lukaszh said:
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
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, {a
n} is "Cauchy" if and only if the sequence {a
n- a
m} 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 n
th term is the decimal expansion of [itex]\pi[/itex] to n places, is Cauchy because the n
th and m
th 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 [itex]\pi[/itex] and [itex]\pi[/itex] is not a rational number.