Showing that (yet another) sequence is a cauchy sequence

We work in the metric space $(\mathbb{R}, d_E)$ where $d_E$ is the usual (euclidian) metric.

In a certain proof, there is stated:

$\forall p,q \in \mathbb{N}:\quad(p < q \implies |a_p - a_q| \leq 2^{-p}$) hence $(a_n)_n$ is a cauchy sequence.

How does this follow? Is this the reasoning?

My attempt:

Let $\epsilon > 0$ and choose $p$ such that $2^{-p} < \epsilon$. Then, for $m,n > p$, we have:

$$|a_m - a_n| = |a_m - a_{p} + a_{p} - a_n| \leq |a_m - a_{p}| + |a_n - a_{p}| = 2^{-p} + 2^{-p} = 2.2^{-p} < 2 \epsilon$$

hence the sequence is cauchy.

• The first equality is good. The first inequality is good. The second equality should be an inequality and only holds if $p < m$ and $p < n$, which can be done by choosing by taking $n, m \geq N$ where we choose $N > p$ large. The basic idea is correct. – amarney Oct 30 '17 at 18:46
• I did take $p < m$ and $p < n$? – user370967 Oct 30 '17 at 18:53
• You explicitly say "then, for $m, n > p$", which is equivalent to saying "then, for $p < m$ and $p < n$, so yes. I missed that part when I wrote my first comment. – amarney Oct 30 '17 at 18:54
• So except the inequality there was no mistake? – user370967 Oct 30 '17 at 18:55
• Yes, I don't see anything wrong with anything you did besides that, your proof is correct. – amarney Oct 30 '17 at 18:55

Let $\epsilon > 0$ be given, and choose $p$ such that $2^{-p} < \epsilon$. Then for $n, m \geq N$ where $N > p$, we have \begin{align*} |a_m - a_n| = |a_m - a_p + a_p - a_n| \leq |a_m - a_p| + |a_p - a_n| \leq 2 \cdot 2^p < 2 \epsilon. \end{align*}
It is basically correct yes, as you were told in the comments. If you don't mind, I will add another approach. Let $\varepsilon>0$. Pick $p\in\mathbb N$ such that $2^{-p}<\varepsilon$. Then, if $m,n\in\mathbb N$ are such that $m,n\geqslant p$, there are three possibilities:
• $m=n$: then $|a_m-a_N|=0<\varepsilon$;
• $m>n$: then $|a_m-a_n|\leqslant 2^{-n}\leqslant2^{-p}<\varepsilon$;
• $m<n$: then $|a_m-a_n|\leqslant 2^{-m}\leqslant2^{-p}<\varepsilon$.