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Proving Dedekind cut Multiplication is closed

  1. May 28, 2015 #1
    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. jcsd
  3. May 29, 2015 #2

    wabbit

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    Gold Member

    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
  4. May 29, 2015 #3
    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##
     
  5. May 29, 2015 #4

    wabbit

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    Gold Member

    Right - but you can also start by proving that ## \forall \alpha>0^*, \exists r\in\alpha|r>0 ##, which simplifies the whole argument.
     
  6. May 29, 2015 #5
    Ah, that is a great point! Formulating and proving a lemma would indeed make the proof more efficient.
     
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