Let $(x_n)$ and $(y_n)$ be sequences such that $\lim y_n = 0$. Suppose that for all $k \in \Bbb N$ and all $m ≥ k$ we have
$|x_m − x_k| ≤ y_k$.
Show that $(x_n)$ is Cauchy.
I need a little guidance on how to approach the problem. As I see this is the same definition of Cauchy sequences. But I do not see how to connect everything in a logic sequence in order to have a rigorous proof.
My attempt of reasoning
I started first defining the $\lim$ of $y_n$.
For every $\varepsilon>0$ exists $N$ s.t. $n>N$ $|y_n|<\varepsilon$ for all $n>N$
Then I see that all terms of $y_n$ get smaller and smaller as $n$ gets larger.
So distance between $x_m$ and $x_k$ gets smaller as the terms get bigger. But one thing that puts me off is that
$m ≥ k$ and $| x_m − x_k| ≤ y_k$ why are they $\leq$?
Thanks for help in advance'
Best Answer
I think the way you started out is great. We need to show the following:
Given $\epsilon > 0$, there exists $N$ such that $n, m \geq N$ implies $|x_{n} - x_{m} | < \epsilon$.
But, as you correctly noted, given $\epsilon > 0$, $\exists N$ such that $n \geq N$ implies $|y_{n}| < \epsilon$.
But we have the inequality that for all $m \geq n$, $|x_{m} - x_{n} | \leq |y_{n}|$ (actually, this inequality also holds for all $n$, and in particular, for all $n \geq N$), which means $|x_{m} - x_{n} | < \epsilon$ for all $m \geq n$ (and all $n \geq N$). In other words, for all $n, m \geq N$, $|x_{m} - x_{n}| \leq |y_{n}| < \epsilon$.
So, given $\epsilon > 0$, we found our $N$ (the $N$ that made $|y_{n}| < \epsilon$ if $n \geq N$) such that $n, m \geq N$ implies $|x_{n} - x_{m}| < \epsilon$, which is exactly the definition of the sequence being Cauchy.