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If $\Omega \in \mathbb{C}$ is open, show that any of its subsets without limit points is at most countable.

I have seen answers here on how to prove that no limit points implies at most countable: Need a hint: show that a subset $E \subset \mathbb{R}$ with no limit points is at most countable.

But, I was wondering if the following argument holds: Assume a set A is uncountable and has no limit points. Then, we can find an $r>0$ and $x \in A$ such that $N_{r}(x)$ contains infinitely many elements of $A$ because $A$ is uncountable [EDIT: I now realize this only works if $\Omega = \mathbb{C}$]. But then, the closure of this region is compact, so the set does have a limit point.

Jon Hillery
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Your argument :

" Assume a set $A$ is uncountable and has no limit points. Then, we can find an $r>0$ and $x∈A$ such that $N_r(x)$ contains infinitely many elements of $ A $ because $A$ is uncountable. But then, the closure of this region is compact, so the set does have a limit point."

Does not work, because the set $A$ in uncountable and there is no need for some point $x$ in $A$ to have infinitely many elements in $N_r(x)$ to make it uncountable.

Mohammad Riazi-Kermani
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The problem here is that you don't necessarily have $N_r(x) \subseteq A$, so even if $N_r(x)$ has a limit point, that point doesn't need to be a limit point of $A$.

However, your argument can be fixed with some work: Given $r > 0$, let $A_r \subseteq A$ be the set of all points such that the neighborhood of radius $r$ around this points contain a point of $A$ other than that point, i.e. $$A_r = \{x \in A : |N_r(x) \cap A| \geq 2\}.$$ (should work just as well with infinitely many points)

Then because $A$ is uncountable, we get $A_r \neq \emptyset$ for all $r > 0$. Furthermore, we have that $$r < s \Rightarrow A_r \subseteq A_s.$$ Using these two properties, you can argue that $$\lim_{r \to 0} A_r$$ contains at least one limit point of $A$.

You might need to use Zorn's Lemma though...

Dirk
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  • Why does $A$ being uncountable imply that $A_{r} \neq \emptyset$ for all $r>0$? – Jon Hillery Mar 02 '18 at 10:25
  • Fix one $r > 0$. Then you can cover all of $\mathbb{C}$ with a countable number of balls of radius $r$. In this way, $A$ would be a subset of a countable set, and thus itself countable. – Dirk Mar 02 '18 at 10:40