0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Having uncountably many distinct neighborhoods does not contradict second countability. Consider $\mathbb{R}$ for example: for every $a$, $(-a,a)$ is a neighborhood of $0$, which is uncountably many distinct neighborhoods. However, if you can find uncountably many disjoint neighborhoods, you'll be in business. $\endgroup$ – Nate Eldredge May 1 '14 at 3:36
1
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.