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I am wondering if the following can be proved in ZFC:

For every infinite cardinal $\kappa$, there exists a totally ordered abelian group G of cardinality $\kappa$ with at least $\kappa^+$ many convex subgroups.

I'm inclined to think the answer is "yes" since it can be shown in ZFC that for infinite $\kappa$, there is a chain of subsets of $\kappa$ of size $\kappa^+$. References would be appreciated as well as this question lies outside my research area. Thanks!

Greg Oman
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  • Considering the ordered-group of Hahn series $\mathbb{Z}((\Omega))$ with coefficients in $\mathbb{Z}$ and value (totally-)ordered set $\Omega$, we can see that the question reduces to the existence for each infinite cardinal $\kappa$ of a totally ordered set $\Omega_{\kappa}$ with at least $\kappa^+$ final segments (if $I$ is such a final segment, then $\mathbb{Z}((I))$ is a convex subgroup of $\mathbb{Z}((\Omega_{\kappa}))$.), which is the same as having a Dedekind completion of cardinal at least $\kappa^+$. This doesn't seem to hard a problem, but I have to say I can't find a solution. – nombre Jan 06 '17 at 15:06
  • @nombre By searching a little bit, this answer shows that GCH would give you: for every $\kappa$, there is a linear order of size $\kappa$ such that the Dedekind completion of $X$ has size $2^\kappa$. – Pedro Sánchez Terraf Jan 07 '17 at 16:38
  • @PedroSánchezTerraf Yes, this is the example of surreal subfields. Here in absence of GHC, the probelm is that their cardinal can be way too high. What I wondered when I thought about this problem is whether the subfield of $No(\kappa)$ generated by the ordinals $< \kappa$ is a suitable example, if one does not want to go up to $2^{\kappa}$ but will accept $\kappa^+$. But I really can't find a way to exhibit $\kappa^+$ initial / final segments of it. What do you think? – nombre Jan 08 '17 at 11:12
  • @nombre I do not have a fair grasp on surreal numbers to have intuition on this. But from what I read, $No(\omega)$ already has the cardinality of the continuum, so it's too big (i.e., $No(\kappa)$ does not have cardinality $\kappa$ in general). – Pedro Sánchez Terraf Jan 08 '17 at 16:02
  • @Pedro Sánchez Terraf: Well if $\kappa$ is a cardinal, the ordered field $No(\kappa)$ is just the complete binary tree $2^{<\kappa}$ ordered lexicographically, with a field structure. Here, I was mentioning a subfield of it which is hard to describe in terms of maps $\alpha < \kappa \rightarrow {0;1}$ so I don't see a way to make it easier to grasp for you. $No(\omega)$ in particular is countable, and is isomorphic to the ring of dyadic rationnals, but in general the cardinal of $No(\kappa)$ depends on "combinatorial properties" of $\kappa$ – nombre Jan 08 '17 at 16:40

1 Answers1

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The answer is yes. To see this, for any given infinite cardinal $\kappa$, let $\lambda$ denote the least cardinal such that $2^{\lambda} > \kappa$.

Let $2^{<\lambda}$ denote the set of maps $\alpha \rightarrow \{-1;1\}$, $\alpha < \lambda$, ordered lexicographically taking $-1$ to be strictly inferior to nothing, which is strictly inferior to $1$. It is a total order, and incidently it can be endowed with an interesting ordered field structure, seeing it as a subfield of $No$, the Field of surreal numbers. Its Dedekind completion is $2^{<\lambda+1}$, which contains the set $\{-1;1\}^{\lambda}$, and has therefore a cardinal $\geq \kappa^+$. $2^{<\lambda}$ however, is the union $\bigcup \limits_{\alpha < \lambda} \{-1;1\}^{\alpha}$, and has cardinal $\sup \limits_{\alpha < \lambda} 2^{|\alpha|} = \kappa$ by minimality of $\lambda$ (if this supremum was $< \kappa$, then $2^{\lambda}$ would be $\leq \kappa$).

Now consider the group $G:=\mathbb{Z}^{(2^{<\lambda})}$ of maps $f: 2^{<\lambda} \rightarrow \mathbb{Z}$ whose support $s(f):=f^{-1}(\mathbb{Z} - \{0\})$ is finite. For $0 \neq f \in \mathbb{Z}^{(2^{<\lambda})}$, define $d^{\circ} f := \max(s(f))$, and for $f\neq g \in \mathbb{Z}^{(2^{<\lambda})}$ say that $f \prec g$ when $0 < (g-f)(d^{\circ}(g-f))$. $(G,\prec)$ is an ordered group of cardinal $\max(|\mathbb{Z}|,|2^{<\lambda}|) = \kappa$.

For any element $x \in 2^{<\lambda+1}$, consider the subgroup $G_x:= \mathbb{Z}^{(]-\infty;x[_{2^{<\lambda}})}$. $G_x$ is a convex subgroup of $G$. Indeed, for $0 \prec f \in G_x$, and $0 \prec g \prec f$, $s(g)$ cannot contain an element $y \in 2^{<\lambda}$ strictly superior to $x$, otherwise $d^{\circ} g$ would be out of $s(f)$ and so $(f-g)(d^{\circ}(f-g))$ would be $-g(d^{\circ}g)$, which is strictly negative because $g$ is strictly positive: this contradicts $g \prec f$. So $g \in \mathbb{Z}^{(]-\infty;x[_{2^{<\lambda}})}$ . This and the stability of $\mathbb{Z}^{(]-\infty;x[_{2^{<\lambda}})}$ by taking opposites suffices to prove that it is convex.

By density of $2^{<\lambda}$ in its Dedekind completion, for any two $x < z \in 2^{<\lambda + 1}$, there is $y \in 2^{<\lambda}$ such that $x < y < z$, and so $\chi_{\{y\}} \in G_z - G_x$, so $2^{<\lambda+1} \ni s \longmapsto G_s$ is injective in the set of convex subgroups of $G$.

Therefore, $G$ is an ordered group of cardinal $\kappa$ with at least $\kappa^+$ convex subgroups.

nombre
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  • I think you have found "the book" solution. Thank you to everyone that engaged in working on my question which led to such a beautiful argument! – Greg Oman Jan 13 '17 at 20:55
  • After going through the solution, there's a slight error (but this can be fixed). Suppose CH fails, and let $\kappa=\aleph_1$. then $\lambda=\aleph_0$, but $2^{<\aleph_0}=\aleph_0<\kappa$. If the group $G$ you constructed had size less than $\kappa$, then simply take an ordered group $H$ of size $\kappa$ and consider the lex. order on $G\times H$. For a convex subgroup $G_x$ of $G$, consider $G_x\times H$. This gives you at least $\kappa^+$ convex subgroups of a group of size $\kappa$. It's late, so apologies if any of this is nonsense. – Greg Oman Jan 14 '17 at 06:43
  • @G.O. Well noticed, I'll edit when I'm on my computer. – nombre Jan 14 '17 at 08:49