I'm trying to solve the following exercise.
Let $K$ be an ordered field and $(x_n)_{n\in \mathbb{N}}$ be an Cauchy sequence in $K$. Show that there exists an increasing Cauchy sequence $(y_n)_{n\in \mathbb{N}}$ and a decreasing Cauchy sequence $(z_n)_{n\in \mathbb{N}}$ with
$$x_n=y_n+z_n.$$
If the field were complete, it would be easier. But I'm not sure how to construct $(y_n)_{n\in \mathbb{N}}$ and $(z_n)_{n\in \mathbb{N}}$ without an existing limit.
Best Answer
The statement seems to be wrong. Let's consider $K= \mathbb{R}$ (which is an ordered field) and $x_n=(-1)^n/n$. The sequence $(x_n)_n$ converges (to $0$) and thus is a Cauchy sequence. Assume that the exists a monotone increasing sequence $(y_n)_n\subseteq \mathbb{R}$ and a monotone decreasing sequence $(z_n)_n\subseteq \mathbb{R}$ such that $x_n= y_n + z_n$. Then we get
$$ x_{n+1} -x_n = (y_{n+1}-y_n) +(z_{n+1}-z_n). $$
As $(z_n)_n$ is decreasing, we get that $z_{n+1}-z_n\leq 0$ and hence, we have
$$ y_{n+1}-y_n \geq x_{n+1}-x_n. $$
Again, as $y_{k+1}-y_k \geq 0$ we get
\begin{align*} y_{2n} &= y_1+\sum_{k=1}^{2n-1} (y_{k+1}-y_{k}) \geq y_1+ \sum_{j=1}^{n} (x_{2j}-x_{2j-1}) \\ &= y_1+ \sum_{j=1}^n \left( \frac{1}{2j} + \frac{1}{2j-1} \right) \geq y_1 + \frac{1}{2} \sum_{j=1}^n \frac{1}{j}. \end{align*}
Hence, we get that $\lim_{n\rightarrow \infty} y_{2n} = \infty$ and $(y_n)_n$ cannot be a Cauchy sequence in $\mathbb{R}$.