# Uncountable set with exactly one limit point

Is there any uncountable subset of $\mathbb{C}$ with exactly one limit point?

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As it ends up: no –  Omnomnomnom Aug 16 '13 at 2:30

Assuming the axiom of choice, the answer is no. Without the axiom of choice, the question is still open to my knowledge.

Suppose that $A$ is an uncountable subset of $\Bbb C$ and $a$ is its only accumulation point. Consider $D_n=\{x\in A\mid |x-a|\geq\frac1n\}$. If there is some $D_n$ which is uncountable, then we can find an accumulation point of $D_n$ which is not $a$, and therefore $A$ has two accumulation points. Therefore $D_n$ is countable for every $n\in\Bbb N$. However $A\setminus\{a\}$ is the union of all the $D_n$'s, and therefore a countable union of countable sets.

All that is left is to show that indeed every uncountable set has at least one accumulation point. Suppose that $A$ is an uncountable set. Let $D$ be a countable dense subset of $\Bbb C$. Consider the sets $B(d)=\{x\in\Bbb C\mid |x-d|\leq\frac12\}$, then $\Bbb C=\bigcup_{d\in D}B(d)$. By the uncountability of $A$ we have that for some $d\in D$, $A'=B(d)\cap A$ is uncountable (otherwise it's the countable union of countable sets). In particular $A'$ is an infinite subset of a compact set, and thus has an accumulation point $a$. It's not hard to show that $a$ is an accumulation point of $A$.

The axiom of choice was used when we said that the countable union of countable sets is countable. Whether or not this is true when the axiom of choice fails is still open, to the best of my knowledge.

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Axiom of countable choice? –  Pedro Tamaroff Aug 16 '13 at 2:33
@Peter: That would be sufficient. In fact even "countable union of countable sets of real numbers is countable" should suffice. It's easier to just say "axiom of choice", though. –  Asaf Karagila Aug 16 '13 at 2:34
OK. ${}{}{}{}{}$ –  Pedro Tamaroff Aug 16 '13 at 2:34
@AsafKaragila: Is countable union of countable sets of real numbers is countable weaker than countable choice? –  Michael Albanese Aug 16 '13 at 4:09
@Michael: Much weaker. –  Asaf Karagila Aug 16 '13 at 4:12

No, any uncountable subset of $\mathbb{C}$ must have more than one limit point. Here is a relatively straightforward argument to see that such a set must have at least two limit points. I'm fairly certain that one actually gets infinitely many limit points.

Suppose that $E \subseteq \mathbb{C}$ is uncountable.

First, you can see that there is some $R > 0$ so that $E \cap B(0,R)$ is uncountable. (If not, then $E$ is the countable union of the countable sets $E \cap B(0,k), k \in \mathbb{N}$, contradicting uncountability of $E$).

Because of this, you can see that $E$ must have a limit point in $\overline{B(0,R)}$, say $L$. (This uses compactness of $\overline{B(0,R)}$ )

Now choose an open set $U$ containing $L$ sufficiently small so that $(E \cap B(0,R)) \setminus U$ is uncountable (by using uncountability of $E \cap B(0,R)$).

We then see that there is another limit point of $E$ somewhere in $\overline{B(0,R) \setminus U}$ which must necessarily be distinct from $L$.

EDIT: I should have stated that $B(z_0,R) = \{z\in \mathbb{C} : |z-z_0| < R\}$

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I have an improvement for @Asaf Karagila's answer. Since the rational numbers in [0,1] is countable. All the rational numbers is also countable,which shows that countable unions of countable sets is still countable. Or we can arrange all of the points in a infinite large matrix,then we can easily line the up.I don't think axiom of choice is necessary.// By the way, as in @Asaf Karagila's proof, we can show that there is at least countable limit points,since 1 is the same as finite in this case.

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No, this is false. It is consistent without the axiom of choice that the real numbers themselves are a countable union of countable sets. –  Asaf Karagila Oct 27 '13 at 13:47
Just because some countable union of countable sets can be shown to be countable (without Choice) does not mean that all countable unions of countable sets are countable. (To do so usually requires choosing a surjection from the natural numbers for each of the countable sets.) –  Arthur Fischer Oct 27 '13 at 13:48
Since the countable sets can be lined up, suppose $U_i=\{a_{ij}\}_{j=1}^\infty$, then we can easily line all $U_i$ up by the following:$a_{11},a_{12},a_{21},a_{13},a_{22},a_{31},a_{14},a_{23},a_{32},a_{41}‌​,...,a_{1n},a_{2,n-1},...,a_{n1},...$ //If you want to find any element in $U_i$ for any $i$, I can tell you the precise position of it, am I done?What's the problems in it? –  user103567 Oct 28 '13 at 3:00
There is a difference between a countable set and a countably enumerated set (one with a particular indexing assigned to it). It is true that a countable union of countably enumerated sets is countable, and the proof is basically done in exactly the way you describe. It is also true that every countable set can be countably enumerated. However, without choice, we can only countably enumerate a finite number of countable sets, because we need to choose an enumeration of each set. –  Cameron Buie Oct 28 '13 at 6:18
If $A$ is a countably infinite set then there is a continuum of different enumerations that enumerate it. Since you have infinitely many sets, how do you choose the enumerations? –  Asaf Karagila Oct 28 '13 at 15:29