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Let $S$ be a set. I define

  • $a$ as a limit point of $S$ if there a sequence $(a_n) \subset S$ such that $(a_n) \to a$.
  • $S$ is closed if it contains all of its limit points.

I know how to prove every subset of $\mathbb{N}$ (where the ambient space is $\mathbb{N}$ instead of $\mathbb{R}$) is closed using the usual metric (i.e. the absolute value distance). In fact, I can show that it's open as well. However, how can I prove that $\mathbb{N}$ is closed with any metric function?

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  • $\begingroup$ $\Bbb N$ is sequentially closed in $(\Bbb N,d)$ because for all $a\in \Bbb N$ (the second I've mentioned) such that there is some $(a_n)\subseteq \Bbb N$ (the first I've mentioned) such that $a_n\to a$, $a\in \Bbb N$ (the first I've mentioned). This because "the first I've mentioned" = "the second I've mentioned" $\endgroup$ – Gae. S. Jul 14 at 16:08
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    $\begingroup$ Of course there are some distances $d$ on $\Bbb R$ such that $\Bbb N$ is not closed in $(\Bbb R,d)$. $\endgroup$ – Gae. S. Jul 14 at 16:11
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    $\begingroup$ "Closed" is not a sensible term to apply to an entire space (the whole space is always closed in itself); "complete" is the obvious analogue. If you want to show that every subset of $\Bbb{N}$ is closed in $\Bbb{N}$ with respect to any metric, you can't. If $f : \Bbb{N} \to \Bbb{Q}$ is a bijection, you can define a metric on $\Bbb{N}$ by $d(x, y) = |f(x) - f(y)|$, which will generate plenty of non-closed subsets of $\Bbb{N}$. $\endgroup$ – Theo Bendit Jul 14 at 16:13
  • $\begingroup$ You need to clarify the question. The constant sequence $(a)_{n\in\mathbb{N}}$ converges to $a$, hence by your first definition any $a\in\mathbb{N}$ is a limit point of $\mathbb{N}$. $\endgroup$ – Chrystomath Jul 18 at 7:38

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