Show that every Cauchy sequence is bounded

Question

I am supposed to prove, that every Cauchy sequence is bounded.

From the definition of Cauchy sequence:$\left ( \forall \varepsilon > 0 \right )\left ( \exists n_{0} \in\mathbb{N} \right ) \left ( \forall m,n \in\mathbb{N}:m,n\geq n_{0} \right )d\left ( x_{n},x_{m} \right )< \varepsilon $.

We can fix $ \varepsilon$: $\varepsilon=1$. Then $m\geq n$, so that $d\left ( x_{n},x_{m} \right )< \varepsilon $. Let p be point in a space...

I do not know how to continue, or if it is correct. I want to somehow include triangle inequality, but I am not really sure, if that is correct.

Any help?

What is $d(x_n-x_m)$? It's usually $d(x_n,x_m)$ or $\vert x_n-x_m\vert$. — Keen-ameteur, Feb 23 '20 at 13:53

score 1 · Accepted Answer · answered Feb 23 '20 at 13:54

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You are on the right track. I will use $|x-y| $ for $d(x,y)$ so $|x| =d(x,0)$.

As you said, for $\epsilon = 1$ we have $n_0$ s.t. for all $n,m\ge n_0$ we have $|x_n-x_m| <1$ in particular for all $n\ge n_0 , |x_n-x_{n_0}|<1$.

So , $|x_n| < |x_{n_0}|+1$ for $n\ge n_0$.

Do you see how to continue?

answered Feb 23 '20 at 13:54

infinity

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So then we can say, that : $d\left ( p,x_{n} \right )\leq d\left ( p,x_{n0} \right )+d\left ( x_{n0}+x_{n} \right )< d\left ( p,x_{n} \right )+1$? – Feb 23 '20 at 14:32
What is $p$ ? @Emanuel – infinity Feb 23 '20 at 14:33
p is any point in the space – Feb 23 '20 at 14:34
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No, just take $M= max{|x_1|,..,|x_{n_0}| , |x_{n_0}|+1}$ and see that $|x_n| \le M$ for every $n$ @Emanuel – infinity Feb 23 '20 at 14:35
So I do not need any triangle inequality? – Feb 23 '20 at 14:38
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@Emanuel i usedto get $|x_n| < |x_{n_0}| +1$ from $|x_n-x_{n_0}| < 1$ – infinity Feb 23 '20 at 14:38
Thank you so much – Feb 23 '20 at 14:47

Show that every Cauchy sequence is bounded

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