Field bigger than $\mathbb{R}$ 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.
 A: 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 $\varepsilon=|b-|a||$.
A: 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 $u$.
