A totally ordered set is Dedekind-complete if any subset which has an upper bound also has a least upper bound. Now any two ordered fields which are Dedekind-complete are order-isomorphic as well as field-isomorphic. Another way of phrasing this is that the second-order theory of real numbers is categorical, i.e. it has a unique model upto isomorphism.

But I’m interested in what happens if you drop the field structure. My question is, are any two dense, totally ordered, Dedekind-complete sets order-isomorphic? In other words, is the real number system the unique dense totally ordered Dedekind complete set?


2 Answers 2


No, not at all. First of all, there are some trivial examples: you could take the empty set, or a singleton set, or a closed (or half-open) interval in $\mathbb{R}$. To eliminate such trivialities, you should additional ask for your ordered set to be nonempty and have no greatest or least element.

However, there are also nontrivial examples that cannot be ruled out so easily. Indeed, every totally ordered set has a Dedekind-completion which can be constructed just like $\mathbb{R}$ as the completion of $\mathbb{Q}$. So for instance, if you start with any dense ordered set $X$ of cardinality greater than $2^{\aleph_0}$ with no greatest or least element, its Dedekind-completion is a Dedekind-complete dense ordered set with no greatest or least element which cannot be isomorphic to $\mathbb{R}$ since it does not even have the same cardinality.

(For an explicit example of such an $X$, note that if $Y$ is any totally ordered set (e.g., an ordinal), then $Y\times\mathbb{Q}$ with the lexicographic order is a dense totally ordered set with no greatest or least element.)

Incidentally, you might wonder why you can't construct counterexamples for fields in the same way. After all, if you have any ordered field $X$, you can take its Dedekind-completion as an ordered set. You could then try to imitate the construction of the field structure on $\mathbb{R}$ from that of $\mathbb{Q}$ to get a field structure on the completion of $X$. However, it turns out that the verification of some of the field axioms on $\mathbb{R}$ uses more than just that it is the completion of an ordered field $\mathbb{Q}$--it also uses the fact that $\mathbb{Q}$ is Archimedean. So, you can similarly construct a completion of any Archimedean ordered field, but that will always just give you $\mathbb{R}$.

  • $\begingroup$ Thanks for your answer. So does that mean that it’s not possible to put a field structure on, say, the Long Line? $\endgroup$ Sep 30, 2018 at 4:00
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    $\begingroup$ That's correct (assuming you want the field structure to be compatible with the order), because the long line is Dedekind complete but not order-isomorphic to $\mathbb{R}$. $\endgroup$ Sep 30, 2018 at 4:03
  • $\begingroup$ Your edit is interesting. Do you know what part of the proof uses the Archimedean property? $\endgroup$ Sep 30, 2018 at 4:10
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    $\begingroup$ Well, it depends exactly how you formulate the definition of the field operations. But for at least one formulation, the existence of additive inverses uses the Archimedean property. (You can get around this by defining addition using separate cases for positive and negative numbers, but then associativity of addition will require the Archimedean property.) $\endgroup$ Sep 30, 2018 at 4:13
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    $\begingroup$ @Anguepa: No; for instance, the (open) long line is another such order which is not isomorphic to $\mathbb{R}$. $\endgroup$ Oct 16, 2018 at 17:34

I liked Eric Wofsey's answer, but wanted to give two examples of Dedekind complete ordered sets.


$\quad L = \{ \frac{1}{n} \mid n \in \Bbb N \}$

$\quad A = \Bbb Z \cup L$

It is easy to see that $(A, \lt) \subset (\Bbb R, \lt)$ is Dedekind complete. Moreover, the subset

$\quad P = \{x \in A \mid x \gt 0\}$

of $A$ is bounded below by $0$ but $P$ has no minimal element.

Let $a \in A$ be any element. Set

$\quad B = A \setminus \{a\}$

Exercise: Show that there exists an order preserving isomorphism between $(A, \lt)$ and $(B, \lt)$.


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