Prove that $(x_{n})_{n\in\mathbb{N}}$ is Cauchy iff $(y_{n})_{n\in\mathbb{N}}$ is Cauchy.

Question

Exercise

Let $(X,d_{X})$ be a metric space and let $(x_{n})_{n\in\mathbb{N}}$ and $(y_{n})_{n\in\mathbb{N}}$ be two sequences such that $d(x_{n},y_{n}) < \frac{1}{n+1}$ for every $n\in\mathbb{N}$.

Prove that $(x_{n})_{n\in\mathbb{N}}$ is Cauchy iff $(y_{n})_{n\in\mathbb{N}}$ is Cauchy.

My attempt

Suppose that $x_{n}$ is a Cauchy sequence. This means that \begin{align*} (\forall\varepsilon > 0)(\exists N\in\mathbb{N})(\forall m,n\in\mathbb{N})(m\geq n\geq N \Rightarrow d_{X}(x_{m},x_{n}) < \varepsilon). \end{align*}

I do not know to proceed, but I think the following inequality may be useful:

\begin{align*} d_{X}(y_{m},y_{n}) & \leq d_{X}(y_{m},x_{m}) + d_{X}(x_{m},x_{n}) + d_{X}(x_{n},y_{n})\\ \end{align*}

Can anyone suggest me how finish the proof?

Answer 1

Let's start here:

$$\begin{align*} d_{X}(y_{m},y_{n}) & \leq d_{X}(y_{m},x_{m}) + d_{X}(x_{m},x_{n}) + d_{X}(x_{n},y_{n})\\ \end{align*}$$

• Since $(x_n)$ is Cauchy, the middle term goes to zero as $n,m \to \infty$
• By assumption, we have that $$d(x_{m},y_{m}) < \frac{1}{m+1} \qquad d(x_{n},y_{n}) < \frac{1}{n+1}$$ and hence $d(x_m,y_m),d(x_n,y_n) \to 0$ as $n,m \to \infty$

Consequently, $d(y_m,y_n) \to 0$ as $m,n \to \infty$.

I'll let you fill in the details as to why it works; this is just the rough sketch proof of the matter.