Let $X$ have a countable basis and $A\subset X$ is uncountable. Would you help me to prove that uncountably many points of A are limit points of A.
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This question is similar to: Uncountably many points of an uncountable set in a second countable are limit points. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Lucenaposition Aug 20 '24 at 23:14
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HINT: Let $\mathscr{B}$ be a countable base for $X$. Let $\mathscr{B}_0=\{B\in\mathscr{B}:A\cap B\text{ is countable}\}$, and let $$C=\bigcup_{B\in\mathscr{B}_0}(A\cap B)\;.$$
- Show that $C$ is countable.
- Show that if $a\in A\setminus C$, then $a$ is a limit point of $A$. (You can prove more here: every open neighborhood of $a$ actually contains uncountably many points of $A$.
Brian M. Scott
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