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

Question

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.

Answer 1

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.

For (2) you’re actually almost there: you have

$$\bigcap_{n\ge 1}B_d^f\left(n,1+\frac1{2n}\right)=\bigcap_{n\ge 1}\{m\in\Bbb N:m\ge n\}\;,$$

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?