4
$\begingroup$

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?

  • 0
    @GEdgar: How does one prove a definition?2012-08-15

1 Answers 1

11

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 $0$.

  • 0
    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