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June 17th, 2014, 06:10 PM  #1 
Member Joined: May 2014 From: None of your business Posts: 31 Thanks: 0  Showing R is an ordered set with cuts
I started reading "Baby Rudin", and I got to Theorem 1.19 in the appendix of chapter 1. I think I've got this right, but I'm not quite sure, because my logic trick maybe a fake. To start things, the members of the set R, are subsets of Q called "cuts" (Dedekind cuts, I think?). A cut is any set alpha, a subset of Q, the with the following properties: 1) alpha is not the empty set and alpha is not the set Q (alpha not equal to Q) 2) If p is in alpha, q is in Q, and q<p, then q is in alpha 3) If p is in alpha, then p<r for some r in alpha The Rudin says 2) implies two different logically equivalent statements, which I checked: i) If p is in alpha and q is not in alpha, then p < q (very useful) ii) If r is not in alpha and r<s, then s is in alpha (I haven't used this one) Onwards to the heart of my longwinded post. The Rudin says you can define "alpha < beta" to mean alpha is a subset of beta. The Rudin also says you can prove that at least one of the three relations hold, (alpha < beta) or (beta < alpha) or (alpha not equal to beta) I followed along with Rudin's consise statements that you can show if (alpha not < beta) and (alpha not = beta) (Rudin actually never wrote (alpha not = beta) as that is redundant), then (beta < alpha). Fair enough. I said this was the heart of the post a few paragraphs back, so this is the left atrium of the heart of the post. I'm trying to prove that if (alpha not < beta) and (beta not < alpha) then (alpha = beta). I start with a proof by contradiction. (My reasoning is in italics) Suppose there is some alpha and beta in R such that [(alpha not < beta) and (beta not < alpha)] and (alpha not = beta). Since alpha is not < beta, alpha is not a proper subset of beta. So there is some p in alpha and p is not in beta. Likewise, beta is not < alpha, so there is some q in beta and q is not in alpha. Now using statement i) from the top, since p is in alpha and q is not in alpha, p<q. Likewise, since q is in beta, and p is not in beta, q<p. So we have p<q and q<p. A contradiction. So the supposition is wrong and for all alpha and beta in R, if (alpha not < beta) and (beta not < alpha), then (alpha = beta). This seemed like a trick. Is this proof right? Please help 
June 17th, 2014, 08:08 PM  #2 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
You would be far better off to use the proper notation, as what you are trying to say isn't clear. So what I think you are trying to show is: $\alpha \not\subset \beta$ and $\beta \not\subset \alpha \implies \alpha = \beta$ Note that this isn't true for sets in general, so there must be something special about cuts that makes it true. The first thing to show is that: if $\alpha \not\subset \beta$ then $\beta \subseteq \alpha\ (\ast\ast)$ So assume that $\alpha \not\subset \beta$, so that we have $q \in \alpha  \beta$ (this is a set difference, not subtraction. It is important to note that we HAVE to assume that $\alpha,\beta$ are nonempty to be sure $q$ exists). Now let $t \in \beta$. We need to show that $t \in \alpha$. Since $q,t \in \Bbb Q$, we can use the ordering on $\Bbb Q$ to consider 3 cases: a. $q < t$ b. $q = t$ c. $t < q$. If (a), then by (2) we have $q \in \beta$, contradicting our choice of $q$. A similar consideration holds for (b). So (c) must hold, in which case by (2) again, since $q \in \alpha$, we must have $t \in \alpha$. Now from $(\ast\ast)$, we know that if $\alpha \not\subset \beta$ one (or both) of the following statements holds: $\alpha = \beta$ $\beta \subset \alpha$. So if we stipulate that $\alpha \not\subset \beta$ AND $\beta \not\subset \alpha$, the only possibility is $\alpha = \beta$. Your proof is "probably right", but you need to be clearer about the distinction between $\subset$ and $\subseteq$ (which amounts to the distinction between $<$ and $\leq$ in $\Bbb Q$). ******************** A shorter proof if you know about partial orders: inclusion defines a partial order on the set of rational cuts. The proof of $(\ast\ast)$ above shows that the trichotomy rule holds on the set of these cuts, and is thus a total order. ******************** The reason why your proof may seem like a "fake" is this: what we really mean by the rational cut $\alpha$ is: $\{q \in \Bbb Q: q < \alpha\}$, where in the setbuilder definition $\alpha$ is a real number. Of course, this won't do, we can't use a real number to define itself, it's circular logic. Fortunately the finepatterned order structure of the reals is very closely related to the pattern of the rationals. So we USE this, to "fill in the gaps". Last edited by Deveno; June 17th, 2014 at 08:14 PM. 
June 17th, 2014, 08:15 PM  #3 
Member Joined: May 2014 From: None of your business Posts: 31 Thanks: 0 
How do I make the symbols you do in your posts?

June 17th, 2014, 08:27 PM  #4 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
This forum uses a plugin called MathJax, which is a LaTex rendering system. To render LaTex, you enclose symbols within dollar signs for inline rendering, and double dollar signs for display style, on a separate line, centered below. LaTex has it's own system of commands and syntax, which I can't explain very well in a short post. But I can give you some common expressions: a \in A yields: $a \in A$ a_0 + a_1x + a_2x^2 yields: $a_0 + a_1x + a_2x^2$ some other common commands: \leq = $\leq$ \subseteq = $\subseteq$ \emptyset = $\emptyset$ \neq = $\neq$ \frac{a}{b} = $\frac{a}{b}$ \sqrt{x} = $\sqrt{x}$ \to = $\to$ It's very useful, and I urge you to take some time to try to learn it. 
June 18th, 2014, 01:40 PM  #5 
Member Joined: May 2014 From: None of your business Posts: 31 Thanks: 0 
Ok, what I was trying to prove was: If $\alpha \nsubseteq \beta$ and $\beta \nsubseteq \alpha$ , then $\alpha = \beta$, for any $\alpha , \beta \in R$. But since you've proven, albeit differently, that if $\alpha \nsubseteq \beta$, then $\beta \subseteq \alpha$. It follows that $\alpha \subseteq \beta$ and $\beta \subseteq \alpha$, so by definition of set equality, $\alpha = \beta$. 
June 18th, 2014, 06:41 PM  #6 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
The negation of an or statement is an and statement. Now $A \not\subseteq B$ means: "$A$ is neither a proper subset nor the same set as $B$", or: "$A \not\subset B$" AND "$A \neq B$". I would stay away from compound negations, they get really confusing to work with. For real numbers, if $\alpha \not\leq \beta$ and $\beta\not\leq\alpha$, we have an impossible situation, so I don't think that's really what you want to say. ****************** A word about Dedekind's construction: we commonly model the real numbers as an (oriented) line, picking "more positive" (greater than) to mean "to the right". So the idea is, choosing any point $x$ on the line (any real number) divides the real numbers into two disjoint sets: $(\infty,x)$ and $[x,\infty)$ Since the rationals are DENSE (between any two real numbers there is a rational number), and their ordering is Archimedean (this essentially means $\Bbb Q$ has no "infinitely large" or "infinitesimally small" elements), we can think of $x$ as being: $\sup((\infty,x))$ Now again, this involves using $x$ to define $x$, but the density of rationals, means we can define: $\sup(\{q in \Bbb Q: q \in (\infty,x)\})$ and this is the same supremum. WE still have the problem, though, that we're using $x$ to define $x$. So this is the clever thing Dedekind did: he abstracted the order properties of an interval $(\infty,x)$, and if we use instead the interval $(\infty,x) \cap \Bbb Q$, now we have a description of a set which uniquely determines a real number which involves only rational numbers. In essence, then, what Dedekind did is invoke the TOPOLOGY of the real numbers (specifically, the ORDER topology). 

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