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I've just started on Principles of Mathematical Analysis by Walter Rudin (Third Edition) and came across the following definition for the least-upper-bound property:

Definition: An ordered set $S$ is said to have the least-upper-bound property if the following is true:

If $E\subset S$, $E$ not empty, and $E$ is bounded above, then sup $E$ exists in $S$.

My question is the following: Does this mean that open bounded sets of the form $(a,b)$ do not have the least-upper-bound property, since the supremum of the subset $(\frac{a+b}{2},b)$ is $b\notin (a,b)$ ?

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The property is defined relative to the set $S$, not to any potential superset of $S$. The example you give has $S=(a,b)$ and the subset you chose as $E$, namely $((a+b)/2,b)$ is not bounded above in $S$ (it is bounded above in $\mathbb R$, but this doesn't matter).

In fact, you can easily check that $S=(a,b)$ has the least upper bound property, the point being that if $E\subseteq S$ is nonempty and bounded above (in $S$), then its supremum exists in $\mathbb R$, and therefore in $S$, since this supremum is larger than $a$ and strictly smaller than $b$.

For a more dramatic example, consider $S=(0,1)\cup\{2\}\cup(3,4)$. This set also has the least upper bound property. For instance, $\sup(0,1)=2$. Of course, once we consider $S$ as a subset of the reals, this changes but, again, the property is defined as something intrinsic to $S$, independent of what happens in any potential larger sets containing $S$.

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  • $\begingroup$ Ah I see; So if I may clarify further, whenever we say a set $E$ is 'bounded above', it is understood to be 'bounded above' relative to some set $S$ where $E\subset S$? $\endgroup$ – Sean Lee Jan 6 at 18:12
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    $\begingroup$ Yes, being bounded above is relative to whatever the "ambient" universe $S$ is being considered. $\endgroup$ – Andrés E. Caicedo Jan 6 at 18:13
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Assuming you mean $S=(a,b)$: Actually yes, $(a,b)$ does have the least upper bound property. True, if $E=((a+b)/2,b)$ then there is no $\sup E$ in $S$. But that doesn't matter, because (relative to $S$) the set $E$ is not bounded above!

Look back at the definition: If $E\subset (a,b)=S$ is bounded above in $S$ there exists $c\in S$ such that $x\le c$ for every $x\in E$. Saying that $b$ is an upper bound for $((a+b)/2,b)$ doesn't matter, since we're only talking about elements of $S$.

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