Is a Cauchy sequence bounded in a metric space if the space is not complete

Question

I don't think a metric space needs to be complete for a Cauchy sequence to be bounded. Being complete means the sequence converges to a point in the space, but the standard proof for Cauchy sequence being bounded does not require you to know what the sequence converges to. For that reason, I think it doesn't matter whether the metric space is complete or not.

Answer 1

If $(x_n)_{n\in\Bbb N}$ is a Cauchy sequence in a metric space $(X,d)$, then there is some $N\in\Bbb N$ such that $n\geqslant N\implies d(x_m,x_n)<1$. Therefore$$\{x_n\mid n\geqslant N\}\subset B_1(x_N).$$Now, let $M=\max\{d(x_N,x_k)\mid k<N\}$; it exists, since it is the maximum of a finite set. Then$$\{x_n\mid n<N\}\subset B_{M+1}(x_N)$$and therefore, since $M+1>1$,$$\{x_n\mid n\in\Bbb N\}\subset B_{M+1}(x_N).$$