# Clarification for the least upper bound property

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 $$b\notin (a,b)$$ ?

## 2 Answers

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 $$S$$.

• 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 $E\subset S$? – Sean Lee Jan 6 at 18:12
• Yes, being bounded above is relative to whatever the "ambient" universe $S$ is being considered. – Andrés E. Caicedo Jan 6 at 18:13

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 $$S$$.