- #1
SrVishi
- 75
- 15
Homework Statement
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.
Homework Equations
None
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.