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.
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.
Best Answer
${\bf 1.}$ 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 $x'\sim x$.