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 $m\neq n$.
- Show that: $(\mathbb{N},d)$ is complete
- 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.