Complete metric space

1. Sep 3, 2009

lukaszh

Hello,

why the set of all real numbers is complete metric space with euclidean metric? I know, that metric space is complete iff all sequences in it converges. But 1,2,3,4,... diverges.

Thanx

2. Sep 3, 2009

wofsy

Complete is if all Cauchy sequences converge, not any sequence

3. Sep 4, 2009

HallsofIvy

Staff Emeritus
It's what we "know" that gets us into trouble!

As wofsy said, a metric space is complete iff all Cauchy sequences converge, not "all sequences".

And a sequence, {an} is "Cauchy" if and only if the sequence {an- am} converges to 0 as m and n go to infinity independently. It is easy to show that any Cauchy sequence converges. The rational numbers are not "complete" because there exist Cauchy sequences that do not converge. For example, the sequence, {3, 3.1, 3.14, 3.1415, 3.14159, 3.141592, ...}, where the nth term is the decimal expansion of $\pi$ to n places, is Cauchy because the nth and mth term are identical to the min(n, m) place and so their difference goes to 0 as m and n go to infinity. But the sequence does not converge because, as a sequence in the real numbers, it converges to $\pi$ and $\pi$ is not a rational number.

Last edited: Sep 4, 2009