Textbook Proof

The textbook offers the following proof:

First we notice that ${a_{j}}$ is bounded: say that ${a_{j}}\leqslant M$ for every j.

Let $\epsilon > 0$

Choose an integer $N > 0$ so that $\left|{a}_{j} - \alpha \right| < \frac{\epsilon }{\left ( 2M + 2\left | \beta \right | \right )}$ when $j > N$

Also choose an integer $\tilde{N}$ such that $\left|{b}_{j} - \beta \right| < \frac{\epsilon }{\left ( 2M + 2\left | \beta \right | \right )}$ when $j > \tilde{N}$

Then for $j > max\left \{ N , \tilde{N} \right \} $

$\left|{a}_{j}{b}_{j} + \alpha\beta \right| = \left|{a}_{j}\left({b}_{j} - \beta \right) + \beta\left({a}_{j} - \alpha \right) \right| \leq \left|{a}_{j}\left({b}_{j} - \beta \right)\right| + \left| \beta\left({a}_{j} - \alpha \right) \right|$

$\leq M\cdot \frac{\epsilon }{2M + 2\left | \beta \right |} + \left|\beta\right|\cdot \frac{\epsilon }{2M + 2\left | \beta \right |} \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

So the sequence $\left\{a_{j}b_{j}\right\}$ converges to $\alpha\beta\:\blacksquare$


What I'm having trouble with is understanding how the author knew to set $\left|{a}_{j} - \alpha \right| < \frac{\epsilon }{\left ( 2M + 2\left | \beta \right | \right )}$ and $\left|{b}_{j} - \beta \right| < \frac{\epsilon }{\left ( 2M + 2\left | \beta \right | \right )}$

Why and how did they author choose $\frac{\epsilon }{\left ( 2M + 2\left | \beta \right | \right )}$

  • $\begingroup$ What are the hypothesis on $a_j, b_j $? $\endgroup$ – user8469759 Sep 16 '15 at 22:35
  • $\begingroup$ That they converge to $\alpha,\beta$ respectively, I would think. $\endgroup$ – uniquesolution Sep 16 '15 at 22:36

Regarding the question why did they choose to divide $\epsilon$ by this particular number, the answer is: they first wrote this line: (note that you mistakenly wrote $+$ instead of $-$)

$$|a_jb_j-\alpha\beta|=|a_j(b_j-\beta)+\beta(a_j-\alpha)|\leq |a_j(b_j-\beta)|+|\beta(a_j-\alpha)|$$

Then they said something like, "OK, we know $a_j$ is bounded by some $M$, so we can write:

$$|a_j(b_j-\beta)|+|\beta(a_j-\alpha)|\leq M|b_j-\beta|+|\beta||a_j-\alpha|$$

Now they said "Great. Now we can make $|b_j-\beta|$ and $|a_j-\alpha|$ small if $j$ is large enough, but at the end we would like that everything will be smaller than some pre-given $\epsilon>0$. Say we knew that $|b_j-\beta|< X$,and $|a_j-\alpha|<X$. Then we would get: $$|a_jb_j-\alpha\beta|<MX+|\beta|X$$

How should we choose $X$ so that the sum will be smaller than $\epsilon$? Well, we would need to choose


So let's choose just one half of the right hand side, i.e. $X={\epsilon\over 2M+2|\beta|}$." This is more or less the reasoning.

  • $\begingroup$ I really appreciate the breakdown. Thank you very much. $\endgroup$ – speespa Sep 17 '15 at 0:31

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