Prove that (, ) is a complete metric space

Question

On the space of the codes $X = \{1,2,3, ..., N-1\}^n$ is defined the distance between two points $x,y \in X$ with $$d(x,y) := \sum_{i=1}^n \frac{|x_i - y_i|}{(N+1)^i}.$$

Prove that each pair $(x,y)\in X^2$ is at a finite distance and that $(X, d)$ is a metric space. Prove that $(X, d)$ is a complete metric space.

I can't figure out how to prove this, please help.

Answer 1

1. $d(x,y)$ is finite for $x,y\in X$: Since we have finitely many summands and the summands are finite numbers, $d(x,y)$ is finite for every $x,y\in X$

2. $(X,d)$ is a metric space:

   • $d(x,y)=0$ iff $x=y$: Since all summands are nonnegative, $d(x,y)=0$ iff for all $i\in \{0, \dots n\}$ $\frac{|x_i - y_i|}{(N+1)^i} = 0$ iff for all $i$ one has $|x_i-y_i| = 0$ iff for all $i$ one has $x_i = y_i$ iff $x=y$
   • $d(x,y) = d(y,x)$: follows directly, since $|x_i - y_i| = |y_i - x_i|$
   • $d(x,z) \leq d(x,y) + d(y,z)$: follows directly from $|x_i - z_i| = |x_i - y_i + y_i - z_i| \leq |x_i - y_i| + |y_i - z_i|$ (and standard inequality rules for sums and products)

3. $(X,d)$ is complete: As Mindlack commented, this follows since $X$ is finite.