Can we guarantee that there exists an $\epsilon' > 0$ such that holds for this inequality?

Question

I am currently trying to prove the multiplicative limit law:

let $(a_n)^{\infty}_{n=m}, (b_n)^{\infty}_{n=m}$ be convergent sequences of real numbers, and $X, Y$ be the real numbers $X = \lim_{n\to \infty}a_n$ and $Y = \lim_{n\to \infty}b_n$. $$ \lim_{n \to \infty}a_nb_n = \left(\lim_{n\to \infty}a_n\right) \cdot \left(\lim_{n\to \infty}b_n\right) $$

Since both $(a_n)^{\infty}_{n=m}$ and $(b_n)^{\infty}_{n=m}$ are convergent to X and Y respectively, We know that $|a_n - X| \leq \epsilon'$ and $|b_n - Y| \leq \delta$.

We also know, by some lemma we proved earlier in the book, that $|a - b| \leq \epsilon \land |c - d| \leq \delta \implies |ac - bd| \leq \epsilon \cdot |c| + \delta \cdot |a| + \epsilon \delta$.

This is perfect, as I can use it to show that $|a_nb_n - XY| \leq \epsilon$ for some arbitary $\epsilon > 0$, as long as I show that there exists $\epsilon' * |Y| \leq \frac{\epsilon}{3}$ and that there exists some $0 < \delta < 1$ such that $\delta \cdot (|X| + \epsilon') \leq \frac{2}{3}\epsilon$

I could prove the first part using the Archimedean property of the reals, but I am not so sure about the second part. The second part feels like it should work since we can choose an arbitrarily small $\delta$, but I can't prove that it does. Am I doing something wrong? is it possible to change this proof a bit to make it work?

Answer 1

If $a_n \to a, b_n \to b$ then there is some $M$ such that $|a|,|b_n| \le M$.

Then $|a_nb_n -ab| = |a_nb_n -a b_n + a b_n -ab| \le |a_n-a| |b_n| + |a| |b_n-b| \le M (|a-a_n|+ |b-b_n|)$.

Now choose $N$ big enough so that $|a-a_n|, |b-b_n| < {\epsilon \over 2 M}$.