# Thread: All Cauchy Sequences are bounded in a metric space

1. ## All Cauchy Sequences are bounded in a metric space

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 $a \in A$ and K such that $d(x_n,a) < K \ \ \ \ \ \forall n$

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

Consider the following:

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

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

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

we would have
$d(x_n,x_{N+1}) < 1 + d(x_n,x_{N+2})$

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

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!

I need to prove that it is bounded, meaning that I have to an element $a \in A$ and K such that $d(x_n,a) < K \ \ \ \ \ \forall n$
Now, since the sequence is Cauchy I can set $\epsilon = 1 \exist N$ such that $\for n,m > N$, we have $d(x_n,x_m)<1$
Now you have that $N$. BTW it is $m,n \ge N$.
Let $M = \sum\limits_{1 \leqslant j < k \leqslant N} {d(x_j ,x_k )}$. Your bound is now $M+2$.