# 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.

Proof:

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 $${a_nb_n} sim {u_nv_n}$$.