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Can we find a set other the $\mathbb{R}$ satisfying all the field axioms, order properties and completeness axiom?

By another set I mean, it differs from $\mathbb{R}$ may be in terms of topology, cardinality, etc,.

Edit

I am just curious to see some structure which evidently differ from $\mathbb{R}$, yeah topologically we can find, but topological difference doesn't quit a apparent difference in some sense for me. Yeah I understand my question is vague... but I think I let the reader to get the point.

Saravanan
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    You can just define another topology on $\Bbb R$... topology is not inherent in a set. – Kenny Lau May 11 '17 at 06:19
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    Related: isomorphism of Dedekind complete ordered fields. – dxiv May 11 '17 at 06:24
  • @KennyLau Yeah fine nice, I think I have to edit my question. I am just curious to see some structure which evidently differ from $\mathbb{R}$, yeah topologically we can find, but topological difference doesn't quit a apparent difference in some sense for me. Yeah I understand my question is vague... but you get the point – Saravanan May 11 '17 at 06:36
  • @dxiv so we can't find a set which differs in cardinality – Saravanan May 11 '17 at 06:44
  • We speak of "the" reals because if $(R,+,0,\times, 1,<)$ and $(R',+',0',\times',1',<')$ are order-complete ordered fields then they are isomorphic: There will be a (unique) field-isomorphism between them, and it will also be an order-isomorphism. – DanielWainfleet May 11 '17 at 08:55
  • We can embed any ordered field $F$ into a strictly larger ordered field $G$ of the same ( or arbitrarily larger) cardinality, but if $F=\mathbb R$ then $G $ cannot be order-complete. – DanielWainfleet May 11 '17 at 08:59
  • so we cant have a bigger order complete field? – Saravanan May 11 '17 at 09:06
  • @DanielWainfleet This is not true if we allow one to be non-separate. – Stefan Mesken May 28 '17 at 15:46
  • @Stefan. What is "non-separate"? – DanielWainfleet May 28 '17 at 16:17
  • @DanielWainfleet The absence of a countable dense set. – Stefan Mesken May 28 '17 at 16:18
  • @Stefan.What precisely is wrong? – DanielWainfleet May 28 '17 at 22:12
  • @StefanMesken No, DanielWainfleet is right - $\mathbb{R}$ is the unique (up to isomorphism) Dedekind complete ordered field. Looking at nonseparable fields doesn't help. The proof has two steps: (1) show that no non-Archimedean ordered field is complete (HINT: think about the set of infinite elements). (2) Show that if an ordered field is Archimedean, then it embeds into $\mathbb{R}$ via a unique map (HINT: to each element of the field we may assign the set of rationals $\le$ it ...). – Noah Schweber Aug 09 '17 at 06:24
  • @NoahSchweber Thanks for clearing that up for me! – Stefan Mesken Aug 09 '17 at 14:27

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We can find a set that is not $\mathbb{R}$ which satisfies the field axioms and is a totally ordered set, however we cannot find one that is Dedekind complete. Up to isomorphism $\mathbb{R}$ is the only Dedekind complete ordered field, and this is actually its defining property in a significant sense.

The study of real-closed fields may be of some interest to you; it is the first widely recognized generalization of the study of $\mathbb{R}$, looking at totally ordered fields with the same first order theory as the real numbers. This means that any statement $\phi$ which is true about elements of the real numbers must also be true for the elements of this other field, $\mathbb{F}$.

Completeness is a topological property of $\mathbb{R}$ and consequently part of its second order theory, the theory of subsets of $\mathbb{R}$. There are many real-closed fields which are not $\mathbb{R}$ -- the smallest among them is $\overline{\mathbb{Q}}$, the algebraic closure of the rational numbers, and the largest real-closed field is the Surreal numbers $N_0$. We also have that $\overline{\mathbb{Q}}\subset\mathbb{R}\subset N_0$, in the sense that there is a strict subfield of $N_0$ which is isomorphic to $\mathbb{R}$ and a strict subfield of $\mathbb{R}$ which is isomorphic to $\overline{\mathbb{Q}}$.

If you would like to see more examples I construct a proper class of non-isomorphic real-closed fields in this paper https://arxiv.org/abs/1706.08908, the smallest one appearing in here being $\mathbb{R}$ and the largest being $N_0$ -- this part of the construction begins on page $53$.

Alec Rhea
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