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While I have a basic understanding of what this definition states, I've been running into trouble interpreting the results when I test it with certain sets other than the traditional examples used to show that the Rationals do not have this property.

$\textbf{Definition 1.10:}$

"An ordered set is said to have the least-upper-bound property if the following is true: $$\text{If }\subset\text{, }E \text{ is not empty, and } \text{ is bounded above, then } \sup() \text{ exists in } ."$$ $\\$

I am interpreting this as a series of nested conditional statements: $$\Big[ \big( \left( E\subset S \right) \land \left( E \neq \varnothing \right) \land \left( E \text{ bounded above}\right) \big) \implies (\ \sup(E) \text{ exists in }S)\ \Big] \implies S \text{ has LUB property}$$

From here, we usually let $S=\mathbb{Q}$ and $E=\{q\in\mathbb{Q}:q^2<2 \}$ to show that $\mathbb{Q}$ does not have the LUB property. This makes sense to me because while the antecedent within brackets is true, the consequent is false. Thus the bracketed conditional is false, forcing the entire definition to be true.

But what about if (for example) $S=\mathbb{Q}$ and $E=\{q\in\mathbb{Q}:q^2<-1\}$ ?

In this case, $E=\varnothing$, thus the antecedent within the brackets is false. This renders the entire conditional in brackets true. Then the whole definition can only be true if $\mathbb{Q}$ has the LUB property - which is clearly wrong.

Can someone help me track down my misunderstanding?

Mittens
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CPmkI
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    $S$ has the LUB property if all non-empty bounded-above subsets have a supremum. – Martin R Oct 24 '22 at 05:26
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    You should have an extra universal quantifier for $E$. $[\text{for all $E$, } [\text{your big square brackets’ stuff}]]\implies \text{$S$ has LUB property}$. I know Rudin doesn’t state explicitly the quantifier over $E$, but it is something the reader is expected to ‘know’ is there by context. – peek-a-boo Oct 24 '22 at 05:27
  • It was really that simple! Thank you both for pointing it out! – CPmkI Oct 24 '22 at 08:34

2 Answers2

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There is something wrong with your series of conditionals: $E$ is a free variable. That means that the statement depends on both $S$ and $E$.

But "$S$ has the least upper bound property" doesn't depend on $E$, so we need to make $E$ a bound variable by using a universal quantifier:

$$\Big[ \forall E : \left\{ \big( \left( E\subset S \right) \land \left( E \neq \varnothing \right) \land \left( E \text{ bounded above}\right) \big) \implies (\ \sup(E) \text{ exists in }S)\ \right\} \Big] \leftrightarrow S \text{ has LUB property}$$

Joe
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David Lui
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    It was really that simple! Thank you for pointing it out! – CPmkI Oct 24 '22 at 08:34
  • @SuzuHirose Thanks for the guidelines (I'm new here). In fact, the first of the answers that solved my problem were given as replies to the initial question. I can't give them green checks, so I've just green checked this one. – CPmkI Oct 25 '22 at 07:00
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You found a set which has a least upper bound in $\mathbb Q$. But to satisfy the LUB property, there must be a least upper bound for EVERY set $E$, not just one that you choose.

dmh
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