# Let $\{a_n\}$ be a Cauchy Sequence in $M$. Prove that the set $\{a_n\}$ is bounded.

Let $\{a_n\}$ be a Cauchy Sequence in $M$. Prove that the set $\{a_n\}$ is bounded.

Proof: let $\{a_n\}$ be a Cauchy sequence in $M$, then there exists a $N \in \Bbb N$ such that $n,m \geq N$ implies $d(a_n,a_m) < \varepsilon$.

Thus, $a_n$ is bounded by $N$.

Is this correct?

• No, it is not. Read what your statements say carefully. $N$ has nothing to do with an upper or lower bound of the values of the $a_n$, but is a lower bound on the index $n$. – Simon S Nov 16 '15 at 17:48
• The bound has to do with $\varepsilon$, not with $N$. – Alex M. Nov 16 '15 at 18:43

## 3 Answers

Hint:

For $\epsilon = 1$ there exists an $N$ such that for all $n, m > N$, $|a_n - a_m| < 1$ and in particular $|a_{N+1} - a_m| < 1$ for all $m > N$.

Can you now find an upper and lower bound on all $a_n$?

But is not $\mathbb{R}$, i.ea Suppose $(\forall \epsilon > 0)(\exists N> 0)(n, m> N \Rightarrow d(x_n,x_m)<\epsilon)$.Prove$(\exists x\in x) (\epsilon> 0)( \forall n> 1)d(x_n, x)<\epsilon$. Any hint? For your solution.

Cauchy sequences converge. Thus the tail of such sequences are bounded between (L-epsilon,L+epsilon). But then the whole sequence is bounded because any finite set (the first k terms) is also bounded.