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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?

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

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

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Hint: there are finitely many terms $a_n$ with $n<N.$

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