Talk:Completeness Criterion (Metric Spaces)

Is Proof 1 vaild?
In proof 1, the assumption of the existence of such an $M$ seems to be invalid. For example, for all $m, n \in \N$, let $x_n = 0$ and $\displaystyle y_{n,m} = \frac n m$. Comments? Abcxyz 09:59, 10 March 2012 (EST)


 * It is not too hard to show that one can construct $y_{n,m}$ with the desired properties. However, as you correctly pointed out, not every sequence $y_{n,m}$ satisfies the desired property (but by defining $y'_{n,m} = y_{n,m+p}$ for suitable $p \in \N$, this may be overcome). The validity of the proof isn't threatened, but comments need to be made on this subtleties. --Lord_Farin 10:16, 10 March 2012 (EST)