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

user90533
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  • As it ends up: no – Ben Grossmann Aug 16 '13 at 02:30
  • @Omnomnomnom sir, is same true for $\mathbb{R}$, I mean there exists uncountable subset of $\mathbb{R}$ having exactly one limit point? According to me yes! For eg: $\mathbb{R}-{1}$ Am I correct? – Akash Patalwanshi Jun 14 '18 at 16:04
  • Sorry I think I gave wrong example, because $\mathbb{R}-{1}$ has uncountable limit points. Sir can you give me example if exists? – Akash Patalwanshi Jun 14 '18 at 16:08
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    @AkashPatalwanshi The same is true for $\Bbb R$ (assuming the axiom of choice). If there were an uncountable subset of $\Bbb R$ with exactly one limit point, then this subset could be embedded in $\Bbb C$, contradicting Asaf's result below. – Ben Grossmann Jun 14 '18 at 16:21

3 Answers3

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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.

Asaf Karagila
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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\} $

Jake
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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.

Wei
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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
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    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.) – user642796 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? – Wei Oct 28 '13 at 03:00
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    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 06:18
  • 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
  • Oh I see,thanks. I haven't ever noticed this problem before. – Wei Oct 30 '13 at 08:35
  • Should I delete my false answer? – Wei Oct 30 '13 at 08:37