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I have the least upper bound property as follows

For a subset $E$ of $S$, if $S$ has the least upper bound property then the supremum of $E$ is in $S$

What this seems to be saying to me, is that if I take something $E$ less than or equal to something $S$, then the largest element in $E$ will be in $S$.

Which seems quite obvious, which makes me think I'm not appreciating what's really being said here.

Perhaps some counter examples would help me to fully appreciate this property.

I reviewed some of the other questions.

Proving that ℝ satisfies the Least Upper Bound property discussed metric spaces and proving the property.

Is Wikipedia wrong about the least-upper-bound property? spoke about something specific relating to wikipedia.

edit - adding definition from text

This is the definition from the text that I'm using :

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baxx
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  • The least upper bound is not (necessarily) the largest element. As its name reveals, it is the smallest of the upper bounds for the elements of $E$. These upper bounds have in general no reason to be in $E$. – Bernard May 10 '18 at 09:16
  • How could there be an upper bound of $E$ which wasn't greater than or equal to all elements in $E$ – baxx May 10 '18 at 09:18
  • You misunderstood my comment: I meant it is not necessarily the largest element of $E$, simply 'cause $E$ has not necessarily a largest element. Of course, if it has one, it is the l.u.b. – Bernard May 10 '18 at 09:21
  • yes @Bernard that makes sense. For example if $E = [1,2]$ then $4$ would be an upper bound (it just wouldn't be the least upper bound). – baxx May 10 '18 at 09:22
  • And $(0,1)$ has no largest element, but $1$ is its least upper bound. – Bernard May 10 '18 at 09:25
  • that also makes sense @Bernard. – baxx May 10 '18 at 09:26
  • I hope the book didn't neglect to define "greatest-lower-bound property". – DanielWainfleet May 10 '18 at 16:09

2 Answers2

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Let $S$ be a set equipped with a linear order $<$.

Then order $\langle S,<\rangle$ has by definition the least upper bound property if every non-empty subset $E$ that is bounded above has a least upper bound (which is called the supremum of $E$).

This for instance is the case in $\langle\mathbb R,<\rangle$.

But it is not the case in e.g. $\langle\mathbb Q,<\rangle$.

Take subset $E=\{q\in\mathbb Q\mid q<\pi\}$. Then $E$ is not empty and bounded above (any $b\in\mathbb Q$ with $\pi<b$ serves as upper bound of $E$) but there is no least upper bound (note that $\pi\notin\mathbb Q$).

drhab
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    Btw the fact that $\mathbb{Q} $ doesn't have lub property needs to be demonstrated by using only the properties of $\mathbb{Q} $ and not using irrational numbers like $\pi$. That gives a true understanding of the limitation of $\mathbb{Q} $. Thus for example consider the set of all positive rationals whose square is less than $3$. The proof is slightly longer for this example. – Paramanand Singh May 11 '18 at 05:50
  • @ParamanandSingh Yes. That is a fair point. – drhab May 11 '18 at 06:44
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"What this seems to be saying to me", namely that if I take something $E$ less than or equal to something $S$, then the largest element in $E$ will be in $S$ does not describe the least upper bound property of $S$.

The least upper bound property of $S$ says the following: If $E$ is a nonempty subset of $S$, and the set ${\rm upb}(E)\subset S$ of all upper bounds of $E$ is nonempty then ${\rm upb}(E)$ has a least element $\sigma$. This $\sigma$ is uniquely determined, and is denoted by $\sup E$. Sometimes $\sup E\in E$, in which case $\sup E$ is called the maximal element of $E$, and is denoted by $\max E$.

Christian Blatter
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