- #1

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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.