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I am told that the following proposition (II) is true by definition. I then attempt to prove the two implications that follow from (II) being true.

The members of $R$ will be certain subsets of $Q$, called cuts. A cut is, by definition, any set $\alpha \subset Q$ with the following three properties.

(I) ...

(II) If $p \in \alpha$, $q \in Q$, and $q < p$, then $q \in \alpha$.

(III) ...

The letters $p, q, r, ...$ denote rational numbers, and $\alpha, \beta, \gamma, ...$ will denote cuts.

(II) implies ...

If $p \in \alpha$ and $q \not\in \alpha$, then $p < q$.

If $r \not\in \alpha$ and $r < s$, then $s \not\in \alpha$.

Proof of Theorem 1.19, Step 1, Principles of Mathematical Analysis By Walter Rudin.


First Proof

Let $p \in \alpha$ and $q \not\in \alpha$.

$\alpha = \{ p,q \in Q : q < p \}$ From (II).

$q \not\in \alpha \rightarrow q \in \{ p,q \in Q : q > p \}$ Since an object is a member of a set if and only if it satisfies the defining property of the set.

$Q.E.D.$

Second Proof

Let $r \not\in \alpha$ and $r < s$.

$\alpha = \{ p,q \in Q : q < p \}$ From (II).

Since we had $r \not\in \alpha$ in the hypothesis, we can say that there is no $q \in Q$ such that $q < r$.

Therefore, $q > r \ \forall q \in Q, r \not\in \alpha$

Let $s = q$.

Since we have $r \not\in \alpha$, $s \in \{ s, r \in Q : s > r \}$

Therefore, we can conclude that $s \not\in \alpha = \{ s, r \in Q : s < r \}$

$Q.E.D.$


I would greatly appreciate it if people could please take the time to review my proofs for correctness.


New Second Proof

Let $r \not\in \alpha$ and $r < s$.

$\{ q \in Q : q < p \} \subseteq \alpha$ This is from (II).

$r \not\in \alpha \rightarrow r \not\in \{ q \in Q : q < p \} \subseteq \alpha$

$r \in \{ q \in Q : q > p \}$ Since if $q = p$, then $q = r \in \alpha$.

Since we have that $r < s$, $r \in \{ q \in Q : s > q > p \}$

But how can I conclude from here that $s \not\in \alpha$?

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Well, in both cases, you should note that your definition of cut $\alpha$ is not proper since: $$\alpha=\{p,q\in \mathbb{Q}:q<p\}$$ is not clear enough. By that, I mean that, when describing a set, at left you note the elements of this set of which "greater" set are elements, hence, when you say that $\alpha$ consists of $p,q\in\mathbb{Q}$, it is not clear whether you intend that both $p,q$ belong to $\alpha$ or not. It would be more proper to mention that: $$\{q\in\mathbb{Q}:q<p\}\subseteq\alpha\tag{rule}$$

Now, we can use the following rule from mathematical logic: $$(A\Rightarrow B)\Leftrightarrow(\neg B\Rightarrow\neg A)$$ Now, for your first proof, let $p\in\alpha$ and $q\in\mathbb{Q}$. We know that: $$q<p\Rightarrow q\in\alpha$$ From our rule, we have that: $$(q<p\Rightarrow q\in\alpha)\Leftrightarrow(q\not\in\alpha\Rightarrow q\geq p)$$ Now, since $p\in\alpha$ and $q\not\in\alpha$, it is clear that $p\neq q$, so, we have that $q>p$.

Can you, now, use these thoughts and your own ideas to alter you second proof?

Edit: As far as why $\alpha_p:=\{q\in\mathbb{Q}:q<p\}\subseteq\alpha$ and not neccessarily $\alpha_p=\alpha$, it is not guaranteed by property $(II)$ that $\alpha$ even represents a rational number, so no-one can ensure us that $p$ is the $\max\alpha$ or the $\sup\alpha$. So, there might exist $s>p$ such that $s\in\alpha$, which means that $\alpha_p\subset\alpha$.

Further edit: As for the second proof, note at first that, for two sets $A,B$ if $A\subseteq B$ does not mean that $B\subseteq A^c$, where with $A^c$ we note the complementary set of $A$. So, it is not true that: $$\alpha\subset\{q\in\mathbb{Q}:q\geq p\}$$ Moreover, as for $s$, it cannot belong in the set $\{q\in\mathbb{Q}:s>q>p\}$ (can you see why?). Try to mimic the proof of the first claim using the first claim and the rule described.

As a further explanation to the rule, what it is stated is that, you can say that "if $A$ is true, then $B$ is true" if and only if you can say that "if $B$ is not true, then $A$ is not true".

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    $\begingroup$ I' ve added some clarification over that, in a new edit. $\endgroup$ Dec 31, 2017 at 10:47
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    $\begingroup$ Nice! Also note that $\alpha_p$ as defined in this case is exactly the cut that defines the rational number $p$. $\endgroup$ Dec 31, 2017 at 10:56
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    $\begingroup$ Every real number is defined as a set, not as a number. So, every cut defines a number. More specifically, every real number is defined as the set of all the rational numbers than are lesser than the number itself. $\endgroup$ Dec 31, 2017 at 12:55
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    $\begingroup$ Well, you can prove it by contradiction. Suppose $s\in\alpha$ and, taking into account that $r<s$ you can make use of property $(II)$. But, now, you also see that the previous steps where not needed. In general, in this "primitive" stage of calculus, probably the most valuable tool you have is proving by contradiction and/or using logical "tricks" as the one mentioned in the first proof (this second proof can be done in both ways). $\endgroup$ Dec 31, 2017 at 15:55
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    $\begingroup$ Yeap! That's allright! :) Can you, now also prove it - just for excercise reasons; your proof is correct and nicely short - without using, in any manner, contradiction? $\endgroup$ Dec 31, 2017 at 16:22
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Note that $\alpha$ is not given explicitly as a set, it's given that it's a set with specific properties. Your approach to describe $\alpha$ as a set using $p, q$ is wrong. The implications can be proved easily using contradiction (assume hypotheses and negative of conclusion).

Thus to prove the first implication assume that $p\in\alpha, q\notin\alpha$ (hypotheses) as well as $q\leq p$ (negative of the conclusion). Now if $q=p$ then $q\in\alpha$ because $p\in\alpha$. And if $q<p$ then by the property (II) $q\in\alpha$. This is the contradiction which proves the implication. You can try to prove the other implication as well in a similar manner.

Moreover you should try to go beyond the symbols and think about their meaning in your natural language. Thus if $p\in\alpha$ then by second property we see that all numbers smaller than $p$ are also in $\alpha$. So whatever is not lying in $\alpha$ must necessarily be greater than $p$.


You may ask why we can't describe the set $\alpha$ in the manner you are trying. Well, the properties of $\alpha$ are meant to describe a general cut and not any specific chosen cut. All the cuts satisfy those properties and using those properties alone you can't describe a specific cut.

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  • $\begingroup$ Thanks for the response. Ahh, I see. But we can do as @Βασίλης Μάρκος did and describe our set as a subset of $\alpha$? This makes sense to me. $\endgroup$ Dec 31, 2017 at 10:54
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    $\begingroup$ @ThePointer : yes but then you are only describing subsets of $\alpha$ and this does not characterize $\alpha$. But if that approach helps in understanding you should definitely go for it. Also try to think of these in your natural language. You will find them very trivial. The symbolism makes such trivialities very hard to understand. $\endgroup$
    – Paramanand Singh
    Dec 31, 2017 at 10:57

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