Why we choose $\epsilon_1 <\min \left(1, \frac{\varepsilon}{2 M}\right)$ and $\epsilon_2<\min \left(1, \frac{\varepsilon}{2 (|a|+1)}\right)$

Question

I am going through the proof of the product of convergent sequences. Given that $a_n \to a$ and $b_n \to b$, we need to prove that $a_nb_n \to ab$

We have $\forall \epsilon_1 >0, \exists N_1 \in \mathbb{N}$ such that $$|a_n-a|<\epsilon_1$$ Also We have $\forall \epsilon_2 >0, \exists N_2 \in \mathbb{N}$ such that $$|b_n-b|<\epsilon_2$$ Now since $\left\{b_n\right\}$ is bounded, $\exists M>0$ such that $|b_n|\leq M,\:\forall n \in \mathbb{N}$

Now by $\Delta$ inequality we have $$|a_nb_n-ab|\leq |b_n||a_n-a|+|a||b_n-b|$$

Now for a given $\epsilon >0$, the author chooses

$\epsilon_1 <\min \left(1, \frac{\varepsilon}{2 M}\right)$ and $\epsilon_2<\min \left(1, \frac{\varepsilon}{2 (|a|+1)}\right)$

Is there any reason to choose epsilons using min function? I guess it is to make $\epsilon_1,\epsilon_2$ as small as possible.

Answer 1

You want to look at the worst case scenario in your sum. We know $|b_n|<M$ and $|a_n-a|<\frac \epsilon {2M}$. Multiplying those gets you to a product that is less than $\frac \epsilon 2$ Similarly, multiplying the second product gets you to less than $\frac \epsilon 2$. Adding them gets you to your desired $\epsilon$. Normally the minimum of 1 is put on to avoid pathological cases where a too high initial choice messes things up and you are always allowed to use a smaller value than needed. Not sure if there's actually a need for it here

Answer 2

Addendum added that shows how I would prove that
$a_n \to a, ~b_n \to b \implies (a_nb_n) \to (ab).$

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For ease of analysis, assume $a,b > 0$, and any $a_n, b_n > 0$.

Edit
See the other Edit remark below, for a justification of these assumptions.

You know that:

$a - \epsilon_1 < a_n < a + \epsilon_1$ and
$b - \epsilon_2 < b_n < b + \epsilon_2$ and

This implies that
$$ab - a\epsilon_2 - b\epsilon_1 + \epsilon_1\epsilon_2 < a_nb_n < ab + a\epsilon_2 + b\epsilon_1 + \epsilon_1\epsilon_2.\tag1 $$

Edit
To a certain extent, I am pulling a fast one, because I am glossing over the fact that one or both of $(a - \epsilon_1), (b - \epsilon_2)$ could be negative. However, the point of this answer is simply to explain the thinking behind the $\min(...)$ constraints, rather than rigorously proving that these constraints work.

What you want is that: $$ab - \epsilon < a_nb_n < ab + \epsilon.\tag2 $$

So, within the parameters of the above Edit remark, the following analysis verifies that the $\min(...)$ constraints guarantee that (for example) if the RHS inequality in (1) above is satisfied, then the RHS inequality in (2) above is satisfied.

Analysis of the LHS inequalities should be similar.

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The $\epsilon_1 = \min(...)$ constraint guarantees that
$b\epsilon_1 < (\epsilon/2).$

Also, the $\epsilon_1 = \min(...)$ constraint guarantees that
$\epsilon_1\epsilon_2 < \epsilon_2$.

The $\epsilon_2 = \min(...)$ constraint guarantees that
$a\epsilon_2 + \epsilon_1\epsilon_2 < (a+1)\epsilon_2 < (\epsilon/2).$

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Addendum
Personally, judging by the OP's presentation of the question, I think that the underlying proof, from whatever textbook is being used, was rather obtusely written.

This is how I would approach it:

Given any $\delta > 0$ there exists $N_1, N_2$ that satisfy the following:

• for all $n \geq N_1, |a - a_n| < \delta.$
• for all $n \geq N_2, |b - b_n| < \delta.$

Let $N$ denote $\max(N_1, N_2)$. Then, for all $n \geq N$, you have that both of the above inequalities are satisfied.

It is desired to find a relationship between $\delta$ and $\epsilon$ such that the above inequalities guarantee that

$$|ab - a_nb_n| < \epsilon. $$

If $b = 0 = a$, then $|ab - a_nb_n| = |a_nb_n| < \delta^2$.
Then, simply take $\delta = \sqrt{\epsilon}.$

If $b = 0 \neq a$, then $|ab - a_nb_n| < |a_n| \delta.$

Then, add the constraint that $\delta \leq |a|$.
This will imply that $|a_n| < |2a|$.
This implies that $|ab - a_nb_n| < |2a|\delta.$ So, specify that

$$\delta = \min\left( ~|a|, \frac{\epsilon}{|2a|}\right).$$

Therefore, without loss of generality, $b \neq 0.$

By virtually identical analysis, without loss of generality, $a \neq 0.$

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Now, I will use the triangle inequality: $$|ab - a_nb_n| \leq |ab - ab_n| + |ab_n - a_nb_n|.$$

I know that $|ab - ab_n| = |a|\times|b - b_n| < |a|\delta.$

I know that $|ab_n - a_nb_n| = |b_n|\times|a - a_n| < |b_n|\delta.$

Therefore, I know that $|ab - a_nb_n| < \{~|a| + |b_n| ~\}\delta.$

Now I will add the artificial
Constraint-1 $~\displaystyle \delta \leq |b|.$

Since $|b - b_n| < \delta \leq |b|$,

I know that $\displaystyle |b_n| < |2b|$.

Putting this all together, I have that

$$|ab - a_nb_n| < \left\{ ~|a| + |2b| ~\right\}\delta.$$

Therefore, I want that
$\displaystyle \left\{ ~|a| + |2b| ~\right\}\delta \leq \epsilon.$

Thus, I simply need to combine the above statement with Constraint-1, as follows:

$$\delta = \min\left( ~|b|, ~\frac{\epsilon}{|a| + |2b|} ~\right).$$