# Problems The Real Field Definition (Step 1) Theorem 1.19 Baby Rudin

Step 1 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.

• $$\alpha$$ is not empty, $$\alpha \neq Q$$.
• If $$p\in\alpha$$, $$q\in Q$$, and $$q < p$$, then $$q\in\alpha$$.
• If $$p\in\alpha$$, then $$p < r$$ for some $$r\in\alpha$$.

Well, I'm confused in the second item.

Experiment: Let $$\alpha=\{p: p\in Q \land p <0\}$$. If we take $$q\in\alpha$$ and then we analyse (II) for all elements of $$\alpha$$, we have to find that $$q for all $$p\in\alpha$$ because the other statements are true ($$q\in Q$$ and obviously $$p\in\alpha$$ and $$q\in\alpha$$). However $$p=q\in\alpha$$ doesn't satisfy $$q, then we can find a contradiction. I'm wrong? Why yes or why no?

• The condition $q < p$ is a hypothesis of the second condition. If $p = q$, this hypothesis isn't satisfied. Nov 1, 2015 at 17:39

The second property says that if $p$ is an element of $\alpha$ and if $q$ is another rational number that happens to be less than $p$ then $q$ must also be an element of $\alpha$.

The idea is that each point of $\alpha$ is less than each point of $Q\cap\alpha^c$. Example: $\alpha=\{x\in Q: x\le 0$ or $x^2<2\}$.

You are mistaking a statement for its converse. The statement is:

For all $p \in \alpha$, every $q \in Q$ which is less than $p$ is in $\alpha$.

You are thinking of:

For all $p \in \alpha$, every $q \in \alpha$ is less than $p$.

You're right that the second statement doesn't make sense, because if $p\in\alpha$, one could take $q=p$. Since $p < p$ is never true, this is a contradiction unless $\alpha$ is empty. But that contradicts the first condition.

But that's not what the original condition is. In general, a statement and its converse are not logically equivalent. "All Ferraris are red" does not imply that "All red cars are Ferraris."

• Aparently, I have a confusion with that kind of statements. I thougth that "If $a$, $b$, and $c$ are true then $d$ is true" is equivalent to say " $d$ is true if $a$ and $b$ and $c$ also are true and $d$ is false if any of $a$,$b$, or $c$ are false". The latter means for me $d= a\cdot b\cdot c$. I interpret about you said that these types of statements only "tall" about a part of the truth and not whole truth. I means, if $a$ or $b$ or $c$ are false then I can't say that $d$ is false or true. ¿I'm rigth? Nov 1, 2015 at 18:44
• @Fonseca Yes, you seem to have pinpointed your confusion. The statement "if $a,b,c$ are true then $d$ is true" only means the first part of what you thought is equivalent to it. It does not mean that, if one (or more) of the hypotheses fails then $d$ can't be true. Nov 1, 2015 at 20:31

The second condition says that $\alpha$ is downward closed: if $p \in \alpha$ then $(-\infty, p) \cap \mathbb{Q} \subseteq \alpha$. I.e. for any $p$ in $\alpha$, everything less than $p$ is also in $\alpha$.

• @egreg No it's not. The first is that $\alpha$ is not empty and is not all of $\mathbb{Q}$. Nov 1, 2015 at 21:22
• Sorry, I disregarded it. :( Nov 1, 2015 at 21:34

It' easiest if you just put these into common language:

i) a cut isn't empty. a cut isn't the whole field Q.

ii) if a cut contains an element, it contains all the rationals less than that element as well.

This does not mean that every element is less than every other element. (Which wouldn't make any sense.) It just means once a cut contains an element then it will contain all those below as well.

iii) a cut doesn't have a largest element. (for any element in the cut you can find a larger in the cut.

Unstated conclusion: If q isn't in a cut then all r > q are also not in the cut.

CHEATING BY LOOKING AHEAD: A cut is simply the segment $(-\infty, r)$, restricted to the rationals, for any real r . That's all.

EXCEPT we can't SAY that because we haven't defined what a real number is yet. So we ABSOLUTELY can not say that. Although that is precisely what we are getting at.

A cut is all the rationals up to ... whenever the rationals stop being in a cut. For any cut then for any rational q, either q is in the cut, or q is too big for the cut.

Cuts basically "cut" the rationals into two parts, those small enough to be in the cut, and those too big to be in the cut.

All $(-\infty, q); q\in Q$ (restricted to Q) are cuts but these are aren't the only cuts. In $(-\infty, q)$ $q$ is the rational least upper bound of the cut. Some cuts don't have rational least upper bounds. The set $\{q| q \in Q, q^2 < 2\}$ is a cut without a least upper bound. It will be the cut, $(-\infty, \sqrt 2) \cap Q$ but we can't say that because $\sqrt 2$ is utterly undefined in Q.

The problem is that I thougth that "If $a$, $b$, and $c$ are true then $d$ is true" is equivalent to say " $d$ is true if $a$ and $b$ and $c$ also are true and $d$ is false if $a$, $b$, or $c$ are false". The latter means for me $d=a⋅b⋅c$. But this is not the meaning, the meaning is just the first part of what I thought is equivalent to it.

In this manner, for any $q\in Q$ and for any $p\in\alpha$ such tha $q> p$, the statement (II) can't say something about if $p\in\alpha$ is true or false.