Is a Cauchy sequence bounded in a metric space if the space is not complete? 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.
 A: 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 $M+1>1$,$$\{x_n\mid n\in\Bbb N\}\subset B_{M+1}(x_N).$$
A: A more conceptual way to see it: Let $X$ be a metric space, $x_n$ a Cauchy sequence in $X$, and let $\overline X$ be its completion. We can isometrically embed $X$ into $\overline X$, and the image $\overline x_n$ of the sequence $x_n$ will still be a Cauchy sequence, since its elements still have the same distance from each other. But this sequence converges, since we're in a complete space, so it must be bounded. And since the distances between the $x_n$ are the same as the ones between the $\overline x_n$, $x_n$ must also be bounded.
A: Yes it is.
Let $(x_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in a metric space $(E,d)$. By definition, there exists $N \in \mathbb{N}$ such that for every $p \geq N$, you have
$$d(x_p,x_N)\leq 1$$
Let
$$M = \max \lbrace d(x_i,x_N), 0 \leq i \leq N-1 \rbrace + 1$$
You have then that for every $i \in \mathbb{N}$,
$$d(x_i,x_N) \leq M$$
so for every $i \in \mathbb{N}$,
$$x_i \in B(x_N, M)$$
(where $B(x_N, M)$ denotes the ball centered at $x_N$ of radius $M$) ; i.e. the sequence is bounded.
A: Hint:For fixed $y$ the sequence ${d(x_n,y)$ is  a Cauchy sequence of real numbers. Hence it is bounded.
