# If $\lim\limits_{n\to \infty}|x_n|= 0$ then $\sup\limits_{n\geq 1}|x_n|<\infty$

I'm self-studying Functional analysis and in one of the proofs, I have the following conclusion which I don't quite understand.

If $$\lim\limits_{n\to \infty}|x_n|= 0$$ then $$\sup\limits_{n\geq 1}|x_n|<\infty$$

I suppose $$\lim\limits_{n\to \infty}|x_n|= 0$$ implies that $$|x_n|<\epsilon,\;\;\forall\,n\geq N,$$ for some $$N$$. So, taking sup over $$\{x_n\}_{n\geq N},$$ we have \begin{align} \sup\limits_{n\geq N}|x_n|\leq\epsilon<\infty.\end{align} So, how come the result? Could someone explain?

• Mike.You forgot the $1\le n \lt N$ elements |x_n|.Does this change anything? Nov 29 '18 at 15:35
• @Peter Szilas: Didn't realize that! Thanks! Nov 29 '18 at 15:36
• Small matter:)Welcome. Nov 29 '18 at 15:39

Since $$x_n \to 0$$ then exists $$\bar n$$ such that forall $$n \ge \bar n$$

$$0\le |a_n|\le 1$$

now take

$$a_{max}=\max\{|a_n|:n=1,2,\ldots,\bar n\}$$

and indicate with $$M=\max\{1,a_{max}\}$$ then

$$0\le |a_n|\le M$$

Hint: there are finitely many terms $$a_n$$ with $$n