- #1
GcSanchez05
- 17
- 0
I am just posting this because my prof sent me an email with weird letters and I want to see if i can read it here...Please disregard unless you can tell me where I can go to copy and paste this so it makes sense
WLOG assume both secuences are bounded by the same number M > 0. Then, choose \epsilon' = (\epsilon)/(2M). For \epsilon' there is n_1, and n_2 such that for
n, m > n_1 ---> |x_n - x_m|< \epsilon' (the sequence <x_n> is Cauchy)
as well as
n, m > n_2 ---> |y_n - y_m| < \epsilon' (<y_n> is Cauchy)
But then for n_0 = max {n_1, n_2} we have
|x_ny_n - x_my_m| < M(|x_n - x_m| + |y_n - y_m|) < M(\epsilon' + \epsilon')
= M((\epsilon)/(2M) + (\epsilon)/(2M)) = \epsilon.
Done.
WLOG assume both secuences are bounded by the same number M > 0. Then, choose \epsilon' = (\epsilon)/(2M). For \epsilon' there is n_1, and n_2 such that for
n, m > n_1 ---> |x_n - x_m|< \epsilon' (the sequence <x_n> is Cauchy)
as well as
n, m > n_2 ---> |y_n - y_m| < \epsilon' (<y_n> is Cauchy)
But then for n_0 = max {n_1, n_2} we have
|x_ny_n - x_my_m| < M(|x_n - x_m| + |y_n - y_m|) < M(\epsilon' + \epsilon')
= M((\epsilon)/(2M) + (\epsilon)/(2M)) = \epsilon.
Done.