# Proving product of two reals is real.

I am supposed to prove the following proposition from Tao's Analysis 1.

Proposition 5.3.10 (Multiplication is well defined). Let $$x=\operatorname{LIM}_{n\to\infty}a_n$$, $$y=\operatorname{LIM}_{n\to\infty}b_n$$, and $$x'=\operatorname{LIM}_{n\to\infty}a'_n$$ be real numbers. Then $$xy$$ is also a real number. Furthermore, if $$x=x'$$, then $$xy=x'y$$.

For context, a real number is defined to be an object of the form $$\operatorname{LIM}_{n\to\infty}a_n$$, where $$(a_n)_{n=1}^\infty$$ is a Cauchy sequence of rational numbers, and two real numbers are said to be equal if they have equivalent Cauchy sequences.

Therefore, to prove that $$xy$$ is a real number, I want to show that for all rational $$\varepsilon>0$$, there exists some $$N$$ such that $$|a_jb_j-a_kb_k|\leq\varepsilon$$ for all $$j,k\geq N$$. Since $$x$$ and $$y$$ are real, that means that

(a) For all rational $$\delta>0$$, there exists some $$M$$ such that $$|a_j-a_k|\leq\delta$$ for all $$j,k\geq M$$.

(b) For all rational $$\delta'>0$$, there exists some $$M'$$ such that $$|b_j-b_k|\leq\delta'$$ for all $$j,k\geq M'$$.

(c) We can therefore define $$N:=\max{(M,M')}$$.

I know that $$|a_jb_j-a_kb_k|\leq\delta|b_k|+\delta'|a_j|+\delta\delta'$$, and my initial idea was to somehow define $$\delta$$ and $$\delta'$$ such that $$\delta|b_k|+\delta'|a_j|+\delta\delta'=\varepsilon$$, but this approach does not seem to be working, as I cannot find satisfactory values for $$\delta$$ and $$\delta'$$. I am not sure what else I can do to the inequality to remove the $$|b_k|$$ and $$|a_j|$$ terms inside it, and do not know how to proceed.

Any help with this would be greatly appreciated, thanks for your time!

• You probably need to use the result that every Cauchy sequence is bounded to get a uniform bound on the $b_k$ and $a_j$ to then connect your $\delta$s to $\varepsilon$. Dec 16, 2019 at 13:09
• @yvesDaoust It is said that they are Cauchy sequence of rational numbers. Dec 16, 2019 at 13:12
• @Jeanba: ok, thanks.
– user65203
Dec 16, 2019 at 13:13

$$|a_jb_j-a_kb_k|=|b_j(a_j-a_k)+a_k(b_j-b_k)|\le |b_j||a_j-a_k|+|a_k||b_j-b_k|\le A\delta+B\delta'$$ where $$A,B$$ are upper bounds on $$|a_j|$$ and $$|b_k|$$ for $$j>M$$ and $$k>M'$$ (say $$|a_M|+\delta$$ and $$|b_{M'}|+\delta'$$).
To achieve a given $$\delta''$$, you can take $$\delta=\dfrac{\delta''}{2A},\delta'=\dfrac{\delta''}{2B}$$.