# How to prove that $\mathbb{N}$ is closed using any metric function

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?

• $\Bbb N$ 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" – Gae. S. Jul 14 at 16:08
• Of course there are some distances $d$ on $\Bbb R$ such that $\Bbb N$ is not closed in $(\Bbb R,d)$. – Gae. S. Jul 14 at 16:11
• "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 $\Bbb{N}$. – Theo Bendit Jul 14 at 16:13
• 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 $\mathbb{N}$. – Chrystomath Jul 18 at 7:38