0

If $X$ is a finite set and d is an arbitrary metric. Prove that $(X, d)$ is complete.

My solution:

Let $X =$ {$x_1, x_2, ... , x_n|n \in \mathbb{N}$}

$\exists N\in \mathbb{N}$ s.t. $x_n = x$ for some $x\in X$

$\therefore$ {$x_n$}$^\infty _{n = 1}\rightarrow x$

$\implies d(x_n, x)\leq \epsilon$, $\forall n>N$

$\therefore$ $x_n\rightarrow x$ as $n \rightarrow \infty$

Shows every cauchy sequence in X converges in X therefore it is complete. Not sure if this is entirely correct.

Also how do I show any two metrics on X are equivalent? What are the possible metrics on X?

Asaf Karagila
  • 393,674
WD-40
  • 39
  • Your reasoning doesn't show any relation to the fact that $X$ is finite. Since $X$ is a finite metric space and metric spaces are Hausdorff, $X$ must carry the discrete topology. –  May 20 '14 at 11:51

1 Answers1

2

its a finite metric space, therfore there's a minimum over $d(x,y)$ where $x \ne y$ so use the fact that in a cauchy series you can get the elements of the series close as you want, and show that every cauchy series is constant from some place N, and then obviously converges to that constant.

Snufsan
  • 2,125