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 $y_{n, m} = \dfrac n m$. 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)