# A question on metric space $(\mathbb{N},d)$, where $d: \mathbb{N} \times \mathbb{N} \mapsto \mathbb{R^+}$

Consider metrics $d: \mathbb{N} \times \mathbb{N} \mapsto \mathbb{R^+}$, where $d(m,n)=0$ if $m=n$ and $d(m,n)=1+\frac{1}{m+n}$ if $m\neq n$.

1. Show that: $(\mathbb{N},d)$ is complete
2. Let $B^f_d(n,1+\frac{1}{2n})$ be the closed ball. What is $\bigcap\limits_{n=1}^{\infty} B^f_d(n,1+\frac{1}{2n})$?

For 1, take a Cauchy sequence $(x_n)$ in $\mathbb{N} \Rightarrow\forall\epsilon>0, \exists K \in \mathbb{N}$ such that $d(x_m,x_n)<\epsilon$, for EVERY $m,n\geq K$. This is true when $m=n$ (as $d(m,n) = 0$). But when $m \neq n,d(x_m,x_n)=1+\frac{1}{x_m+x_n}$ can never be smaller than $\epsilon$, for $0<\epsilon \leq 1$. So actually there is NO Cauchy sequence in $\mathbb{N}$ wrt metrics $d$ (?!) Did I make mistakes somewhere?

For 2, I found $B^f_d(n,1+\frac{1}{2n})=\{{m \in \mathbb{N}: m\geq n}\}$, which forms a decreasing sequence of closed balls ($B_n \subset B_{n-1}...\subset B_1$). Then I stuck here.

Your reasoning for (1) was correct until you got to your conclusion. The correct conclusion is that the Cauchy sequences in $\Bbb N$ are precisely the sequences that are eventually constant. Of course these converge (to the point at which they are eventually constant), so the metric space is complete.
$$\bigcap_{n\ge 1}B_d^f\left(n,1+\frac1{2n}\right)=\bigcap_{n\ge 1}\{m\in\Bbb N:m\ge n\}\;,$$
so you need only evaluate the righthand side. Let $k\in\Bbb N$; can you find a set $\{m\in\Bbb N:m\ge n\}$ that does not contain $k$? Can $k$ belong to the intersection?
• For every $k \in \mathbb{N}$, we can always find a set that does not contain $k$, for example: $\{m \in \mathbb{N}: m \geq k+1\}$ . So actually the whole intersection is just $\emptyset$ Oct 28, 2015 at 20:49
• @SiXUlm: It’s clear from the definition of $d$ that in this question $\Bbb N$ represents the set of positive integers: if $0$ were included, $d(0,n)$ would be undefined for $n>0$. (In my usage $n\in\Bbb N$, but that’s clearly not the case here.) Thus we needn’t consider $k=0$, and as you say, the intersection is empty. Oct 28, 2015 at 20:52