real analysis – If ${a_n} sim {u_n}$ and ${b_n} sim {v_n}$, then ${a_nb_n} sim {u_nv_n}$.

I want to prove that if ${a_n} sim {u_n}$ and ${b_n} sim {v_n}$, then ${a_nb_n} sim {u_nv_n}$, where each sequence is a Cauchy sequence of rationals.

Here is my attempt.


Since all the given sequences are Cauchy, they are bounded by some number. Choose M to be the max of these bounds, that way it bounds all the sequences.

Both ${a_n-u_n}$ and ${b_n-v_n}$ converge to zero. In other words, for all positive rational epsilon, there exists positive natural numbers $N_1$ and $N_2$, for all $nin mathbb{N}^+$, such that if $n geq max{N_1, N_2}$, then ${a_n}$ and ${b_n}$ are $frac{epsilon}{2M}$ close to ${u_n}$ and ${v_n}$, respectively. So we get,

$|a_nb_n – u_nv_n| = |a_nb_n -a_nv_n +a_nv_n – u_nv_n|$

$|a_nb_n – u_nv_n| = |a_n(b_n-v_n) + v_n(a_n – u_n)|$

$|a_nb_n – u_nv_n| leq |a_n||b_n-v_n| + |v_n||a_n-v_n|$

Now, since ${a_n}$ and ${v_n}$ are bounded by $M$ it follows that:

$|a_nb_n – u_nv_n| leq M|b_n-v_n| + M|a_n-v_n|$, and we know that the sequences are eventually $frac{epsilon}{2M}$ close so…

$|a_nb_n – u_nv_n| leq M|b_n-v_n| + M|a_n-v_n| leq Mfrac{epsilon}{2M} + Mfrac{epsilon}{2M} = epsilon$.

Therefore ${a_nb_n-u_nv_n}$ converges to zero and ${a_nb_n} sim {u_nv_n}$.