Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Metric space question?
Let {a_n} (n=1 to infinity) and {b_n} (n=1 to infinity)
be Cauchy sequences in a metric
space (M; rho): Prove that
{rho(a_n; b_n)} (n=1 to infinity) is Cauchy in R
:
I understand what it means to be cauchy in this instance. Since each sequence is cauchy, it means that the distance to any other point in that particular sequence is less than epsilon. And I need to prove that the distance from any point in one sequence within epsilon of any point of the other sequence.
So, I think, I just need to define the given sequences to be less than epsilon/something (probably 2). take that rho(a_n,b_n) a and expand say |a_n - b_n| = |a_n +/- something - b_n|, then use the triangle inequality, get epsilon, and we are done.
Just not sure what I'm not seeing in order to finish this up.
1 Answer
- ?Lv 710 years ago
Let epsilon > 0.
Since {a_n} and {b_n} are Cauchy, there exists N_1 and N_2 such that
rho(a_m, a_n) < epsilon/2 for all m, n > N_1, and
rho(b_m, b_n) < epsilon/2 for all m, n > N_2.
Thus if we let N = max(N_1, N_2),
rho(a_m, a_n) < epsilon/2 for all m, n > N, and
rho(b_m, b_n) < epsilon/2 for all m, n > N.
Thus for all m, n > N,
|rho(a_m, b_m) - rho(a_n, b_n)|
= |rho(a_m, b_m) - rho(a_n, b_n)|
= |rho(a_n, b_m) - rho(a_m, b_m) + rho(a_n, b_n) - rho(a_n, b_m)|
<= |rho(a_n, b_m) - rho(a_m, b_m)| + |rho(a_n, b_n) - rho(a_n, b_m)| from the triangle inequality
<= rho(a_m, a_n) + rho(b_m, b_n) from a consequence of the triangle inequality
< epsilon/2 + epsilon/2
= epsilon
Therefore, {rho(a_n, b_n)} is Cauchy in R.
Lord bless you today!
