# Mulitplication of Cauchy sequences is well defined

Is the Following Proof Correct? My Proof concerns the implication $x = x'\implies xy = x'y$ in the proposition below.

Proposition Let $x = \text{lim}_{n\to\infty}a_n$, $y = \text{lim}_{n\to\infty}b_n$, and $x' = \text{lim}_{n\to\infty}a_n'$ be real numbers. Then $xy$ is also a real number. Furthermore, if $x = x'$, then $xy = x'y$.

Proof. Assume that $x'=x\in\mathbf{R}$ and $y\in\mathbf{R}$. Let $\epsilon>0$, by hypothesis $(a_n)_{n=0}^{\infty}$ and $(a_n')_{n=0}^{\infty}$ are equivalent and $(b_n)_{n=0}^{\infty}$ is Cauchy-Sequence therefore for some $N\in\mathbf{N}$ we have $\forall n\ge N(|a_n-a_n'|\leq\frac{\epsilon}{M})$ where $M\in\mathbf{Q}^+$ is a bound for $(b_n)_{n=0}^{\infty}$ whose existence is guarteed by Lemma $\textbf{5.1.15}$. Now given an arbitrary $k\in\{N,N+1,N+2,\dots \}$ we have $|a_n-a'_n|\leq\frac{\epsilon}{M}$ and so $$|a_nb_n-a_n'b_n| = |b_n|\cdot|a_n-a_n'|\leq |b_n|\cdot\frac{\epsilon}{M}$$ but $|b_n|\leq M$ which implies $0<|b_n|\cdot\frac{1}{M}\leq 1$ and so $|b_n|\cdot\frac{\epsilon}{M}\leq\epsilon$ we may therefore conclude that $|a_nb_n-a_n'b_n|\leq\epsilon$ indicating that the sequences $(a_nb_n)_{n=0}^{\infty}$ and $(a_n'b_n)_{n=0}^{\infty}$ are equivalent we may therefore surmize that $xy=x'y$.

$\blacksquare$

Note:

• This problem takes place in the context of defining real numbers as limits of Cauchy-Sequences.
• Lemma $\textbf{5.1.15}$ says that given a cauchy sequence $(a_n)_{n=0}^{\infty}$ there exists a positive rational $M$ such that $|a_i|\leq M,\forall i\in\mathbf{N}$.
• I know a little bit of analysis but I am not familiar with the Cauchy sequence construction of the real numbers. If you haven't defined the real numbers yet, what does $\lim_{n \to \infty} a_n$ mean if this sequence tends to an irrational number? – Ovi Jun 20 '18 at 1:58
• Where you say that this is in the context of defining real numbers as limits of Cauchy sequences, I think you mean defining real numbers as equivalence classes of Cauchy sequences? After all, the whole point of the construction is that the Cauchy sequences don't have limits before the construction, so it's unclear what defining real numbers as their limits would mean. – joriki Jun 20 '18 at 2:11
• Although the OP should explain the point raised in the comments, one possibility is that this question might be written in the context of Terence Tao's "Analysis I", which uses a formal notation $LIM a_n$ to represent the equivalence class of a Cauchy sequence $(a_n)$. – Lee Mosher Jun 20 '18 at 3:20
• My Apologies to all the participants here for my tardiness and slopiness. Special thanks to @LeeMosher. Indeed We define the real number $x = \text{LIM}_{n\to\infty}a_n$ to be representative of the equivalence class of all equivalent sequences of rationals. My Apologies for the confusion. Thankyou all for your patience. – Atif Farooq Jun 20 '18 at 5:25