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I can prove the contrapositive:

$x_n$ does not tend to $0$ implies either:

  1. $x_n$ diverges (does not converge), in which case neither does $\|x_n\|$, or
  2. $x_n$ converges to $x \neq 0$ which implies $\|x_n\|$ converges to $\|x\|\neq 0$.

Okay, but is there a simpler way to do this?

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    Write down what it means to have $x_n \to 0$!2012-05-30
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    In some works, $\|x_n\| \to 0$ would be the *definition* of $x_n \to 0$. What is your definition of $x_n \to 0$?2012-05-30
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    Also, your statement 1 is wrong. It is possible that $x_n$ diverges but $\|x_n\|$ converges. Consider, in $\mathbb{R}$, the sequence $x_n = (-1)^n$.2012-05-30
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    Or worse, $x_n=e^{in}$ as a sequence of complex numbers...2012-05-30
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    Statement 1 should be "In which case ||x_n|| does not tend to 0" obviously2012-05-30
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    Oh by the way, my definition comes in the form " x_n --> x in V means that ||x_n-x|| --> 0 in F "2012-05-30
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    I don't see how in the original proof you show the claim 1 (with "$\|x_n\|$ does not tend to $0$").2012-05-30
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    My original argument is completely horrifically wrong on lots of levels...2012-05-30
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    Maybe it would be instructive to prove the more general fact: $x_n \to y$ implies $\|x_n-y\| \to 0$.2012-08-15
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    @GEdgar: How does one prove a definition?2012-08-15

1 Answers 1

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You can prove this directly:

Recall that $x_n\to x$ if for every $\epsilon>0$ there is some $n_0$ such that for all $n>n_0$ we have $\|x_n-x\|<\epsilon$.

If $\|x_n\|\to 0$ this means that $\|x_n-0\|$ tends to zero and therefore $x_n$ converges to $0$.

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    Yeah it was that last line I couldn't see. The reason this is not completely obvious to me is that you can only garuntee this works for 0. For all other points it isn't true...2012-05-30
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    @Adam: Usually writing down the definitions solves a lot of problems.2012-05-30
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    I might vote myself down for not seeing that trick - it is too easy and I should spot it even though I have only seen it the other way round before... edit: you're not allowed to vote on yourself!2012-05-30
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    Asaf: From my perspective, I have been prolongued to the continuity of ||.||, i.e. the ||x_n - x||--> 0 implies | ||x_n|| - ||x|| | --> 0 implies ||x_n|| --> ||x||. It is funny that although the other argument only works for 0 and so I would expect it to be some lengthy argument involving the Kernel etc, it is really simple... but yeah...what can I say lol2012-05-30