9

Is there any field containing $\mathbb{R}$ for which every non-empty subset has an infimum and a supremum in that field?

I am trying to understand whether $\overline{\mathbb{R}}$ (which is not a field) is the best possible.

jpv
  • 2,011
  • 1
    Surely you mean every nonempty subset. This sort of question has to be phrased just so, because minor changes will affect whether the question is even reasonable. – Carl Mummert Sep 28 '11 at 10:50
  • @Carl: I don't see why it matters whether we allow the empty set or not. Already the answer to the question is -- as you have already pointed out! -- negative: in fact for any ordered field $F$, the ordering is not bounded above or below, since for any $x \in F$, $x+1 > x$ and $x-1 < x$. It would change the question (although still not the eventual answer...) to require only bounded subsets to have infima and suprema, since at least $\mathbb{R}$ has this property. – Pete L. Clark Sep 28 '11 at 14:37
  • @Pete: That was the sort of thing I was suggesting, very gently, in my comment. The question has to be phrased just so. As it stands, even after the edit, there is no ordered field at all that has the desired property, much less one containing $\mathbb{R}$. – Carl Mummert Sep 28 '11 at 15:17
  • @Carl: okay, agreed. – Pete L. Clark Sep 28 '11 at 18:18
  • @jpv: Why not just rephrase the question thus: Is there any ordered field that has $\mathbb{R}$ as a subfield and in which every nonempty set with an upper bound has a supremum? (I suspect my answer posted below is about as straightforward a way of answering that question as can be found.) – Michael Hardy Sep 29 '11 at 19:00

2 Answers2

14

Since you mention ‘supremum’ and ‘infimum’ I will assume you mean to ask about ordered fields. Suppose there were such an ordered field. Let $\omega$ be the supremum of $\{ 0, 1, 2, \ldots \}$. Since $0 < 1$, we must have $\omega < \omega + 1$. By transfinite induction we find that this field must contain a well-ordered subclass isomorphic to the class of ordinal numbers; but no set can contain every ordinal number. So we have a proper class—which, for technical reasons, means that there is no such field.

Edits: However, if we allow class-sized fields, there is a field in which every subset has an upper bound (though not necessarily a supremum), namely Conway's surreal numbers. Indeed, we may define $\omega$ to be the surreal number of earliest birthday such that $n < \omega$ for every finite $n$. But as GEdgar observes in the comments, $\omega - 1, \omega - 2, \ldots$ will all also be upper bounds for $\{ 0, 1, 2, \ldots \}$, so this set, while bounded above, does not have a supremum.

Also, as Carl Mummert observes in the comments, we could also consider $u$, the supremum of all the elements of the field. Since $0 < 1$, we would have $u < u + 1$, which contradicts the maximality of $u$.

Zhen Lin
  • 90,111
  • 22
    I think there is an easier argument. If every subset has a supremum, then the entire field has a supremum. But then what happens when you add 1 to the supremum? – Carl Mummert Sep 28 '11 at 10:49
  • Zhen: Thanks for the answer. I will have to learn what transfinite induction is. – jpv Sep 28 '11 at 10:55
  • Carl: Thanks for the answer. So, one has to live with $\overline{\mathbb{R}}$ :(. – jpv Sep 28 '11 at 10:56
  • @Carl: Good catch. I was thinking about that but got sidetracked by the non-continuity of addition. – Zhen Lin Sep 28 '11 at 11:08
  • Zhen, I don't get why by transfinite induction it will contain every ordinal. – Asaf Karagila Sep 28 '11 at 11:58
  • @Zhen: If you can always to $+1$ and you can always do supremum, what happens with the first ordinal that you miss? – GEdgar Sep 28 '11 at 12:32
  • 5
    The Conway note is wrong. In the surreals, the set ${0,1,2,\dots}$ has no supremum. Indeed, if $x$ is an upper bound for that set, then so is $x-1$. – GEdgar Sep 28 '11 at 12:35
  • 1
    @GEdgar: Ah. I completely forgot to verify that $\omega$ actually is a supremum. I'll fix it. – Zhen Lin Sep 28 '11 at 13:47
  • Zhen: Correct me if I'm wrong on this. The ordinals can be defined within $No$, however the surreal numbers are defined as cuts (games). Then $\omega$ as a set is bounded, and has a supremum; however it is not $\omega$ (the ordinal) which is the bound of it. – Asaf Karagila Sep 28 '11 at 14:05
  • 1
    In Conway's system, $\omega$ is a certain element of the field, not a subset of the field. So talk about ${0,1,2,\dots}$ if you want to talk about that subset. – GEdgar Sep 28 '11 at 16:09
  • Carl Mummert's point is obvious, and I assumed of course that the question meant that every nonempty set that has an upper bound within the field has a least upper bound within the field. But certainly there's a simpler proof, and I posted it. This proof is needlessly exotic. And.....could it even be that its validity depends on issues of set theory that aren't really relevant to the question? The question can be answered without those considerations. – Michael Hardy Sep 29 '11 at 18:55
7

If it's an ordered field with $\mathbb{R}$ as a proper subfield, then it's not hard to show that it has nonzero infinitesimals, i.e. numbers $\varepsilon > 0$ such that $$ \underbrace{\varepsilon + \cdots +\varepsilon}_{n\text{ terms}} < 1 $$ no matter how big a finite cardinal number $n$ is. So what is the least upper bound of the set of all infinitesimals? Call it $c$. Is $c$ itself an infinitesimal? If so, then so is $2c$ (that's an easy exercise that takes a second or two), but that contradicts the fact that $c$ is an upper bound. But if $c$ is not an infinitesimal, then neither is $c/2$ nor anything between $c$ and $c/2$ (another easy exercise), and that contradicts the "leastness" of the least upper bound.

Therefore every ordered field with $\mathbb{R}$ as a proper subfield fails to have the least upper bound property.

Later edit: How does one show that which I said in the first paragraph is not hard to show? Let's suppose $a$ is a member of this larger field. Then either $|a|$ exceeds all positive reals, or is smaller than all positive reals, or some positive reals are less than $|a|$ and some greater. In the first case let $\varepsilon = 1/|a|$; in the second let $\varepsilon = |a|$. In the third case the set of reals $\le |a|$ has a least upper bound (within the real field) $b$. Then let $\varepsilon=|b-|a||$.

Michael Hardy
  • 1
  • The above appears to be excerpted from the Wikipedia article on the Archimedean property, viz. the section on "Archimedean property of the real numbers". Our policy here is to cite sources when posting such excerpts. Doing so also has the added benefit of giving the reader links to further sources. – Bill Dubuque Sep 29 '11 at 19:33
  • 1
    I did not get it from a Wikipedia article. More likely, I am the author of that passage in that Wikipedia article. I have been one of the most prolific contributors to Wikipedia mathematics articles since 2002. – Michael Hardy Sep 29 '11 at 20:17
  • 9th of July, 2003. I was in Toledo, Ohio: http://en.wikipedia.org/w/index.php?title=Archimedean_property&oldid=1130647 In those days you could create a new Wikipedia article while not logged in. I did hundreds of those. – Michael Hardy Sep 29 '11 at 20:20
  • Well, in any case, now we have a link that may prove helpful for readers. – Bill Dubuque Sep 29 '11 at 20:26
  • 1
    I'd probably have added such a link if I'd recalled that that article exists. – Michael Hardy Sep 29 '11 at 20:36
  • I probably learned this argument from lecture notes in a course taught by Gebhard Fuhrken. He had a rather formal style of lecturing and sometimes he was hard to follow, but then I'd go to the library and look at the notes from a previous time when he taught the course, and they were almost verbatim what I'd heard in class, and they were perfectly comprehensible. – Michael Hardy Sep 29 '11 at 20:40
  • It's a well-known argument in valuation theory (as applied to asymptotics, e.g. google "hardy rosenlicht field"). – Bill Dubuque Sep 29 '11 at 20:57