Wrong inequality in simple proof Cauchy $\implies$ Boundedness.

Question

These notes from Oxford University contain an apparently very simple proof that Cauchy sequences (real or complex) imply boundedness.

I understand the Cauchy condition $|a_m - a_n| < \epsilon$, and that the proof assigns $\epsilon=1$ as an arbitrary value.

Question: I can't understand how the following inequality is derived from the triangle inequality:

$$|a_m| \leq 1 + |a_N|$$

My Attempt: I have tried using the reverse triangle inequality with no success:

$$|a_m| - |a_N| \leq |a_n -a_N| < 1$$

And so,

$$|a_m| < 1 + |a_N|$$

Here the inequality is $<$ and not $\leq$ as per the reproduced notes.

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For convenience, the proof is reproduced below.

Answer 1

There is nothing wrong with writing $a\leq b$ when you can prove that $a<b.$

For example, we can say:

$$1+x^2\geq 0,$$ even though we know $1+x^2>0.$

Saying that, for all $x,$ $$f(x)\leq g(x)$$ does not mean that for some $x_0,$ $f(x_0)=g(x_0).$ It also doesn’t mean that for some $x_1,$ $f(x_1)<g(x_1).$

It means literally “less than or equal to.”

That is, for every $x,$ either $f(x)<g(x)$ or $f(x)=g(x).$

It could be $<$ for every $x.$ It could be $=$ for every $x.$ Or it could be a mix. Or we might have no way of telling.

Answer 2

Edit: Assuming your sequence converges: In case you're interested in a different, relatively-straightfoward argument: Say your sequence converges to $L$. Then all-but-finitely-many terms will be within $\epsilon$ of $L$. All these are contained in the finite ball $B(L, \epsilon)$ . Now you just need to consider the first $N$ terms; $N$ of course finite, so that the set $\{ a_1, a_2,..., a_N \}$ is bounded, say by $M$...Then you can find a ball of finite radius that containes all terms in the sequence.