if d(xi,xi+1)<1/2i then xi is a Cauchy sequence
Lemma 1.
Suppose x1,x2,…, is a sequence in a metric space.
If for some N≥1, we have d(ai,ai+1)<1/2i for all i≥N,
then {xi} is a Cauchy sequence.
Proof.
Let us denote by d the metric function. If ε>0, then for some N∈ℕ we have 1/2N<ε. Thus, if N<m<n we have
d(xm,xn) | ≤ | d(xm,xm+1)+⋯+d(xn-1,xn) | ||
= | (12)m+⋯+(12)n-1 | |||
= | (12)m-1n-m∑i=1(12)i | |||
= | (12)m-11-(12)n-m1-12 | |||
< | (12)m | |||
< | (12)N | |||
< | ε, |
where we have used the triangle inequality and the
geometric sum formula (http://planetmath.org/GeometricSeries).
∎
Title | if d(xi,xi+1)<1/2i then xi is a Cauchy sequence |
---|---|
Canonical name | IfDxiXi112iThenXiIsACauchySequence |
Date of creation | 2013-03-22 14:37:31 |
Last modified on | 2013-03-22 14:37:31 |
Owner | matte (1858) |
Last modified by | matte (1858) |
Numerical id | 5 |
Author | matte (1858) |
Entry type | Result |
Classification | msc 26A03 |
Classification | msc 54E35 |