Let $ x = \lim_{n\rightarrow\infty}a_n, y = \lim_{n\rightarrow\infty}b_n$, and $ x' = \lim_{n\rightarrow\infty}a'_n$ be real numbers. Then $xy$ is also a real number. Furthermore, is $x=x'$, then $xy = x'y$.

Here is my attempt.

We need to show that $xy = \lim_{n\rightarrow\infty}a_n b_n$ is a real number.

from the hypothesis we know that $a_n$ and $b_n$. are eventually $\delta$ -steady sequences for every $\delta> 0 $. so we can choose $a_n$ and $b_n$ to be eventually $\sqrt{\epsilon}$-steady sequences.

since $(a_n)_{n=1}$ is eventually $\sqrt{\epsilon}$-steady, we know there is an $N \geq 1$ such that $d(a_n,a_m) < \sqrt{\epsilon}$ for every $n,m > N$. By a similar argument we can say the same for $(b_n)_{n=1}$.

by proposition 4.3.7 (h).

$d(a_nb_n,a_mb_m) < \sqrt{\epsilon}|b_n| + \sqrt{\epsilon}|a_n| + \epsilon $

From here i am not sure where to go, I am pretty sure i am going the right way, as this sort of resembles the definition of a cauchy sequence that is eventually $\sqrt{\epsilon}|b_n| + \sqrt{\epsilon}|a_n| + \epsilon $-close. I think I require something that just involves epsilon.

The second part of the question I havent attempted yet, but I thought I'd include it just in case anyone would like to lend a helping hand.

by the way propostion 4.3.7 (h). states that if x and y are $\epsilon$-close and z and w and $\delta$ close then xz and yw are $(\epsilon|z| + \delta|x| + \epsilon \delta$)-close.


A real number is defined to be an object of the form $ x = \lim_{n\rightarrow\infty}a_n$, where $a_n$ is a cauchy sequence of rational numbers.

  • $\begingroup$ The capital LIM as oppose to lim is used as a crutch to introduce limits, sorry for not accepting edits. $\endgroup$ – user197848 Jan 24 '15 at 17:34
  • 1
    $\begingroup$ Then you'll need to give a context for what LIM means. $\endgroup$ – Thomas Andrews Jan 24 '15 at 17:39
  • $\begingroup$ If you're using this definition, then actually a real number is an equivalence class of Cauchy sequences of rational numbers. $\endgroup$ – Math1000 Jan 24 '15 at 18:53
  • $\begingroup$ Thanks, would any of you guys happen to know where i have messed up? $\endgroup$ – user197848 Jan 24 '15 at 19:03
  • $\begingroup$ $ \lim n \rightarrow \infinity a_n b_n $ $\endgroup$ – user197848 Jan 29 '15 at 13:18

${\bf 1.}$ Let $a.$ and $b.$ be two Cauchy sequences of rational numbers. Then $$a_mb_m-a_nb_n=a_m(b_m-b_n)+b_n(a_m-a_n)$$ and therefore $$|a_mb_m-a_nb_n|\leq |a_m|\>|b_m-b_n|+|b_m|\>|a_m-a_n|\ .\tag{1}$$

Since Cauchy sequences are bounded there is a common bound $C>0$ for the $|a_n|$ and the $|b_n|$. Let an $\epsilon>0$ be given. Then there is an $N\in{\mathbb N}$ such that the two sequences are ${\epsilon\over 2C}$-steady for $n\geq N$. On account of $(1)$ this implies that

$$|a_mb_m-a_nb_n|\leq C{\epsilon\over 2C}+C{\epsilon\over2C}=\epsilon\qquad(m, \>n\geq N)\ ,$$ which proves that the sequence $c_n:=a_nb_n$ $\>(n\geq1)$ is Cauchy, whence defines a real number.

${\bf 2.}$ A similar argument, using the inequality $$|a_n'b_n-a_n b_n|\leq C\>|a_n'-a_n|\ ,$$ proves that $x'y\sim xy$ when $x'\sim x$.

  • $\begingroup$ Why is every answer being downvoted? $\endgroup$ – Squirtle Feb 3 '15 at 20:11
  • $\begingroup$ Thanks for the clear answer, i never organised a bounty before, i think i have just awarded it to you.... so yeah, thanks! $\endgroup$ – user197848 Feb 4 '15 at 8:52

Multiplication of Cauchy Sequences

Real numbers are being defined as Cauchy sequences rational numbers. We can multiply two sequences of numbers to get another sequence. Will it also be Cauchy?

Lemma: If $a_n$ and $b_n$ are Cauchy, then $a_n b_n$ is also Cauchy

In the case of the real numbers, Cauchy sequences are not just close, they are guaranteed to have a limit. I don't quite remember how the proof goes.

Lemma: If $a_n$ and $b_n$ are Cauchy with limits $x$ and $y$, then $a_n b_n$ is Cauchy with limit $xy$

Finally there is a question of whether two different sequences can represent the same number? What about:

$$ 1- \frac{1}{n} \hspace{0.25in}\text{ vs }\hspace{0.25in} 1 + \frac{1}{n} $$

In your case you just want to subtract the Cauchy sequences, and get $$xy - x'y = (x - x')y = 0\cdot y = 0$$

However $\mathbf{0}$ here means "a Cauchy sequence converging to $0 \in \mathbb{Q}$". So all the steps here need to be checked carefully. Can you do that?

Comment In real life sequences are definitely not Cauchy, e.g. $(-1)^n$. Or in reality $(-1)^{f(n)}$ where $f(n)$ is something totally random and may chance over time.

Think about it for a little bit, for very large $n > M$, then $a_M - \epsilon < a_n < a_M + \epsilon$ and a similar statement for $b_n$. Then:

$$ (a_m - \epsilon)(b_m - \delta) < a_n b_n < (a_m + \epsilon)(b_m + \delta) $$

The error term is like $(\epsilon b_m + \delta a_m) + \delta \epsilon$ which is Tao's proposition. How big is this error?

$$ (\epsilon (b_N + \delta) + \delta (a_M + \epsilon)) + \delta \epsilon = (\delta a_M + \epsilon b_N) + 3\delta \epsilon$$

This is getting rather expensive I suspect $\epsilon^\ast = \delta a_M + \epsilon b_N$.

Summary: to multiply two Cauchy sequences $a_nb_n$ and get to be $\epsilon$-close to $xy$, we must truncate much further than to get $a_n$ to be $\epsilon$-close to $x$ or $b_n$ to be $\epsilon$-close to $y$.


Proof Part 1 - Cauchy Sequences are Bounded

Let $\{a_n\}$ be a Cauchy sequence of rationals. Let $\epsilon>0$. Observe there exists $N \in \mathbb{N}$ such that for all $n,m>N, |a_n - a_m| < \epsilon$. Consider the ball $B_\epsilon(a_n)$ which contains $a_m$ for all $m>N$. Therefore consider the points $\{a_1, a_2, \ldots, a_N\}$. These points possibly lie outside $B_\epsilon(a_n)$. Consider now the set of distances $\{\epsilon, d(a_n,a_1), d(a_n,a_2), \ldots, d(a_n,a_N)\}$ and let $M$ be the maximum element of this set. Then $B_{M+1}(a_n)$ certainly contains all points of the sequence, hence the sequence is bounded.

Proof Part 2 - Product of Cauchy sequences is Cauchy

Let $\{a_n\}$ and $\{b_n\}$ be Cauchy sequences of rationals. By the proof of part 1, we know that each sequence is bounded, therefore let $A = \sup_n \{|a_n|\}$ and $B = \sup_n\{|b_n|\}$. If both $A = B = 0$ then define $A' = B' = 1$, otherwise let $A' = A$ and $B' = B$. We want to check that for all $\epsilon>0$ there exist $N \in \mathbb{N}$ where if $m,n > N$ then $|a_nb_n - a_mb_m| < \epsilon$. There exists $N_1, N_2 \in \mathbb{N}$ where if $n,m > N_1$ then $|a_n - a_m| < \frac{\epsilon}{A'+B'}$ and $m,n > N_2$ yields $|b_n - b_m| < \frac{\epsilon}{A'+B'}$. Let $N = \max\{N_1, N_2\}$. For $n,m > N$ we have

\begin{eqnarray*} |a_nb_n - a_mb_m| & = & |a_nb_n - a_nb_m + a_n b_m - b_nb_m| \\ & \leq & |a_nb_n - a_nb_m| + |a_nb_m - a_mb_m| \\ & \leq & |a_n|\cdot |b_n - b_m| + |b_m|\cdot|a_n - a_m| \\ & < & A' \cdot \frac{\epsilon}{A'+B'} + B \cdot \frac{\epsilon}{A'+B'} \;\; =\;\; \epsilon. \end{eqnarray*}

Proof Part 3 - $a_nb_n \to xy$

The argument is similar to part 2. Let $\{a_n\}$ and $\{b_n\}$ be Cauchy sequences of rationals such that $a_n \to x$ and $b_n \to y$. Then $\{a_nb_n\}$ is a Cauchy sequence of rationals. Let $\epsilon >0$. Let $N_1, N_2 \in \mathbb{N}$ such that for $n>N_1$ then $|a_n - x| < \frac{\epsilon}{A'+|y|}$ and $n>N_2$ yields $|b_n - y| < \frac{\epsilon}{A'+|y|}$, where $A = \sup_n \{a_n\}$ and $A'$ as defined in part 2. Let $N = \max\{N_1, N_2\}$ and $N>N$. We then find

\begin{eqnarray*} |a_nb_n - xy| & = & |a_nb_n - a_ny + a_ny - xy| \\ & \leq & |a_nb_n - a_ny| + |a_ny - xy| \\ & \leq & |a_n|\cdot |b_n - y| + |y|\cdot |a_n - x| \\ & < & A'\cdot \frac{\epsilon}{A' + |y|} + |y|\cdot \frac{\epsilon}{A'+|y|} \;\; =\;\; \epsilon. \end{eqnarray*}

Proof Part 4

Let $\{a_n\}, \{b_n\}, \{a_n'\}$ be rational Cauchy sequences where $a_n \to x, \; b_n \to y,$ and $a_n' \to x'$. Suppose $x = x'$ and consider

\begin{eqnarray*} xy - x'y & = & \lim_{n\to \infty} a_nb_n - \lim_{n\to\infty}a_n' b_n \\ & = & \lim_{n\to\infty} a_nb_n - a_n'b_n \\ & = & \lim_{n\to\infty} (a_n - a_n')b_n \\ & = & (\lim_{n\to\infty}a_n - \lim_{n\to\infty}a_n')\cdot \lim_{n\to\infty} b_n \\ & = & (x-x')y \;\; =\;\; 0. \end{eqnarray*}

  • $\begingroup$ Does someone want to explain to me why this was down voted? $\endgroup$ – Mnifldz Feb 1 '15 at 18:28
  • $\begingroup$ Your answer is impressive. I upvote you. :) $\endgroup$ – user160928 Feb 2 '15 at 0:48
  • $\begingroup$ Thanks Guillaume! It's much appreciated. I do however would like to know people's reasons for down voting. It would be petty if it were merely about the bounty... $\endgroup$ – Mnifldz Feb 2 '15 at 1:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy