# Every Cauchy sequence in a metric space is bounded

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 $x_n\in{B(0,r)}$

• Do you instead mean "Now for $x_n \geq k$ we have $x_n \in B(x_k, \epsilon)$"? Are you sure you mean $B(0,r)$ at the end? – Daniel Freedman Dec 31 '11 at 17:28
• well, I guess you got the idea. But: i) $B(0,r)$ -- what is $0$? ii) clearly $x_n\in B(x_n, \varepsilon)$. You want $x_m\in B(x_k, \varepsilon)$ iii) what is $\max(y), \min(y)$? You seem to know what to do, but the way you put it is not too clean. – user20266 Dec 31 '11 at 17:30
• It's close, but some corrections will be needed. Did you want to write $x_n\in{B(x_k,\epsilon)}$ instead of $x_n\in{B(x_n,\epsilon)}$? Also $\max y$ makes sense, for instance, if $y$ is a real number, but you are working with points of metric space; so it is not clear what you mean by $\max(y)$. – Martin Sleziak Dec 31 '11 at 17:31
• Thanks, yes I mean $x_n\in{B(x_k,\epsilon)}$. If I change $y=\{d(x_0,x_1),d(x_0,x_2),.....,d(x_0,x_n)\}$ and then change my open ball to $B(x_0,r)$ will this work? – hmmmm Dec 31 '11 at 17:41
• I think you mean "Now for $n\geq k$..." – leo Jan 24 '12 at 22:59

You seem to have the gist of it, but some of your details are a bit off.

Firstly, I find your first line a tad confusing. You seem to be defining your Cauchy sequence as a sequence that satisfies some condition based on $\epsilon$. What you should be saying is "given a Cauchy sequence $\{ x_n \}$, fix $\epsilon > 0$ and let $k$ be such that $d(x_n, x_m) < \epsilon$ whenever $n, m \geq k$". Words are often easier to read than symbols and abbreviations like s.t., $\exists$ etc. In any case, you do not (or should not) mean to use "$\forall$". For the argument to work, $\epsilon$ needs to be fixed. Can you see why?

As Thomas points out, "for $x_n \geq k$ we have $x_n \in B(x_n , \epsilon)$" is rather obvious. Maybe you made a typo there, but can you see why it is much more useful to observe that for $x_n \geq k$, we have $x_n \in B(x_k, \epsilon)$?

So far we've trapped all the terms after (and including) the $k$-th in a ball of size $\epsilon$ around $x_k$. It will help here if we think graphically. We'd now also like to trap the first $k-1$ terms in a ball around $x_k$, because then we'd know that all of the terms in the sequence lie within a ball around $x_k$ (which, as you stated, is precisely the definition of boundedness), since we can just take the larger of the two balls.

Well, thankfully, we only have finitely many terms $x_1, ... x_{k-1}$. So we know that (at least) one of them is furthest away from $x_k$ (and that this distance is finite). Call this distance $d$.

So if $D = \mathrm{max} \{\epsilon, d \}$, then everything in the sequence must lie inside $B(x_k, D + 1)$ (or $B( x_k, D + \delta)$ where $\delta > 0$).

• Thanks very much. Sorry about my notation, my first line is supposed to be saying that $\{x_n\}$ is cauchy (from the definition) I should then go on an say "fix $\epsilon$....". I think I might have addressed the main problems in the comment I left but thanks for your answer it is much clearer than mine. – hmmmm Dec 31 '11 at 17:51

The intuition behind what you wrote is undoubtedly right. The actual details, not so good. We give an argument that is in spirit close to what I perceive as your intuition.

Pick a fixed $\epsilon >0$. (Actually, we can choose $\epsilon=1$, or $47$.) Then there is an $m$ such that if $n \ge m$, then $d(x_m,x_n)<\epsilon$. Let $p$ be any point in the space, and let $$k=\max_{i\le m} \:d(p, x_i).$$ The maximum exists, since $\{x_i\colon\; i\le m\}$ is a finite set.

We show that for any $n$, $d(p,x_n)<k+\epsilon$. This is obvious for $n\le m$, since then $d(p,x_n)\le k$. For $n>m$, we have, by the Triangle Inequality, $$d(p,x_n)\le d(p,x_m)+d(x_m, x_n) <k+\epsilon.$$

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).

• +1. Sir thank you so much for explaining why we need to fix $\epsilon$. – Akash Patalwanshi Aug 6 '19 at 2:29

Choose $N$ so that $m, n\ge N\implies d(x_m, x_n) < 1.$ Using the the triangle inequality, it is fairly easy to see that $x_n \in B_2(x_N)$ for $n \ge N$. Consequently, we have all but finitely many of the $x_n$ contained in a bounded subset of the space. Hence, the sequence is bounded.