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I am trying to understand Cauchy sequences a little better and would really appreciated any insight/advice you can offer:

Defition: A sequence $\{x_n\}$ in a metric space $(X,d)$ is called a Cauchy sequence if for every $\epsilon > 0$ there exists a natural number $n_0$ such that if $m,n \geq n_0$ then $d(x_m, x_n) < \epsilon$.

This is the sequence I made:

$\{ 4,3,2, 10, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},....\}$ where renamed $\{ x_1,x_2,x_3,...\}$

Question 1

Is this a Cauchy sequence?

I would say yes.

Question 2

I am trying to understand(by using the definition) how this will apply as a cauchy sequence and what is enclosed in a $\overline{B}(x_{n_0},\epsilon)$.

My reasoning:

Let $ \epsilon = 1$, then if $n_0 \geq 5$, we have $d(x_m, x_n) < \epsilon$ for $m,n \geq n_0$

Only for $ \epsilon = 10$, then for $n_0 \geq 4$, (so we can include the 10 in the sequnce) then $d(x_m, x_n) < \epsilon$ for $m,n \geq n_0$

But if and only if $\epsilon \geq 10$ can we say $\forall n_0 \in \mathbb{N}$, then $d(x_m, x_n) < \epsilon$ for $m,n \geq n_0$

Is this basically how we view cauchy sequences?

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    $\begingroup$ It does not suffice to show that such an $n_0$ exists for some epsilons (such as $\epsilon = 1$ or $ 10$), you must show it for all $\epsilon > 0$. $\endgroup$
    – Martin R
    Commented Apr 28, 2022 at 14:12
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    $\begingroup$ The "for all $\epsilon >0$" condition formalize the intuitive: "however small we choose an interval". Thus, the defining condition reads: however small we choose an interval we can find a suitable term of the sequence such that, from that term on, all terms falls into the interval. This means that the terms of the sequence "converge". $\endgroup$ Commented Apr 28, 2022 at 14:15
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    $\begingroup$ It seems like you understand that the first terms don't matter, only the infinite tail of the sequence matters. That said, if you want to prove that it's cauchy, you can't choose the value of $\epsilon$ yourself. Rather, consider that if $\epsilon>0$ is given, then $|1/n-1/m| < 1/n < \epsilon$ for all $m>n>1/\epsilon$. You can translate this to a specific index $n_0$ in your $x$ sequence. $\endgroup$
    – Milten
    Commented Apr 28, 2022 at 14:17
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    $\begingroup$ @Reuben Yes, you can set $n_0=1$ if and only $\epsilon\ge10$ in this example $\endgroup$
    – Milten
    Commented Apr 28, 2022 at 14:21
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    $\begingroup$ @Reuben But actually, if you want the Cauchy criterion to hold with a specific case of $\epsilon$ and $n_0$, you should consider $B(x_{n_0}, \epsilon/2)$ with half of your radius, to make sure that any to terms later are $\epsilon$-close to each other (this is an easy application of triangle inequality and also pretty much the same idea that proves that converging sequences are cauchy). $\endgroup$
    – Milten
    Commented Apr 28, 2022 at 14:28

1 Answer 1

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To understand the Cauchy sequence, you need to understand the purpose of the Cauchy sequence.

In general, if we want to show the limit that exists, we need to 'guess' the limit and verify by $\epsilon,\delta$ definition or other tools, etc. But sometimes, it is really hard to guess the limit. So we need to use another method to show/describe the convergence of a sequence.

The advantage of using the Cauchy criterion is we don't have to guess the limit. However the assumption to claim the Cauchy criterion implies the existence of a limit, we need the completeness of metric space. For example, consider a sequence $\left\{a_n=\frac{1}{n}\right\}$ in space $\mathbb{R}-\left\{0 \right\}$. It is a Cauchy sequence by definition but the limit does not exist. You can find the proof in Rudin or another standard analysis textbook.

Finally, the intuition of the Cauchy criterion. Assume the metric space is complete. By the definition of limit one. For every $\varepsilon>0$, there exists a $N>0$ such that for $n>N$, then $d(a_n,L)<\varepsilon$.

It means for a sufficient $N$, then all the $a_n$ can lie inside the metric ball $B(L,\varepsilon)$. Let $n>m>N$, so $a_n$ and $a_m$ must lie inside the metric ball also so the shortest distance between $a_n$ and $a_m$ will less than $2\pi \varepsilon$

In laymen's terms, since for sufficient large terms, $a_n$ and $a_m$ are really close to the limit $L$ if it exists, so they will really close together as well.

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  • $\begingroup$ +1. A useful equivalent that students often don't make use of, is that $(a_n)_n$ is Cauchy iff whenever $\epsilon >0,$ there is an open ball $B$ of radius $\epsilon /2$ such that $\{n: a_n\not\in B\}$ is finite. $\endgroup$ Commented Apr 28, 2022 at 20:42

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