Given a Cauchy sequence $(x_n)$ such that $\forall M\in \Bbb{N}$, $\exists k,n\ge M$ such that $x_k<0$ and $x_n>0$. Show that $x_n$ converges to $0$.

Question

Suppose a Cauchy sequence $(x_n)$ is such that for every $M\in \mathbb{N}$, there exists a $k\ge M$ and an $n\geq M$ such that $x_k<0$ and $x_n>0$. How do I show $x_n$ converges to $0$?

I have some vague intuition that since $x_n$ is Cauchy, the difference between terms is getting arbitrarily small, so if there's always at least one positive and one negative term, their difference must be getting arbitrarily small, which means $x_n$ is somehow approaching zero. I'm not sure how to translate this into something formal. I've just been staring at the definition of a Cauchy sequence and trying to think of a way to get from there to $|x_n|<\varepsilon$.

Answer 1

HINT: Every Cauchy sequence in $\Bbb R$ converges. If a sequence converges to $L$, every subsequence of it also converges to $L$. Your condition ensures that $\langle x_n:n\in\Bbb N\rangle$ has a subsequence consisting entirely of negative numbers and another subsequence consisting entirely of positive numbers. Can a sequence of negative numbers converge to a positive number, or vice versa?

Answer 2

From the Cauchy condition, given $\epsilon>0$ there is $N$ such that $|x_n-x_m|<\epsilon$ whenever $m$, $n\ge N$. There are $j$, $k\ge N$ with $x_j>0>x_k$ by your condition. Then $0<x_j-x_k<\epsilon$, and so $\epsilon>x_j>0>x_k>-\epsilon$.

If $n\ge N$, then $|x_n-x_j|<\epsilon$. Therefore $2\epsilon>x_n>-\epsilon$ and so $|x_n|<2\epsilon$. Therefore $x_n\to0$.

Answer 3

Since $(x_n)$ is a Cauchy sequence, For every $\epsilon\gt 0, \exists N_1$ such that for all $i,j\ge N_1$, we have: $|x_i-x_j|\lt \epsilon$
Let $N_1'=\operatorname{sup}\{N_1,M\}$. Hence we have in particular: $|x_n-x_k|\lt \epsilon/2\implies 0\lt x_n-x_k\lt \epsilon/2$, since $x_k\lt 0$, as per given condition.

For $i\ge N_1'$,

$\begin{align}|x_i|=|x_i-0|&\le |x_i-x_n|+x_n\\&\lt \epsilon/2+x_n\\&\lt \epsilon/2+(x_n-x_k)\\&\lt \epsilon/2+\epsilon/2=\epsilon\implies |x_i|\lt \epsilon\implies x_i\to0 \end{align}$