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 $|a_i|\leq M,\forall i\in\mathbf{N}$.