Can the Dedekind completion of the hyperreal numbers be embedded in an ordered field?

Question

This answer shows that the Dedekind completion of the set of hyperreal numbers, endowed with the usual definitions of addition and multiplication of Dedekind cuts, is not an ordered field. But my question is, is it possible to embed this set with its addition and multiplication operations into some ordered field?

I imagine you’d need to add an awful lot of numbers which are greater than all the natural numbers but less than the infinite hyperreal numbers.

@NoahSchweber I meant this answer: https://math.stackexchange.com/a/2011076/71829. But somehow the hyperlink isn’t working. — Keshav Srinivasan, Mar 23 '20 at 03:51

score 7 · Accepted Answer · answered Mar 23 '20 at 05:36

7

No: addition is not cancellative, so it cannot embed in any abelian group. For instance, if $C$ is the Dedekind cut consisting of all elements that are less than some integer, then it is easy to see that $C+1=C$.

answered Mar 23 '20 at 05:36

Eric Wofsey

342,377

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Can you explain why $C+1=C$? – Keshav Srinivasan Mar 23 '20 at 12:47
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It is immediate from the definition of addition...if you add numbers less than $1$ to the elements of $C$, you get the same set. – Eric Wofsey Mar 23 '20 at 13:54
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Oh, I misinterpreted “some integer” as a fixed integer. Thanks, that makes sense. – Keshav Srinivasan Mar 23 '20 at 13:57

Can the Dedekind completion of the hyperreal numbers be embedded in an ordered field?

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