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For all $n\geq N$ For all $n\geq N$For all $n\geq N$For all $n\geq N$For all $n\geq N$

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closed as off-topic by Watson, Math1000, user91500, Cyclohexanol., Claude Leibovici Sep 9 '16 at 8:34

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In addition to Robert Z's answer, you can prove it by contradiction:

Suppose for every $N$ you can find $n \geq N$ sucht that $X_n = 0$, then you can easily (by induction) extract a sequence $X_{\phi(n)}$ which is identically $0$, thus converges to $0$. But all extracted sequences of $(X_n)$ must converge to $a \neq 0$, absurd.

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  • $\begingroup$ This is brilliant! Thanks =) $\endgroup$ – Renzo Mauricio Guzmán Anaya Sep 7 '16 at 15:59
  • $\begingroup$ @RenzoMauricioGuzmánAnaya since this answer seems satisfying to you, you might consider accepting it (clicking on the validation sign under the vote signs). Thanks! $\endgroup$ – P. Camilleri Sep 7 '16 at 17:05
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By definition of limit, for $\varepsilon=\frac{1}{2}\|a\|>0$, there is $N\in\mathbb{N}$ such that for $n\geq N$, $|\|X_n\|-\|a\||\leq \|X_n-a\|<\varepsilon$ which implies that $\|X_n\|>\|a\|-\varepsilon=\frac{1}{2}\|a\|>0.$

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