Terence Tao, Analysis 1. Exercise 5.3.2. Real Numbers and Cauchy Sequences. 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&gt; 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 $d(a_n,a_m) &lt; \sqrt{\epsilon}$  for every $n,m &gt; N$. 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) &lt; \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.

EDIT.

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.
${\bf 1.}$ 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
$$|a_mb_m-a_nb_n|\leq |a_m|\&gt;|b_m-b_n|+|b_m|\&gt;|a_m-a_n|\ .\tag{1}$$

Since Cauchy sequences are bounded there is a common bound $C&gt;0$ for the $|a_n|$ and the $|b_n|$. Let an $\epsilon&gt;0$ 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

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

${\bf 2.}$ A similar argument, using the inequality
$$|a_n'b_n-a_n b_n|\leq C\&gt;|a_n'-a_n|\ ,$$
proves that $x'y\sim xy$ when $x'\sim x$.
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 $n &gt; M$, then $a_M - \epsilon &lt; a_n &lt; a_M + \epsilon$ and a similar statement for $b_n$.  Then:

$$
(a_m - \epsilon)(b_m - \delta) &lt; a_n b_n &lt; (a_m + \epsilon)(b_m + \delta)  $$

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 $y$.