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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.

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    @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.)2011-09-29

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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 $u$.

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    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.2011-09-29
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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 $\varepsilon=|b-|a||$.

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    It's a well-known argument in valuation theory (as applied to asymptotics, e.g. google "hardy rosenlicht field").2011-09-29