Prove that all cauchy sequences are bounded in a metric space.

Suppose that A is a metric space, with d the metric. And suppose that the sequence {Xn} is a cauchy sequence on A.

I need to prove that it is bounded, meaning that I have to an element $\displaystyle a \in A $ and K such that $\displaystyle d(x_n,a) < K \ \ \ \ \ \forall n$

Now, since the sequence is cauchy I can set $\displaystyle \epsilon = 1 $ There exist N such that $\displaystyle \forall n,m > N $, we have $\displaystyle d(x_n,x_m)<1 $

Consider the following:

$\displaystyle | d(x_n,x_{N+1})-d(x_n,x_{N+2}) | \leq d(x_{N+1},x_{N+2}) $ (true by the reserve triangular inequality)

implies that $\displaystyle d(x_n,x_{N+1}) \leq d(x_{N+1},x_{N+2}) + d(x_n,x_{N+2}) $

Now, since $\displaystyle d(x_{N+1},x_{N+2}) < 1 $

we would have

$\displaystyle d(x_n,x_{N+1}) < 1 + d(x_n,x_{N+2}) $

Now, let $\displaystyle K = max \{ d(x_1, x_{N+2}), d(x_2,x_{N+2}), . . . , d(x_{N+2},x_{N+2}) \} + 1 $, therefore we have $\displaystyle d(x_n,x_{N+1}) < K $

But I know something is wrong with this proof because I'm confusing myself with the n, are there anything I can do to clear that up?

Thank you very much!