convergence vs cauchy sequence

Question

I was looking at a proof that every sequence in $\mathbb{R}$ converges if and only if it is a Cauchy sequence.

It starts like this:

Suppose a sequence ($a_n$) converges to $A$, fix $\epsilon > 0$, then there exists an $N$ s.t. for every $n,m \ge N$, the inequalities $|a_n - A| < \epsilon/2$ and $|a_m - A| < \epsilon/2$ hold.

I got hung up on this part:

$$|a_n - a_m| \le |a_n - A| + |a_m - A| < \epsilon.$$

Can someone explain this inequality please?

Answer 1

Simply the triangle inequality. Note $|a-b|=|b-a|$.`

Answer 2

The inequality you provided is a consequence of the triangle inequality https://en.wikipedia.org/wiki/Triangle_inequality.

Starting with what you have:

$$|a_n - a_m| \le |a_n - A| + |a_m - A|$$

Now, given that for every $n, m \ge \Bbb N$, you have $$|a_n - A| < \frac{\epsilon}{2}$$ and $$|a_m - A| < \frac{\epsilon}{2}$$ you can substitute these into the inequality:

$$|a_n - a_m| \le |a_n - A| + |a_m - A| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

So, the inequality $$|a_n - a_m| < \epsilon$$ holds.

This is a common technique used in real analysis and metric space arguments.