$\{r\in \mathbb Q \mid r^2 <2 \}$ has $\{r\in \mathbb Q \mid r^2 >2 \}$ as its set of upper bounds

Question

Let $A=\{r\in \mathbb Q \mid r^2 < 2\} = \{r\in \mathbb Q \mid r^2 \leq 2\}$

I want to prove that $A' = \{r\in \mathbb Q \mid r^2 > 2\}$ is the set of all upper bounds of $A$.

All I could do is showing it via $\mathbb R$: since $A$ in $\mathbb R$ is $\mathbb Q \cap (-\infty, \sqrt{2}]$. We know that the set of upper bounds of $(-\infty, \sqrt{2}]$ is $[\sqrt{2}, \infty)$. Then, by taking the intersection with $\mathbb Q$ (don't know if this is mathematically licit) we get that the set of upper bounds of $A$ in $\mathbb R$ is $A'$.

I am not sure that passing via $ \mathbb R$ is the right way, I pretty sure a more convenable way is without talking about $\mathbb R$.

PS:

• W. Rudin in the beginning of his first chapter (of Principles of Mathematical Analysis) proves that $A$ and $A'$ do not have respectively a greatest and smallest element, but he does not make the connection with the upper bounds of $A$.
• S.D. Abbott in Understanding Analysis claims that $A$ hasn't got a Lower Upper Bound in $\mathbb Q$ (without proving it) in his first chapter $1.3$ then in the following section proves $A$ has a LUB in $\mathbb R$.

Thanks

Answer 1

I think you want to work within the ordered field $\mathbb{Q}$ only, without any thought about $\mathbb{R}.$ So from now on, suppose all the elements being referred (including in sets) to are in $\mathbb{Q}.$ What you want to show is:

Let $A=\{p: p^2 \leq 2\},\ $ let $A' = \{p: p^2 > 2; p\geq 0\}\ $ and let $U:=\{$ all upper bounds of $A\} = \{u: u\geq p\ \forall\ p\in A\}.$ Then $A' = U.$

We need to show two things: $A'\subset U $ and $U\subset A',$ for then we have the desired $A'=U.$

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Let $x\in A',\ y\in A.\ $ Then, $x^2>2\geq y^2.\ $ If $y\geq x,$ then by Proposition $1.18 (b),\ y^2 \geq yx \geq x^2,$ contradicting $x^2 > y^2.$ Since $x\geq 0,$ it follows that $x>y,$ and so $x$ is an U.B. of $A.$ Thus $x\in U,$ and so $A'\subset U.$

Next, let $u\in U.$ By definition of $U,\ u\geq a\ \forall\ a\in A.$ Now suppose further that $u^2 < 2.$ We can find a member of $A,$ for example $\ t:=u - \frac{u^2-2}{u+2},\ $ such that $t^2 > u^2, $ which implies that $t>u,$ contradicting the fact that $u$ is an U.B. of all members of $A.$ This contradiction means $u^2 < 2$ is false, and we therefore have $u^2 \geq 2.$ Rudin already proved that there is no rational $r$ s.t. $r^2=2.$ Therefore, $u^2 > 2.$ Therefore $u\in A',$ and so $U\subset A'.$

Answer 2

I think you are chasing your own tail.

Assuming we have already finished defining the real numbers and showing that the real numbers have the least upper bound property while the rationals do not, then at that point we can show that $\sup A$ exists (because the reals have a least upper bound property). That $\sup A$ is not rational. (because we can show that $B= \{r\in \mathbb Q|r\text{ is an upper bound of } A\}= \{r\in \mathbb Q| r^2 > 2; r> 0\}$ [Note 1: you forgot to take negative rationals into account but that's minor. Note 2: If $r > 0; q> 0$ then $r>q \iff r^2 > q^2$ so that is why those sets are equal. But actually proving the sets are equal doesn't really matter]).

And we can prove $(\sup A)^2 = 2$. That's actually a bit of an excercise I won't go into right now. So taken altogether: i) $A$ is bounded above. ii) $\sup A$ is an positive irrational real. iii) $(\sup A)^2 = 2$ we can deduce $\sup A = \sqrt 2$.

And if for some reason we wanted the set of all real upper bounds of $A$ it be $U= \{x \in \mathbb R| x \ge \sqrt 2\} = [\sqrt 2, \infty)$.

Restricting that set to only rationals will give us less than all upper bounds as there are irrational upper bounds. Restricting it to $Q$ ... takes us back to set $B$, the set of all upper bounds... in $\mathbb Q$ (not $\mathbb R$).

....

Okay. But that is all assuming we have defined the reals and shown/defined the reals to have the least upper bound property.

The entire purpose of this discussion in the text was to show that the rationals do not have the least upper bound property, and then to show we need to extend the rationals into the reals which is a ordered field that does have the least upper bound property.

So the books show that $A$ does not have a rational least upper bound. (By showing via algebra that if $0 < q\in \mathbb Q$ and $q^2$ then we can find a $t\in \mathbb Q$ so that $q < t$ but $t^2 < 2$.) Maybe the books prove that $B=\{r\in \mathbb Q| \text{for all } q \in A, r > q\}$ is equal to $\{r\in \mathbb Q| r^2 > 2; r> 0\}$. But whether they do or don't doesn't really matter. What matters is $A$ does not have a rational least upper bound property.

So... we invent/introduce the reals.

... As for finding the set of all real upper bounds... Well, why bother? It's the least upper bound that matters. But again, as $\sqrt 2$ is the least upper bound. And and $x > $least upper bound, is also an upper bound.... the set of all real upper bounds of $A$ is $[\sqrt 2, \infty)$.