1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $M+1>1$,$$\{x_n\mid n\in\Bbb N\}\subset B_{M+1}(x_N).$$

$\endgroup$
0
$\begingroup$

Hint:For fixed $y$ the sequence ${d(x_n,y)$ is a Cauchy sequence of real numbers. Hence it is bounded.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.