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I need to prove that if $a_k\rightarrow a$ and $b_k\rightarrow b$ in $\mathbb{R}$ then $a_kb_k\rightarrow ab$. So far I have that if $a_k\rightarrow a$ then $\exists$ some number $n_1$ s.t. if $k \geq n_1, |a_k-a|<\epsilon/2$, and $\exists$ some number $n_2$ s.t. if $k\geq n_2$ then $|b_k-b|<\epsilon/2$. \ So suppose I take a number $k>\mathrm{max}\{n_1,n_2\}.$ Then we should have that \ $|a_kb_k-ab|\leq ??\\$ Here is where I get stuck. Is there some trick to this proof I am not seeing?

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    $\begingroup$ Hint: $a_kb_k-ab=a_kb_k-a_kb+a_kb-ab$ and note that $a_k$ is bounded. $\endgroup$ – John B Feb 2 '16 at 1:41
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Let $\epsilon > 0$.

Let $\epsilon_2 = \epsilon/2|a|$ if $|a| \ne 0$ or $\epsilon/2$ otherwise and Let $\epsilon_1 = \epsilon/2(|b| + \epsilon_2)$.

Or don't. The $\epsilon$s are arbitrary so we can tweak them later. The idea is we will get a result bounded by a value expressible in terms of $\epsilon$, $|a|$, and $|b|$ that can be made arbitrarily small. The details aren't worth sweating.

$a_k \rightarrow a$ so for any $\epsilon_1 > 0$ there exists an $N_a$ such that $k > N_a$ implies $|a_k - a| < \epsilon_1$.

$b_k \rightarrow b$ so for any $\epsilon_2 > 0$ there exists an $N_b$ such that $k > N_b$ implies $|b_k - b| < \epsilon_2$.

So $k > \max(N_a, Nb)$ implies

$|a_kb_k - ab| \le |a_kb_k - ab_k| + |ab_k - ab| = |b_k||a_k - a| + |a||b_k - b| < \epsilon_1*|b_k| + \epsilon_2*|a| \le \epsilon_1*(|b_k - b| + |b|) + |a|\epsilon_2 < \epsilon_1(|b| + \epsilon_2) + |a|\epsilon_2 \le \epsilon/2 + \epsilon/2 = \epsilon$

=== 2nd answer (more straight forward)===

I assume we've shown if $a_n \rightarrow a$ then $(a_n + c)\rightarrow (a + c)$ (because $|(a_n + c) - (a +c)| = |a_n - a|$ so $|a_n - a| < \epsilon \implies |(a_n + c) - (a +c)| < \epsilon$)

Likewise $(a_nc) \rightarrow ac$ (if $c = 0$ $|a_nc - ac| = 0 < \epsilon; \forall \epsilon > 0, n$. And if $c \ne 0$, if $|a_n - a| < \epsilon/|c|$ then $|a_nc - ac| < \epsilon$.)

And, of course, if $b_n \rightarrow b$ then $(b_n + a_n) \rightarrow (b + a)$. (because $n > N \implies $|b_n - b| < \epsilon/2; n > M \implies $|a_n - a| < \epsilon/2$ then $n > \max(N,M) \implies |(b_n + a_n) - (b + a)| \le |b_n - b| + |a_n - a| < \epsilon$.)

So now

$(a_nb_n - ab) = (a_n - a)(b_n - b) - 2ab + ab_n + a_nb$

$= (a_n - a)(b_n - b) + a(b - b_n) + b(a - a_n)$.

We know, $b_n \rightarrow b$ so $(a(b - b_n)) \rightarrow 0$. Likewise $(b(a-a_n)) \rightarrow 0$.

And we can figure out: For $\epsilon > 0$, there are $N, M$ such that $k > \max(N,M)$ then $|a_n - a| < \sqrt{\epsilon}$ and $|b_n - b| < \sqrt{\epsilon}$

So $|(a_n - a)(b_n - b) - 0| = |a_n - a||b_n - b| < \epsilon$. So $(a_n -a)(b_n - b) \rightarrow 0$.

So $(a_nb_n - ab)= [(a_n - a)(b_n - b) + a(b - b_n) + b(a - a_n)]$ and $[(a_n - a)(b_n - b) + a(b - b_n) + b(a - a_n)] \rightarrow 0$ so $a_nb_n \rightarrow ab$.

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  • $\begingroup$ Edits needed (many typos) $\endgroup$ – PyRulez Feb 2 '16 at 3:35
  • $\begingroup$ Strictly speaking, it won't receive full-mark if it appears in an undergraduate analysis answer book since it's possible $|b|=0$ or $a=0$, hence the division doesn't make sense and $N_a$ satisfying the condition doesn't exist. $\endgroup$ – John Feb 2 '16 at 3:54
  • $\begingroup$ $$$$Good point. $\endgroup$ – fleablood Feb 2 '16 at 3:55
  • $\begingroup$ The division isn't needed. I only put them because I didn't want "..< |b|\epsilon$ although that would have been acceptable. $\endgroup$ – fleablood Feb 2 '16 at 3:58

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