Let X have a countable basis (I.e. second countable). Suppose A is an uncountable subset. Show A has uncountably many limit points.

I'm not sure where to start with this. A hint will work. This isn't a homework problem, just practice, if that makes you feel better.

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    $\begingroup$ Being sassy and pissing off the users of this website won't help your chances of getting hints. $\endgroup$ – Silvia Ghinassi Oct 27 '15 at 3:34
  • $\begingroup$ I honestly wasn't intending to be sassy or piss anyone off. I have noticed that users don't like to answer if it is for homework, so I wanted to reassure those users. My mistake, I'll be more careful with my wording. $\endgroup$ – YAK Oct 27 '15 at 3:37
  • $\begingroup$ A problem with text only conversations is that it is harder to interpret other people's remarks without being able to hear inflection. Sometimes people are a little too quick to assume a worse interpretation than you intended. $\endgroup$ – Paul Sinclair Oct 27 '15 at 3:42
  • $\begingroup$ However, your question is a duplicate: math.stackexchange.com/questions/17050/… $\endgroup$ – Paul Sinclair Oct 27 '15 at 3:44

Let $\mathcal{B}$ be a countable base for $X$. Suppose $A \subseteq X$ is uncountable.

Let $B \subseteq A$ be the set of points from $A$ that are not limit points of $A$. This means that for each $b \in B$, there is some base element $B_b$ such that $B_b \cap A = \{b\}$ (why?).

Show the $B_b$ are all distinct for different $b \in B$. Now use $\mathcal{B}$ is countable and $A$ is not.


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