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.


1 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?

  • $\begingroup$ 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 $\emptyset$ $\endgroup$
    – SiXUlm
    Oct 28, 2015 at 20:49
  • $\begingroup$ @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. $\endgroup$ Oct 28, 2015 at 20:52
  • $\begingroup$ Got it now! Thanks a lot for your help :) $\endgroup$
    – SiXUlm
    Oct 28, 2015 at 20:54
  • $\begingroup$ @SiXUlm: You’re welcome! $\endgroup$ Oct 28, 2015 at 20:54

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