Arthur's answer; (ZF) Prove 'the set of all subsequential limits of a sequence in a metric space is closed.
Let $\{p_n\}$ be a sequence in a metric space $X$.
Let $B=\{p_n|n\in\mathbb{N}\}$ and $p$ be a limit point of $B$.
It first seemed obvious that there exists a subsequence convergent to $p$, but i realized that
I can construct a 'infinite subset' of $\{p_n\}$ and form a new sequence convergent to $p$, but can't construct a 'subsequence'.
Help me how to construct a subsequence.
Additional Question; What is the precise definition of convergent? I think the definition should mention ordering of $\mathbb{N}$, but every definition i saw doesn't mention this. For example, let $p_n = 1/n$. Say $G$ is the usual ordering of $\mathbb{N}$ (i.e. Well-ordered by $\in$) If i follow this ordering, $\{p_n\}$ is convergent to 0. But if i follow $G^{-1}$, it's convergent to 1.
Or is it defined by the point where is the most, unofficially speaking, 'dense'?