# Proving Dedekind cut Multiplication is closed

1. May 28, 2015

### SrVishi

1. The problem statement, all variables and given/known data
I am reading Rudin's Principles of Mathematical Analysis (3rd edition) and am working through the construction of $\mathbb{R}$. In step 6 of his construction, we must first verify that the field properties of multiplication hold for how he defines multiplication for the positive reals. He defines the positive reals as $\mathbb{R}^{+}=\{\alpha>0^{\ast}\}$ ($\alpha$ is a Dedekind cut, and $>$ is defined as proper superset) where $0^{\ast}=\{p\in\mathbb{Q}|p<0\}$ (the additive identity in the real field). Multiplication for any $\alpha,\beta\in\mathbb{R}^{+}$ is defined as $\alpha\beta=\{p\in\mathbb{Q}|\exists r\in\alpha\wedge\exists s\in\beta(r>0\wedge s>0\wedge p\leq rs)\}$. The first of the field properties is that it is closed under addition. Proving this is where I have trouble. I humor that I have had no problem proving all the steps of the construction on my own, but this is where I mess up.

2. Relevant equations
None

3. The attempt at a solution
To prove that $\mathbb{R}^{+}$ is closed under cut multiplication, I would need to prove that $\alpha\beta$ is both a cut, and greater than $0^{\ast}$. I am stuck on proving that $\alpha\beta$ is nonempty. I would need to find a rational number where I am guaranteed the existence of positive $r\in\alpha\wedge s\in\beta$ such that $(p\leq rs)$. I thought about using 0, since any positive elements of $\alpha$ or $\beta$ would help 0 satisfy $0\in\alpha\beta$. My problem is guaranteeing the existence of any such positive $r$ or $s$. I know that $\alpha,\beta\in\mathbb{R}^{+}$ implies $\alpha>0^{\ast}$ and $\beta>0^{\ast}$. This also implies that $\alpha\supset0^{\ast}$ and $\beta\supset0^{\ast}$ so $\exists r\in\alpha(r\notin0^{\ast})$ and $\exists s\in\alpha(s\notin0^{\ast})$. Since this $r\notin0^{\ast}$ and $s\notin0^{\ast}$, I know that $r\geq0$ and $s\geq0$. These are the only elements of $\alpha$ and $\beta$ I know to work with, but my issue is that they are not guaranteed to be greater than 0, only (greater than or equal to). I considered taking just disregarding using 0 as the element of $\alpha\beta$ to show that it is nonempty and instead just taking the product of the $r$ and $s$, but I still run into the problem of not knowing if they are strictly greater than 0. Any help will be greatly appreciated.

2. May 29, 2015

### wabbit

Just a note, your notation $\mathbb{R}^+$ designates strictly positive reals which is not standard. Also, by a cut I understand you mean a left half line in $\mathbb{Q}$ not containing a largest element. That's fine, just mentionning I am following your definitions here, as I understand them.

To prove that $\alpha\beta\neq\emptyset$, just take any $r>0$ in $\alpha$ and $s>0$ in $\beta$ (check that this is possible by your assumptions). Now set $p=rs$ and prove that it belongs to $\alpha\beta$ .

However, your definition of $\alpha\beta$ is problematic, because it contains positive numbers only, so it is not a cut. This is easy to cure, just add all negative or zero rationals to it - but still, it needs to be cured.

Last edited: May 29, 2015
3. May 29, 2015

### SrVishi

Thanks for the Reply! I just got an epiphany that I don't know why I didn't get earlier! Since, by the inferences I made earlier, I know $\exists r\geq0$ and $\exists s\geq0$. However, I can break this into cases. If $r=s=0$, since $r\in\alpha$ and $s\in\beta$, then I know $\exists x\in \alpha(x>r=0)$ since $\alpha$ is a cut and has no greatest element. Similarly for $\beta$, I can justify that $\exists y\in\beta(y>s=0)$. Thus I would have my desired elements where $xy\in\alpha\beta$. If I assume without loss of generality that $r>0$ and $s=0$, then again $\exists y\in\beta(y>s=0)$ as beta has no greatest element, and the product $ry\in\alpha\beta$. If both are greater than $0$, well, then I can just go ahead and use $rs$. Either way, $\alpha\beta\neq\emptyset$

4. May 29, 2015

### wabbit

Right - but you can also start by proving that $\forall \alpha>0^*, \exists r\in\alpha|r>0$, which simplifies the whole argument.

5. May 29, 2015

### SrVishi

Ah, that is a great point! Formulating and proving a lemma would indeed make the proof more efficient.