# Cauchy sequences are bounded?!

I'm having trouble understanding the proof that Cauchy sequences are bounded, here's the proof I've been given

Let $s_n$ be a Cauchy sequence. We take a concrete value of $\varepsilon$, for simplicity $1$, then there exists a natural number $n_0$ such that $$|s_m-s_n|\leq 1\;\;\;\; \forall m,n\geq n_0$$ Hence (fixing $m=n_0$) we have $$|s_n-s_{n_0}|\leq 1 \;\;\; \forall n\geq n_0$$

And the proof goes on to explain how this is used to bound $|s_n|$ for all $n\geq n_0$ and therefore for $n\leq n_0$ there are a finite number of terms so its just the max of the bounds etc. I understand that. However its the parts in the italics that I can't understand. How can you fix $\varepsilon$ and $s_m$ when they should be variables and the proof should cover $\forall \varepsilon>0$ and $\forall m\geq n_0$?

• In response to the punctuation in the title: Yes!!! They are!!! – Asaf Karagila Apr 7 '14 at 17:21

You do not want to show that something holds for all $\epsilon>0$ and for all $m\ge m_0$, you want to make use of this fact (that is given). For this you are allowed to specialize to a (valid) choice of $\epsilon$ and/or $m$.
• Ok so the bounds of the sequence will change depending on your choice of $\varepsilon$ and $m$? – George1811 Apr 7 '14 at 17:00
in bartle's book, real analysis, he puts it like so: it's a game, in which party A makes the statement $|s_{m} −s_{n} | \leq 1$, $\forall{m,n} ≥n_{0}$ . then, party B comes and challenges that statement by giving party A, a value of epsilon. party A has to come up with a value for "n" such that, any n greater than that value will oblige to the tested hypothesis.