1. ## More Cauchy Sequences

If (Xsubn) and (Ysubn) are Cauchy Sequences, show that (Xsubn + Ysubn) and (Xsubn*Ysubn) are Cauchy Sequences. I am asked to prove this using the definition of a Cauchy Sequence only. Any help is much appreciated.

2. Originally Posted by jkru
If (Xsubn) and (Ysubn) are Cauchy Sequences, show that (Xsubn + Ysubn) and (Xsubn*Ysubn) are Cauchy Sequences. I am asked to prove this using the definition of a Cauchy Sequence only. Any help is much appreciated.
i will do the first (since it is easier ) you do the second

since $\displaystyle \{ x_n \}$ is a Cauchy sequence, we have that for all $\displaystyle \epsilon > 0$ there exists $\displaystyle N_1 \in \mathbb{N}$, such that $\displaystyle n,m > N_1$ implies $\displaystyle |x_n - x_m| < \frac {\epsilon}2$

similarly, we have $\displaystyle N_2 \in \mathbb{N}$ so that $\displaystyle n,m > N_2$ implies $\displaystyle |y_n - y_m| < \frac {\epsilon}2$

Now, choose $\displaystyle N = \text{max} \{ N_1,N_2 \}$. then $\displaystyle n,m > N$ implies

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

so that $\displaystyle \{ x_n + y_n \}$ is Cauchy

(note that the $\displaystyle \le$ follows by the triangle inequality)

now do the second. use the defintion as i did