Uncountable set with exactly one limit point Is there any uncountable subset of $\mathbb{C}$ with exactly one limit point?
 A: 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.
A: 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\} $
A: 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.
