(Quick note: I see there are a lot of Cauchy sequence questions but I did not see this question specifically)

Suppose that the sequence $v_n, n=1,2,3,... $ of elements from an inner product space $V$ converges to $v\in V$. Prove that $v_n$ is a Cauchy sequence.


I use the definition of the Cauchy sequence:

$$|v_n - v_m| = |(v_n-v) + (v-v_m)|$$

Using the triangle inequality,

$$|(v_n-v) + (v-v_m)| \leq |v_n-v| + |v_m-v|$$

By the given, it is clear that $|v_n-v|<\epsilon_0$ given any $n>N$. I'm not really sure what to do with the $|v_m-v|$ part, though. Is it enough to say that $|v_m-v|>0$, and then to call this scalar $\epsilon _1$. Then we can say that $$|v_n-v_m|< \epsilon _0 + \epsilon _1$$ And, since both of these $\epsilon$ values are finite and greater than zero, then their sum is as well, so we set $\epsilon = \epsilon _0+ \epsilon_1$ and see that the definition of the Cauchy sequence is satisfied?

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    $\begingroup$ Any convergent sequence in a metric space is Cauchy. $\endgroup$ – zhw. May 12 '15 at 23:13
  • $\begingroup$ @zhw. I think that's precisely what is intended to be proved. $\endgroup$ – Daniel May 13 '15 at 0:16
  • $\begingroup$ Why then bring up inner products? $\endgroup$ – zhw. May 13 '15 at 1:01

It seems you are a little bit confused. Remember the definition of a Cauchy sequence:

$\{v_n\}_n$ is a Cauchy sequence if for all $\varepsilon>0$ there exists $N\in \Bbb N$ such that for all natural numbers $n,m\geq N$: $|v_n-v_m|<\varepsilon$.

In order to prove that this sequence is Cauchy, you must begin as follows:

Let $\varepsilon>0$...

Now work over this epsilon:

Let $\varepsilon>0$, since $\frac{\varepsilon}{2}>0$, by definition of convergence there exists $N\in \Bbb N$ such that for all $n\geq N: |v-v_n|<\frac{\varepsilon}{2}$. We show that this is the desired $N$ (in the definition of Cauchy sequence). Let $n,m\geq N$, then $|v_n-v_m|\leq |v_n-v|+|v-v_m|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$. Since $\varepsilon>0$ is arbitrary, the desired sequence is a Cauchy sequence.

  • $\begingroup$ Can I work backwards? Given $\epsilon_0 >0$, we know that $|v_n-v|<\epsilon_0$ whenever $n>N$. Not only this, let $\epsilon_1>0$. Then, $|v_m-v| < \epsilon_1$ whenever $m>N$. So, we now consider their sum. As already shown by triangle inequality, the sum is greater than $|v_n-v_m|$ which is less than $\epsilon_0+\epsilon_1$, whenever $m,n>N$. Is this more what you meant? $\endgroup$ – Rellek May 12 '15 at 23:23
  • $\begingroup$ @Rellek Your argument is not precisely good. I've edited my answer so you can take a look to the right argument, however, try to solve it by your own at first with this hint: In the definition of convergence, you choose the epsilon and the property will hold for every $n,m,s,t...\geq N$. $\endgroup$ – Daniel May 12 '15 at 23:30
  • $\begingroup$ @Rellek, there will be an $N_0$ for the $\epsilon_0$ and an $N_1$ for the $\epsilon_1$, so $|v_n-v_m|<\epsilon_0+\epsilon_1$ whenever $n,m\geq \text{max}\{N_0,N_1\}$. $\endgroup$ – Larara May 12 '15 at 23:35
  • $\begingroup$ Ok, I am super rusty on this. Thanks for helping me out on it. $\endgroup$ – Rellek May 13 '15 at 0:19

You know that $\{v_n\}$ converges to $v$. So given $\varepsilon>0$ there exists an $N\in\mathbb{N}$ such that for all $n\geq N$ it holds that $|v_n-v|<\varepsilon$. Now take a $\varepsilon>0$. Let $N\in\mathbb{N}$ be such that for all $n\geq N$ it holds that $|v_n-v|<\varepsilon/2$. So, we have for $n,m\geq N$: $$|v_n-v_m|=|(v_n-v)-(v_m-v)|\leq |v_n-v|+|v_m-v|<\varepsilon/2+\varepsilon/2=\varepsilon$$ So we showed that given $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that for all $n,m\geq N$ $|v_n-v_m|<\varepsilon$, i.e, $\{v_n\}$ is a Cauchy sequence.


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