Is my proof that multiplication is well-defined for reals correct?

[yucheng]

Homework Statement

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

Relevant Equations

N/A

I have referred to this page: https://taoanalysis.wordpress.com/2020/03/26/exercise-5-3-2/ to check my answer.

The way I thought of the problem:
I know $xy = \mathrm{LIM}_{n\to\infty} a_n b_n$ and I know $x'y = \mathrm{LIM}_{n\to\infty} a'_n b_n$. Thus if $xy=x'y$, maybe I can try showing that $|a_n b_n-a'_n b_n| \leq \epsilon$?

My solution:
Given that $x=x'$, we know $\forall\delta>0:(\exists N\geq 1: (\forall n \geq N:|a_n-a'_n|\leq\delta))$. We also know that $y$ is the formal limit of a Cauchy sequence $(b_n)_{n=1}^\infty$, which is bounded by a rational M, i.e. $|b_n| \leq M \implies \frac{1}{|b_n|}\geq \frac{1}{M}$. Let $\delta:=\frac{\epsilon}{M}\leq\frac{\epsilon}{|b_n|}$. From this, we can see that
\begin{align*}
|a_n-a'_n|&\leq\delta\\
&=\frac{\epsilon}{M}\\
&\leq\frac{\epsilon}{|b_n|}\\
\implies |a_n-a'_n|&\leq\frac{\epsilon}{|b_n|}\\
|a_n-a'_n||b_n|&\leq\epsilon
\end{align*}
Thus $|a_n b_n -a'_n b_n|\leq\epsilon$ , $(a_n b_n)_{n=1}^\infty$ is equivalent to $(a'_n b_n)_{n=1}^\infty$, xy is equivalent to x'y.

Is this correct?

P.S. we have shown the laws of algebra and of order holds for rationals, before this.

 

[Office_Shredder]

One thing I think you haven't shown is that $a_n b_n$ and $a'_n b_n$ converge to anything at all. I agree you have shown they are equivalent (if I remember the definition of that correctly) so if they converge to anything they converge to the same number.

I also don't think you have proven that $a_n b_n$ converges to xy to begin with, but that is pretty easy once you have this result (you can pick $a_n=x$ for all n to figure out what all the products converge to for example)

 

[Infrared]

I also don't think you have proven that $a_n b_n$ converges to xy to begin with, but that is pretty easy once you have this result (you can pick $a_n=x$ for all n to figure out what all the products converge to for example)

I don't think it makes sense to prove $a_nb_n\to xy$ in this context. It looks like the author is defining a real number as a equivalence classes of rational Cauchy sequences, and then defining a product $xy$ to be the class of the pointwise product of Cauchy sequences representing $x$ and $y$ separately. So $a_nb_n\to xy$ is a definition rather than a theorem (and the problem is to show that it makes sense)

 

[Office_Shredder]

I don't think it makes sense to prove $a_nb_n\to xy$ in this context. It looks like the author is defining a real number as a equivalence classes of rational Cauchy sequences, and then defining a product $xy$ to be the class of the pointwise product of Cauchy sequences representing $x$ and $y$ separately. So $a_nb_n\to xy$ is a definition rather than a theorem (and the problem is to show that it makes sense)

The model solution linked involves proving that $a_n b_n$ is a cauchy sequence to show that it represents some real number, so I think the concern is real. I agree the way the proposition is defined in the original post it does not assert that xy is a defined limit (which the URL does assert in its statement) so perhaps the OP intentionally has omitted this point.

 

[yucheng]

I also don't think you have proven that anbn converges to xy to begin with, but that is pretty easy once you have this result (you can pick an=x for all n to figure out what all the products converge to for example)

Actually, it was in the proposition, but I omitted that as I actually only managed to prove the last part. So I guess for now, let's just assume anbn converges to xy.

Plus, as what @Infrared said, the author actually defined $x y= \operatorname{LIM_{n\to\infty}}a_n b_n$.

I'll make sure I list out all my assumptions that I took for granted and also some definitions next time. Sorry!

 

[Office_Shredder]

To reiterate my point, xy is defined to be the limit of $a_n b_n$only makes sense if that sequence is cauchy, so you need to show that that holds (iirc,. Real numbers are the limits of cauchy sequences of rational numbers). So the author hasn't defined this to be true, they can only say it's true as long as the limit exists (and then prove it exists)

I agree with the broad strokes of what you have written. I think you have the point exactly right for showing that $a_nb_n - a'_nb_n$ converges to 0.

Edit: deleted a wrong statement. If you pick M big enough of course $b_n$ is bounded, even if you need M to be much bigger than y.