# Least Upper Bound Existence Problem for Real and Rational Numbers

Doubt is regarding the proof of the statement such that there is no Least Upper Bound for the positive rational numbers with definitions such as $$p^2<2$$. Before explaining the conceptual doubt let me mention some the definitions used in this question as follows

Definition 1 Definition 2 Proof of no fixed maximum element for the set $$p^2$$<2

Analyse maximum element q which is less than $$p^2$$. For any p we can always find q that is greater . So it never ends. As a result no maximum element q possible here.  Reasoning for no least upper bound

This is the claim they(Rudin Book) make for the reasoning of "No least upper bound" and is given as follows

Rudin Claim

The set A is bounded above. In fact, the upper bounds of A are exactly the members of B. Since B contains no smallest member, A has no least upper bound in Q.

Doubts

1. As per definition 1 and 2 , members of A also can be Upper Bound . Then how can the Rudin reasoning mentioned above say " In fact, the upper bounds of A are exactly the members of B.".?

Members of B can be upper bound so as members of A. Does the statement "In fact, the upper bounds of A are exactly the members of B." shows correct meaning? Because not just members of B are upper bound candidates, members of A also are eligible provided if we can find a maximum fixed q .

Cant we say it has no least upper bound because we cant find maximum of the $$p^2<2$$ in A?

We can have q which is maximum as mentioned above for any p. Since we cant find fixed q, as it generates new maximum for any p, we say there is no point in finding out the number which is greater than all members of A . Would this point is sufficient enough to say there is no least upper bound than saying Rudin reasoning mentioned above ? If we could find a q which is above of all members of A,it would have been the least upper bound.

My problem is you don't have to go B and find no smallest element there. We can say from the absence fixed maximum in A by seeing the property of q

1. What is $$p^2<2$$ defined on Real numbers than rational? What could be the least upper bound? Would the same issue with q fixing happens here also? But book and lecture says there is least upper bound of R . Then what is that value in this particular case?

NB :: I was following a internet video lecture which is based on Rudin's mathematical analysis 3rd edition book. This doubt is based on page 3-5 of that book. It is evident that there is no $$\sqrt{2}$$ in Q so we can say no least upper-bound. But book explains it based on absence maximum in A and minimum in B. I am bit confused about the proper reasoning

• The least upper bound of set $X=\{q\in\mathbb Q\mid q^2<2\}$ in $\mathbb Q$ doesn't exist, but in $\mathbb R$, it is $\sqrt2$. That's what the Rudin proofs are about. Nov 26 '19 at 6:26
• @DonThousand Thanks.. It is evident that there is no$\sqrt{2}$ in Q so we can say no least upper-bound. But book explains it based on absence maximum in A and minimum in B. I am bit confused about the proper reasoning . Especially when they say on B is candidate for upper bounds before proving there is no minimum possible in B Nov 26 '19 at 6:34
• Imagine the reals weren't a thing. How would you describe there not being a least upper bound in $\mathbb Q$? Nov 26 '19 at 6:35
• What do you mean by gaps? There are no gaps in $\mathbb Q$. Nov 26 '19 at 6:37
• What do you mean by $\sqrt2$? I told you the reals don't exist. You aren't able to think of the rationals by themselves. What does it mean for $X$ not to have a least upper bound (again, the reals do NOT exist). Nov 26 '19 at 6:39

If you're working inside $$\Bbb Q$$, it's clear that $$A$$ has an upper bound (any $$p \in B$$, say $$p=3$$, will do). Suppose it had a least upper bound $$m$$. Then as $$\Bbb Q= A \cup B$$, either $$m$$ must be in $$A$$ or in $$B$$.
But if $$m \in B$$, and $$B$$ has no minimum, there is some $$p \in B$$ such that $$p < m$$ (if not, $$m$$ would be a minimum!) and as all $$p \in B$$ are upper bounds for $$A$$, we would have found a smaller upper bound for $$A$$ than $$m$$, contradiction.
And if $$m \in A$$, we can find $$p \in A$$ with $$p > m$$ (by the argument that $$A$$ has no largest element in $$S$$) and this contradicts the fact that $$m$$ is an upper bound for $$A$$ (!), as it does not "bound" $$p \in A$$ above...
So wherever $$m$$ lies, it cannot be "the least upper bound" of $$A$$. Either it's not "least", or it's not an upperbound. So the inevitable conclusion is that $$A$$ has no lub in $$S$$.
The whole point of constructing $$\Bbb R$$ from $$\Bbb Q$$ is to fill this gap, and "create" a new number $$\sqrt{2}$$ that will be the least upper bound for $$A$$ in $$\Bbb R$$.