Some detailed explanation about how the negation of a Cauchy sequence works?

Question

I know how to write it, but my question is that i dont completly understand what does the definition mean, when it tells you that exist an epsilon, this epsilon is fixed right, but with respect to who? I would really aprecciate a complete explanation of what this definition involves.

Answer 1

In informal terms, a sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ is Cauchy if no matter how small a positive $\epsilon$ you take, you can find a tail of the sequence whose terms are all within $\epsilon$ of one another. Thus, $\sigma$ is not Cauchy if there’s at least one ‘bad’ $\epsilon>0$, an $\epsilon$ such that no matter how far out $\sigma$ you go to take your tail, there are two terms in that tail that are not within $\epsilon$ of each other. Let’s see in more detail what that says about $\sigma$.

The sequence itself is it’s own biggest tail, so there are $k_0,\ell_0\in\Bbb N$ such that $d(x_{k_0},x_{\ell_0})\ge\epsilon$. Let $m_0=1+\max\{k_0,\ell_0\}$. Then $\langle x_n:n\ge m_0\rangle$ is a tail of $\sigma$, so there are $k_1,\ell_1\ge m_0$ such that $d(x_{k_1},x_{\ell_1})\ge\epsilon$. Do it again: let $m_1=1+\max\{k_1,\ell_1\}$, and look at the tail $\langle x_n:n\ge m_1\rangle$ of $\sigma$. It has two terms that are not within $\epsilon$ of each other, so there are $k_2,\ell_2\ge m_1$ such that $d(x_{k_2},x_{\ell_2})\ge\epsilon$. Plainly we can continue this indefinitely to construct indices $k_i,\ell_i$ for $i\in\Bbb N$ such that

$$k_0,\ell_0<k_1,\ell_1<k_2,\ell_2<\ldots\;,$$

and $d(x_{k_i},x_{\ell_i})\ge\epsilon$ for each $i\in\Bbb N$. This gives us two subsequences of $\sigma$, $\langle x_{k_i}:i\in\Bbb N\rangle$ and $\langle x_{\ell_i}:i\in\Bbb N\rangle$ such that $d(x_{k_i},x_{\ell_i})\ge\epsilon$ for each $i\in\Bbb N$. If you step through the first subsequence, moving from $x_{k_0}$ to $x_{k_1}$ to $x_{k_2}$ and so on, while I step through the other subsequence at the same rate, we’ll always be at least $\epsilon$ units apart.

If $\sigma$ were Cauchy, that couldn’t happen: in that case you should be able to prove that if $\langle x_{k_i}:i\in\Bbb N\rangle$ and $\langle x_{\ell_i}:i\in\Bbb N\rangle$ are subsequences of $\sigma$, then $\lim\limits_{i\to\infty}d(x_{k_i},x_{\ell_i})=0$, so no matter what $\epsilon>0$ you pick, from some point on we’ll be less than $\epsilon$ apart.

Answer 2

A sequence $\{x_n\}$ is Cauchy if for all $\epsilon >0$, there exists $N \in \mathbb{N}$ such that for all $m,n>N$, $|x_m-x_n|<\epsilon$.

Negating this, a sequence $\{x_n\}$ is not Cauchy if there exists $\epsilon>0$ such that for all $N \in \mathbb{N}$ there exist $m,n>N$ such that $|x_m-x_n| \geq \epsilon$.

You're right that the $\epsilon$ in "there exists $\epsilon>0$" is fixed. Fixed compared to what? What we mean is that $\epsilon$ doesn't vary with/depend on $N$. No matter what $N \in \mathbb{N}$ we choose, there are two terms in the sequence further out than $N$ that are at least $\epsilon$ apart. The statement would not be very different if we let $\epsilon$ depend on $N$. That would say: "for every $N \in \mathbb{N}$, there exists $\epsilon>0$ and $m,n >N$ such that $|x_m-x_n| \geq \epsilon$," and all that really says is that your sequence is not eventually constant.