# Cauchy

• Sep 25th 2007, 05:17 PM
taypez
Cauchy
Prove that the sum of two Cauchy sequences is Cauchy without using:

Let {x_n} be a sequence of real numbers. Then {x_n} ic Cauchy iff {x_n} converges.
• Sep 25th 2007, 05:29 PM
Jhevon
Quote:

Originally Posted by taypez
Prove that the sum of two Cauchy sequences is Cauchy without using:

Let {x_n} be a sequence of real numbers. Then {x_n} ic Cauchy iff {x_n} converges.

define $\displaystyle \{ z_n \} = \{ x_n \} + \{ y_n \}$. we need to show that: for all $\displaystyle \epsilon > 0$, there exists a $\displaystyle N \in \mathbb {N}$, such that $\displaystyle m,n > N$ implies $\displaystyle |z_n - z_m| < \epsilon$

hint: you will need the triangle inequality here

can you continue?
• Sep 25th 2007, 05:45 PM
taypez
That would lead me to use the Cauchy Theorem which it said not to and since the Cauchy Thm uses Bolzano-Weierstrass, I'm assuming I can't use that either.
• Sep 25th 2007, 05:54 PM
Jhevon
Quote:

Originally Posted by taypez
That would lead me to use the Cauchy Theorem which it said not to and since the Cauchy Thm uses Bolzano-Weierstrass, I'm assuming I can't use that either.

no it would not. did you try it? all you need is the definition of Cauchy and the triangle inequality, nothing else. and that is allowed for this question
• Sep 25th 2007, 06:00 PM
ThePerfectHacker
Quote:

Originally Posted by taypez
Prove that the sum of two Cauchy sequences is Cauchy without using:

Let {x_n} be a sequence of real numbers. Then {x_n} ic Cauchy iff {x_n} converges.

It is a trivial problem.

$\displaystyle |x_n+y_n - x_m - y_m|\leq |x_n-x_m|+|y_n-y_m| < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$.
• Sep 25th 2007, 08:25 PM
Jhevon
Bored.

Let $\displaystyle \{ x_n \}$ and $\displaystyle \{ y_n \}$ be Cauchy sequences.

Define $\displaystyle \{ z_n \} = \{ x_n \} + \{ y_n \}$. We show that $\displaystyle \{ z_n \}$ is Cauchy.

Since $\displaystyle \{ x_n \}$ is Cauchy, for all $\displaystyle \epsilon > 0$, there exists an $\displaystyle N_1 \in \mathbb {N}$ such that $\displaystyle m,n > N_1$ implies $\displaystyle |x_n - x_m|< \frac {\epsilon}2$.

Similarly, since $\displaystyle \{ y_n \}$ is Cauchy, we can find an $\displaystyle N_2 \in \mathbb{N}$ such that $\displaystyle m,n > N_2$ implies $\displaystyle |y_n - y_m|< \frac {\epsilon}2$

Now, fix such an $\displaystyle \epsilon > 0$, and choose $\displaystyle N = \mbox {max} \{ N_1, N_2 \}$. Then $\displaystyle m,n > N$ implies that:

$\displaystyle |z_n - z_m| = |(x_n + y_n) - (x_m + y_m)| = |(x_n - x_m) + (y_n - y_m)|$ $\displaystyle \le |x_n - x_m| + |y_n - y_m| < \frac {\epsilon}2 + \frac {\epsilon}2 = \epsilon$

Thus, $\displaystyle \{ z_n \}$ is a Cauchy sequence

QED.