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Let $\mathbb S$ be an ordered field of cardinality larger than $\mathbb R$. Let $\mathbb S^*$ be the completion of $\mathbb S$ via Dedekind cuts. Now it is well-known that $\mathbb R$ is the unique complete ordered field. So in what was does $\mathbb S^*$ fail to be a complete ordered field, and why?

mbsq
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2 Answers2

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The existence of additive inverses fails in $\mathbb{S}^*.$

A member of $\mathbb{S}^*$ (a Dedekind cut in $\mathbb{S})$ is a subset A of $\mathbb{S}$ such that $A$ has an upper bound, $A$ has no least upper bound, and $(\forall x\in A )(\forall y \lt x )(y\in A).$ If $A$ and $B$ are Dedekind cuts, then $A+B$ is defined to be $\{x \mid (\exists a\in A)(\exists b\in B)(x\lt a+b)\},$ which you can see is a Dedekind cut itself.

You can check that $\{x \mid x \lt 0\}$ is a Dedekind cut, and that it is the additive identity in $\mathbb{S}^*.$

If $\mathbb{R}$ is a proper subset of $\mathbb{S},$ then there is some member of $\mathbb{S}$ greater than all members of $\mathbb{R}$ (proving this uses the ordered field axioms in $\mathbb{S}).$ It follows that $A=\{x\in\mathbb{S}\mid (\exists r\in\mathbb{R})(x\lt r)\}$ is a Dedekind cut.

The element $A$ has no additive inverse in $\mathbb{S}^*,$ which you can see as follows:

Assume $A$ had an additive inverse $B.$ Then there exist $a_0\in A$ and $b_0\in B$ such that $-1 \lt a_0+b_0.$ By the definition of $A,$ there is some $r_0\in \mathbb{R}$ such that $a_0\lt r_0,$ and it follows that $-1\lt r_0+b_0.$ But then we have $r_0+1\in A,$ $b_0$ in $B,$ and the sum of the two is non-negative, which is a contradiction.

Mitchell Spector
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    Well. Most properties fail. Surely you can conclude associativity of addition or multiplication, both fail, probably you can also get the failure of distributivity, multiplicative inverse for non-zero cuts, and maybe even commutativity. – Asaf Karagila Nov 12 '16 at 20:46
  • @AsafKaragila I haven't checked all the details, so maybe I'm missing something, but I think all the usual associative, commutative, and distributive laws still work. – Mitchell Spector Nov 12 '16 at 20:57
  • Asaf is right that associativity fails. I never realized that some care is required to check the usual construction of the reals. – mbsq Nov 12 '16 at 21:30
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    @mbsq \begin{align} A+(B+C)&={x \mid (\exists a\in A)(\exists d\in B+C)(x \lt a+d}) \&={x \mid (\exists a\in A)(\exists b\in B)(\exists c\in C)(\exists d\lt b+c)(x\lt a+d)} \&={x \mid (\exists a\in A)(\exists b\in B)(\exists c\in C)(x\lt a+b+c)}. \end{align} [The hard direction in the last equality is true because $x\lt a+b+c$ implies $x-(a+b)\lt c,$ so, by the density of the ordered field $\mathbb{S},$ there exists $c'$ such that $x-(a+b)\lt c' \lt c.$ So we have $x\lt a+(b+c'),$ and we can take $d=b+c'.$] Similarly, $(A+B)+C$ equals the same set. – Mitchell Spector Nov 12 '16 at 21:57
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    +1: you can also observe that the cut $A$ is an additive annihilator of any $B \le A$, i.e., if $B \le A$, then $A + B = A$. No such $A$ can exist in an ordered field. – Rob Arthan Nov 14 '16 at 20:50
  • @RobArthan Yes, that's a good point. I understood that the OP knew that $\mathbb{S}^*$ wasn't an ordered field, but was interested in seeing where specifically the usual proof that $\mathbb{R}$ is an ordered field breaks down if you try to apply it to other similar Dedekind-cut-style completions. – Mitchell Spector Nov 15 '16 at 00:30
  • I don't understand why you require $A$ to not have a least upper bound. For example, ${x \mid x < 0}$ is surely a Dedekind cut in $\mathbb{Q}$, but it clearly has the least upper bound $0$. Or are you trying to say that there is no least upper bound in $A$? In that case, a better formulation would be that $A$ should have no greatest element. – Smiley1000 Sep 30 '24 at 15:40
  • "If $\mathbb{R}$ is a proper subset of $\mathbb{S},$ then there is some member of $\mathbb{S}$ greater than all members of $\mathbb{R}$ (proving this uses the ordered field axioms in $\mathbb{S}).$" It must also use the Dedekind-completeness of $\mathbb{R}$, otherwise it would apply for example to $\mathbb{Q}$, which it doesn't. ... – Smiley1000 Sep 30 '24 at 17:20
  • ... Here's my proof: Striving for a contradiction, assume that for all $x \in \mathbb{S} \setminus \mathbb{R}$, there are $r_1, r_2 \in \mathbb{R}$ with $r_1 < x < r_2$. Take such an $x$. Consider the set $M = {r \in \mathbb{R} \mid r < x} \subseteq \mathbb{R}$. Since $r_1 \in M$, $M \neq \emptyset$. Since for all $r \in M$, $r < r_2$, $M$ is bounded above. Hence $M$ has a supremum, say $s = \operatorname{sup} M$. ... – Smiley1000 Sep 30 '24 at 17:20
  • ... Consider $y = x - s$ and $z = \frac{1}{|y|}$. By assumption, there is some $t \in \mathbb{R}$ with $z < t$, implying that $|y| > \frac{1}{t}$. Now we either have $s > x$ or $s < x$. ... – Smiley1000 Sep 30 '24 at 17:21
  • ... If $s > x$, we have $s - x > \frac{1}{t}$, meaning that $s > x + \frac{1}{t}$. But then $s - \frac{1}{t}$ is a smaller upper bound for $M$ than $s$. So we obtain a contradiction in the case that $s > x$. Now consider the case $s < x$. Then we have $x - s > \frac{1}{t}$, implying that $x > s + \frac{1}{t}$. But then we have $s + \frac{1}{t} \in M$, contradicting the fact that $s$ is an upper bound for $M$. So we obtain a contradiction in the case that $s < x$. Hence we have a contradiction in all cases. – Smiley1000 Sep 30 '24 at 17:21
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You may be interested in knowing that there is a kind of completion of $\mathbb{S}$ akin to Dedekind completion:

Instead of considering every cut $(A,B)$, take only good cuts: cuts $(A,B)$ where for any $0<\varepsilon \in \mathbb{S}$, there is $(a,b) \in A \times B$ such that $b-a < \varepsilon$.

Then you really get a dense ordered field extension $\widetilde{\mathbb{S}}$ which is complete in the sense that it has no proper dense ordered field extension. This is the same as completing $\mathbb{S}$ using Cauchy $\lambda$-sequences where $\lambda$ is the least cardinal of a subset of $\mathbb{S}$ without bounds. This is also the same as completing $\mathbb{S}$ seen as an ordered field with its canonical uniform structure.

nombre
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  • Is there any where that I could read up more about these sequences and his they complete ordered fields? – Lave Cave Aug 15 '22 at 20:06
  • @LaveCave Regarding the uniform structure, this can be found in the Bourbaki of general topology. For cuts, these are sometimes called Veronese cuts (I found this in one of Philip Ehrlich' papers, can find which one later). Regarding sequences, I'm not sure this is written anywhere, but I have notes about it if you're interested. – nombre Aug 16 '22 at 09:39