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Let $A=\{x\in\mathbb Q\mid x^2\leq 2\}$. Prove that $A$ has no supremum in $\mathbb Q$. I had that to an exam, and I had a grade of $0/15$. Could someone explain me why ?

Proof : Let $M\in\mathbb Q$ s.t. $M=\sup(A)$. Suppose $M<\sqrt 2$. Then, by density of $\mathbb Q$ in $\mathbb R$ there is $x\in \mathbb Q$ s.t. $M<x<\sqrt 2$. Since $x^2\leq 2$, we have that $x\in A$ which is a contradiction.

If $\sqrt 2<M$. Still by density of $\mathbb Q$ in $\mathbb R$, there is $y\in \mathbb Q$ s.t. $\sqrt 2<y<M$. They $y$ is an upper bound of $A$ which contradict that $M$ is the smallest upper bound.

Question : What's wrong in my argument ? (I'll normally see the corrector next week, but I would like to know why this is wrong because to me it looks completely correct).

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    $\begingroup$ One immediate guess: You weren't supposed to use the concept of $\sqrt2$, or any knowledge that $\Bbb Q$ is contained in a bigger set called $\Bbb R$. $\endgroup$
    – Arthur
    Aug 27, 2019 at 10:55
  • $\begingroup$ Your argument is correct. It looks fine to me. Maybe it is what @Arthur wrote. $\endgroup$
    – Malkoun
    Aug 27, 2019 at 10:56
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    $\begingroup$ The question should really explicitly say that they want you to show this within $\mathbb{Q}$, i.e. without embedding $\mathbb{Q}$ into $\mathbb{R}$. If $\mathbb{R}$ wasn't even covered in the course yet then this is obvious and doesn't need to be spelled out, but if you've talked about $\mathbb{R}$ already then they should really be careful about letting you know what you can use in a given problem. $\endgroup$
    – Ian
    Aug 27, 2019 at 11:00
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    $\begingroup$ Your argument is correct if you see $A$ as a subset of $\mathbb R$, but is not valid if you see $A$ as a subset of $\mathbb Q$. By the definition of $A$, I can more or less guess that $A$ has to be seen as a subset of $\mathbb Q$ (otherwise it would be written : $A=\{x\in\mathbb Q\mid x\leq \sqrt 2\}$). $\endgroup$
    – Surb
    Aug 27, 2019 at 11:00
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    $\begingroup$ While giving $0$ is too harsh, I think your approach actually kills the whole beauty / surprise / excitement related to the problem. You should approach the problem by working in $\mathbb {Q} $ and then you will get a firsthand flavor of proofs in analysis. $\endgroup$
    – Paramanand Singh
    Aug 28, 2019 at 1:06

1 Answer 1

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I imagine that full marks would have been avoiding talking about $\mathbb R$ and $\sqrt 2$ and actually constructing rational numbers in the ranges you suggest. What I mean is talking about $m\over n$ as your rational number and saying things like $\left({m\over n}\right)^2<2$ instead of ${m\over n}<\sqrt2$.

Still, I'll take a stand and say that getting 0 marks for this is harsh. A fully correct proof would look very much like this:

Assume $M={m\over n}\in\mathbb Q$ is the supremum of $A$, for the purposes of contradiction. If $\left({m\over n}\right)^2<2$, [insert proof by construction that $M$ is not an upper bound of $A$]. If $\left({m\over n}\right)^2>2$, [insert proof by construction that $M$ is not the least upper bound of $A$]. Obviously, $\left({m\over n}\right)^2=2$ is not achievable in the rational numbers. Therefore, our assumption that $A$ has a supremum in $\mathbb Q$ was erroneous.

The proof shown here is a fair skeleton of that proof. At least one or two points would have recognized that.

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    $\begingroup$ "Still, I'll take a stand and say that getting 0 marks for this is harsh" : I'm not sure about that... I can imagine that the corrector really expected the OP to work in $\mathbb Q$ ignoring that $\mathbb R$ exist. And at the end, he maybe wished that the OP find such rational and not just say that "there is one rational s.t....". In somehow, the argument of the OP is a topological argument, whereas, the corrector expected probably more an algebraic argument... $\endgroup$
    – Surb
    Aug 27, 2019 at 11:07
  • $\begingroup$ @Surb Yeah, I'm certainly not advocating for 10 or even 5 points here. But real analysis does use the topology of the real numbers, so using vocabulary like that doesn't seem awkward to me. It's certainly more verbose than saying "The supremum of $A$ in $\mathbb R$ is $\sqrt2$, but because that isn't a member of $\mathbb Q$ there is no supremum in $\mathbb Q$", which is also fully defensible logically and mathematically. If the problem statement didn't say not to talk about $\mathbb R$ in a real analysis class, how could a student be expected to know not to use it? $\endgroup$
    – user694818
    Aug 27, 2019 at 11:55
  • $\begingroup$ Yes sure, but actually, I wouldn't be surprised this exercise is rather a discrete mathematical exercise than a real analysis exercise. As I remember, when I was a student, I had exactly such an exercise in discrete mathematic class, and the same week I also had this exercise in real analysis class, and I was very confused by the fact that the "analysis proof" were completely wrong in the discrete math by the fact that the tools I was able to use were completely different :-D $\endgroup$
    – Surb
    Aug 27, 2019 at 12:11

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