = (x-\epsilon_x,x+\epsilon_x)$ contains no other points of $S$. The intervals $\{I(x,\epsilon_x/2) | x\in T\}$ 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.

edited Feb 19 '19 at 23:11

answered Feb 05 '14 at 22:13

TonyK

64,559

Why ${I(x,\epsilon/2) | x\in T}$ is disjoint? Thanks – Shweta Aggrawal Feb 19 '19 at 15:27
First before assuming that $x\in T$ you need to show that $T$ is non-empty! – GraduateStudent Aug 08 '19 at 17:58
@SunShine: $T$ doesn't have to be non-empty. For instance, $S$ might be the whole of $\Bbb R$. But then there is nothing more to prove, is there? An empty set is certainly countable. – TonyK Aug 08 '19 at 18:24
I am not sure what you mean by 'each contains a rational number'. How do we know that each element of $S$ is rational? @TonyK – Jac Frall Oct 28 '19 at 00:21
1

@JacFrall: each interval $I(x,\epsilon_x/2)$ contains a rational number, because its length is strictly positive. – TonyK Oct 28 '19 at 00:56
Ok that is what I thought. I did not know if you were referencing $S$. Thank you – Jac Frall Oct 28 '19 at 00:57

score 0 · Answer 2 · answered Feb 05 '14 at 22:14

Assuming a finite number of accumulation points for $S$ would mean that the elements of $S$ that are not accumulation points are isolated, meaning $S$ is at most countable, a contradiction.

Assuming a countable number of accumulation points yields the same problem. The elements of $S$ that are not accumulation points would be isolated, and you still have that $S$ is at most a union of two countable sets which is countable, a contradiction.

score 0 · Answer 3 · answered Feb 05 '14 at 22:57

I think both TonyK's and Darrin's answers are good ones. Here's another, possibly inferior, approach.

First trust the following lemma: for any uncountable set $S$ of real numbers, there exists an interval $(a,b)$ such that both $S\cap(-\infty,a]$ and $S\cap[b,\infty)$ are uncountable. ("You can divide an uncountable set into two with a buffer interval.")

Given the lemma, you can find two uncountable subsets $S_1,S_2$ of $S$ that are separated by an interval. Iterating, you can find four uncountable subsets of $S$ all separated from one another by intervals, then 8, 16, etc. There are uncountably many branches down this repeated-bifurcation tree; each one results in an accumulation point; and the accumulation points can't coincide because of the separating intervals. Thus there are uncountably many accumulation points.

Why is the lemma true? Assume $S$ is contained in $[0,3]$ (an easy reduction), and consider the intersections of $S$ with $[0,1]$, $[1,2]$, and $[2,3]$. If the first and third intersections are uncountable, then we're done. Otherwise, we have an uncountable set in a smaller interval (perhaps length 1, perhaps length 2) and we divide that interval into three equal pieces. We keep doing this until either we find the first and last intersections uncountable (in which case we're done), or else we iterate countably often and end up with all the countable subsets contained in a sequence of nested closed intervals whose lengths tend to $0$. But this latter case can't happen, because then we will have written $S$ as a countable union of countable sets, together with the single point in the intersection of the nested intervals, contradicting the uncountability of $S$.

Must an uncountable subset of R have uncountably many accumulation points?

3 Answers3

Linked

Related