Help with understanding Cauchy sequences? 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?
 A: 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.
