Given a second countable space $X$, show $A \subset X$ has uncountable limit points

Question

Here's my attempt at a solution and I'm wondering if it's correct.

Let $X$ have a countable basis with $A \subset X$ an uncountable set. Show $A$ has uncountably many limit points.

Let $A'$ be the set of all limit points of $A$. Assume $A'$ is countable.

Then $A-A'$ is uncountable. Let $x\in A-A'$, then $x$ is not a limit point. Since $x$ is not a limit point by definition there exists some $U_x$ open s.t. $U_x \cap A = \{x\}$. If we have some $x' \neq x$ where $x' \in A-A'$ then there is some $U_{x'}$ where its intersection with $A$ is similarly $\{x'\}$. Therefore $U_x$ is distinct from $U_{x'}$. Since $A-A'$ is uncountable there are uncountably many distinct neighborhoods in $X$. Therefore $X$ cannot have a countable basis. This is a contradiction so there must be uncountably many limit points.

Answer 1

Nice argument, but I'm not sure it's quite complete, although I think you basically have it.

I'm not sure what you mean when you say that $U_x$ is "distinct" from $U_{x'}$. Perhaps you mean disjoint, but I do not think that one can necessarily choose the $U$'s so that they are all disjoint. What is true is that $U_x \cap (A - A')$ is disjoint from $U_{x'} \cap (A - A')$ since they are distinct singleton sets ($\{x\}$ and $\{x'\}$, respectively). What you have shown is that for every $x$ in $A - A'$, the singleton set $\{x \} = U_x \cap (A - A') $ is open in the subspace topology on $A - A'$. Thus $A - A'$ can be written as the union of open (in the subspace topology) sets $\bigcup_{x \in A - A'} \{x \}$, and these open sets are disjoint in $A - A'$. Since $A - A'$ must be second countable (as a subspace of a second countable space), this shows that $A - A'$ is countable.

(It occurs to me that the real conclusion of this exercise is that $A - A'$ is countable, which is true whether or not $A$ is countable.)