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I read this question about how the Axiom of Countable Choice is both necessary and sufficient to show the following:

If a point $a$ in a metric space $X$ is a limit point of $A\subseteq X$, then there is a sequence of points in $A-\{a\}$ converging to $a$.

Is the Axiom of Countable Choice also enough if we require that the sequence be a sequence of distinct points?

Lilypad
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1 Answers1

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Yes. If $(a_n)$ is a sequence in $A-\{a\}$ converging to $a$, just let $(b_m)$ be the sequence obtained from $(a_n)$ by removing duplicates (so $b_m$ is the $m$th distinct term of $(a_n)$). Then $(b_m)$ still converges to $a$, since it is a subsequence of $(a_n)$.

Eric Wofsey
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