Is there a metric that makes $\mathbb{N} \cup \{\infty\}$ a complete metric space?

Question

Denote the set of extended natural numbers $\mathbb{N} \cup \{\infty \}$ by $X$. Is there any metric $D$ that makes $X$ a complete metric space? In other words a distance function $D: \mathbb{N}\times \mathbb{N} \to \mathbb{R}$ such that

1. $D(n,m)\geq 0$
2. $D(n,m)= 0$ if, only if, $m=n$
3. $D(m,n)=D(n,m)$
4. $D(m,n)\leq D(m,p)+D(p,n)$

My attempt. I tried using the metric $d(m,m)= \frac{|n-m|}{1+|m-n|}$ of $\mathbb{N}$ to produce a metric on $X=\mathbb{N} \cup \{\infty\}$. Necessarily I made some additional definitions like

• $|n\pm\infty|=\infty$ for all $n\in\mathbb{N}$
• $|\infty\pm n|=\infty$ for all $n\in\mathbb{N}$
• $1+\infty=\infty$
• $|\infty-\infty|=0$
• $ \frac{\infty}{\infty}=1$

Define $$D(m,n)=\sum_{k=\min(m,n)}^{\max(m,n)}\frac{1}{2^k}d(\min(m,n),k).$$

Proof of 1.( $D(m,n) \geq 0$ for all $m,n \in X$): Since $d(\min(m,n),k) \geq 0$ for all $k \in \mathbb{N}$, then $D(m,n) \geq 0$.

Proof of 2. ($D(m,n) = 0$ if and only if $m = n$): If $m = n$, then we have $\min(m,n) = \max(m,n) = m$, and $d(m,k) = 0$ for all $k \in \mathbb{N}$. So $D(m,n) = 0$. If $D(m,n) = 0$, then each term in the sum is zero, and since each term is non-negative, this implies that $d(\min(m,n),k) = 0$ for all $k \in \mathbb{N}$ such that $\min(m,n) \leq k \leq \max(m,n)$. In particular, this implies that $\min(m,n) = \max(m,n)$, i.e., $m = n$.

Proof of 3. ($D(m,n) = D(n,m)$ for all $m,n \in X$): Just note that $\min(m,n) = \min(n,m)$ and $\max(m,n) = \max(n,m)$, so $D(m,n) = D(n,m)$.

Now the triangular inequality. But I was not able to prove the triangular inequality. In fact, I think that $D$ may not satisfy the triangle inequality.

Answer 1

I'm not sure whether this $D$ of yours satisfies the triangle inequality or not. It is quite complicated. Still, answering your original question:

Is there a metric that makes $\mathbb{N} \cup \{\infty\}$ a complete metric space?

Sure. The simplest example is the discrete metric $d(x,y)=1$ (whenever $x\neq y$). This will work for any set.

For more sophisticated example consider $Y=\{0\}\cup\{\frac{1}{n}\}_{n=1}^\infty$ (which is compact and thus complete) and any bijection $f:\mathbb{N}\cup\{\infty\}\to Y$. Such bijection exists since both are countable. Then define metric on $\mathbb{N}\cup\{\infty\}$ by transfering from the (Euclidean) metric on $Y$:

$$d(x,y)=|f(x)-f(y)|$$

Answer 2

Define $d(m,n) = |m-n|$ for $m,n$ natural and $d(\infty,n) = 1+|n|$ for $n$ natural.

To see that this is a metric, define the map $i: \mathbb{N} \cup \{\infty \} \to \mathbb{R}^2$ by $i(n) = (n,0)$ and $i(\infty) = (0,1)$, then $d(a,b) = \|i(a)-i(b)\|_1$ for $a,b \in \mathbb{N} \cup \{\infty \}$.

Another example would be $d(m,n) = |\arctan m -\arctan n|$ for $m,n$ natural and $d(\infty,n) = {\pi \over 2} - \arctan n$ for $n$ natural.

Answer 3

Embed your set into $\Bbb R$ by identifying the points with the set $ \{1, \frac12, \frac13, \dots, 0\}$ and then use the usual metric on $\Bbb R$.

In detail: consider the mapping $f:\Bbb N\cup\{\infty\} \to \Bbb R$ defined by $$\begin{align} f(\infty) & = 0 \\\\ f(n) & = \frac1n \qquad \text{for integer $n$} \end{align} $$

and let $$D(a, b) = |f(a) - f(b)|.$$

(This is very similar in concept to copper.hat's suggestion but a little simpler.)