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.

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- Sep 25th 2007, 05:17 PMtaypezCauchy
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 PMJhevon
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 PMtaypez
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 PMJhevon
- Sep 25th 2007, 06:00 PMThePerfectHacker
- Sep 25th 2007, 08:25 PMJhevon
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.