TITLE: Every Cauchy sequence in a metric space is bounded
QUESTION [8 upvotes]: Is the following correct or along the right lines?
Thanks for any help
Question
A sequence $\{x_n\}$ in a metric space is said to be bounded if it is contained in some open ball $B(a,r)$.
Prove that every Cauchy sequence in a metric space is bounded.
Proof 
Put $\{x_n\}\in(X,d)$ s.t. $\forall\epsilon>0\ \exists k\in\mathbb{N}\ s.t. \ d(x_n,x_m)<\epsilon$ whenever $n,m\geq k$.
Now for $x_n\geq k$ we have $x_n\in{B(x_n,\epsilon)}$
Defining the finite sequence $y=\{x_1,x_2,\ldots,x_n\}$.
Take $r=\epsilon+\max\{|\max(y)|,|\min(y)|\}$, and so $x_n\in{B(0,r)}$

REPLY [4 votes]: I have a qualm  with one point in the writing style which I think is worth elucidating (hopefully elucidating):
When you say:
"Put $\{x_n\}\in(X,d)$ s.t. $\forall\epsilon>0\ \exists k\in\mathbb{N}\ s.t. \ d(x_n,x_m)<\epsilon$ whenever $n,m\geq k$."
all you are stating is that $\{x_n\}$ is a Cauchy sequence. You are not fixing a value of $\epsilon$ here, and the statement does not give you a value of $k$ that you can work with.
The statement is passive in this regard: it states that these $\epsilon$'s and $k$'s exist, but it does not provide  specific instances of them by itself.
You have to explicitly state that there is a $k$ that corresponds to a particular and fixed value of $\epsilon$ that you have selected.
So in between the above quoted line and your next line, you'd need to say something like:
"Let $\epsilon>0$"
Now you have a value of $\epsilon$ declared that you can work with. Using the fact that $\{x_n\}$ is Cauchy, you would next say
"since $\{x_n\}$ is Cauchy, there is a $k$ such that..."
(even more nitpicky here, you'd have to add "and we select this value of $k$"; or say something like "since $\{x_n\}$ is cauchy, we may, and do, pick $k$ such that...")
And then you'd proceed with the rest of your proof (with corrections as gleaned from the other fine answers).