Showing that $\{x\in\mathbb{Q}:x≤0 \ or \ x^2≤2\}$ does not have a least upper bound in $\mathbb{Q}$

Asked Jan 27 '22 at 20:34

Active Jan 27 '22 at 20:38

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I want to prove that $C=\{x\in\mathbb{Q}:x≤0 \ or \ x^2≤2\}$ do not have a least upper bound in $\mathbb{Q}$

The definition I'm working with is that a least upper bound, $b$ has to be an upper bound, so $b\in\mathbb{Q}$ where $b>x \ \forall x\in C$ and that its the smallest upper bound, so if $c$ is some upper bound, then $b≤c$

I think the proof is to show that while I can find upper bounds, $b$, I will always be able to find a smaller one. I can prove that upper bound exists by having $b$ be positive and $b^2>2$ but I would like to know how you would always be able to find a smaller one. I think I want to subtract $1/n$ to the bound for a small enough $n$ to create a smaller upper bound. And I would like to do this by avoiding $\sqrt{2}$ if possible

edited Jan 27 '22 at 20:38

asked Jan 27 '22 at 20:34

PurpleFlutter

1

Compare with this post, or similar ones. – Dietrich Burde Jan 27 '22 at 20:36
Is there a way to do this with the approach of finding a small enough $1/n$? – PurpleFlutter Jan 27 '22 at 20:40
I prefer to avoid $\sqrt{2}$ if possible – PurpleFlutter Jan 27 '22 at 20:41
Hint: Try to avoid $\sqrt{4}$ or by looking at ${x\in\mathbb Q:x\le 0\text{ or }x^2\le 4},.$ Now look at ${x\in\mathbb Q:x\le 0\text{ or }x^2\le y}$ with some $y$ such that $\sqrt{y}\in\mathbb Q,.$ Finally look at $\sqrt{y}\not\in\mathbb Q,.$ Can you avoid $\sqrt{y}$? – Kurt G. Jan 29 '22 at 15:42

Showing that $\{x\in\mathbb{Q}:x≤0 \ or \ x^2≤2\}$ does not have a least upper bound in $\mathbb{Q}$

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