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June 17th, 2014, 06:10 PM   #1
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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 long-winded 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
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June 17th, 2014, 08:08 PM   #2
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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 non-empty 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 set-builder 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 fine-patterned order structure of the reals is very closely related to the pattern of the rationals. So we USE this, to "fill in the gaps".
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Last edited by Deveno; June 17th, 2014 at 08:14 PM.
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June 17th, 2014, 08:15 PM   #3
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How do I make the symbols you do in your posts?
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June 17th, 2014, 08:27 PM   #4
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This forum uses a plug-in called MathJax, which is a LaTex rendering system.

To render LaTex, you enclose symbols within dollar signs for in-line 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.
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June 18th, 2014, 01:40 PM   #5
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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, $\alpha = \beta$.
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June 18th, 2014, 06:41 PM   #6
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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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