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!

  • $\begingroup$ 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 $\varepsilon$. $\endgroup$ Dec 16, 2019 at 13:09
  • $\begingroup$ @yvesDaoust It is said that they are Cauchy sequence of rational numbers. $\endgroup$
    – Jeanba
    Dec 16, 2019 at 13:12
  • $\begingroup$ @Jeanba: ok, thanks. $\endgroup$
    – user65203
    Dec 16, 2019 at 13:13

1 Answer 1


$$|a_jb_j-a_kb_k|=|b_j(a_j-a_k)+a_k(b_j-b_k)|\le |b_j||a_j-a_k|+|a_k||b_j-b_k|\le A\delta+B\delta'$$ 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 $\delta=\dfrac{\delta''}{2A},\delta'=\dfrac{\delta''}{2B}$.


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