# Must an uncountable subset of R have uncountably many accumulation points? [duplicate]

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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.

## marked as duplicate by Asaf Karagila♦, AlexR, user61527, egreg, M TurgeonFeb 5 '14 at 23:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• You might like this blog post. – David Mitra Feb 5 '14 at 22:05
• Your second proof assumes that the accumulation points of $S$ belong to $S$. This isn't true unless your set is closed. – EuYu Feb 5 '14 at 22:05
• There are many duplicates of this question. – Asaf Karagila Feb 5 '14 at 22:15
• @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 5 '14 at 22:45
• @Alex: See Brian's answer there. – Asaf Karagila Feb 5 '14 at 22:46

## 3 Answers

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.

• Why $\{I(x,\epsilon/2) | x\in T\}$ is disjoint? Thanks – StammeringMathematician Feb 19 at 15:27
• First before assuming that $x\in T$ you need to show that $T$ is non-empty! – SunShine Aug 8 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 8 at 18:24

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.

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 $S$.