I'm studying ordered fields, and a specific notion regarding ordered fields that I will denote here by their "height".

If $k$ is an ordered field, and $\alpha$ is a non-empty ordinal, a ruler of length $\alpha$ over $k$ is a map $r: \alpha \rightarrow k$ satisfiying:

-$r(\varnothing) = 0$.

-$r$ is an increasing map.

-$\forall \beta \in \alpha(\beta + 1 \in \alpha \rightarrow r(\beta + 1) = r(\beta) + 1$).

For instance, the only ruler of length $\omega_0$ over $k$ is $n \mapsto n.1_k$.

Rulers over $k$ are injective, so their length cannot be greater than or equal to $|k|^+$. There is a supremum to the set of length of rulers over $k$. Let's call this supremum the height of $k$, $h(k)$.

With little work, I have proven that if $cf(k)$ is the cofinality of $k$,

-$(i)$: $\forall \alpha(\alpha \in h(k) \longleftrightarrow \alpha$ is the length of a bounded ruler over $k)$.

-$(ii)$: There may or may not be a ruler of length $h(k)$ over $k$.

-$(iii)$: $h(k)$ is a limit ordinal.

-$(iv)$: $h(k) = \omega_0$ iff $k$ is archimedean.

-$(v)$: $h(k) \geq cf(k)$.

-$(vi)$: $\forall \alpha \in h(k), {\alpha}^{cf(k)} \leq h(k)$.

-$(vii)$: There is an ordinal $\lambda$ either limit or equal to $1$, such that $h(k) = {cf(k)}^{\lambda}$.

-$(viii)$ If $cf(k) > \omega_0$, then $h(k)$ is a cardinal.

I think $h(k)$ might be a cardinal even when $cf(k) = \omega_0$, but I still haven't found a proof or a counterexample of that.

I also wonder if $h(k)$ need be regular. When it is, it is the least ordinal such that every bounded monotone $h(k)$-sequence is Cauchy. If not, $h(k)^+$ is. I have no idea how to prove or disprove this result. I also have doubts that $(viii)$ should be of any help to prove that $h(k)$ is regular.

Does anyone know if $h(k)$ need be a cardinal in general, and if so, if it need be a regular one?

Since nobody has answered yet, I will give more details so as explain how I got these results:

$(i)$: this is because $h(k)$ is a supremum and because restrictions of rulers are rulers.

$(ii)$: for instance, there is a ruler of length $\omega_0$ over $k_1$* but no ruler of length $\omega_1$ over $k_2$* (because $|k_2| = \omega_0$).

$(iii)$: if $r: \beta + 1 \rightarrow k$ is a ruler, then $r \cup \{(\beta+1,r(\beta) + 1)\}$ is a bounded ruler so $\beta + 2 \in h(k)$.

$(iv)$: $h(k) \geq \omega_0$ is always true, and $k$ is archimedean iff the ruler of length $\omega_0$ is cofinal, so with $(i)$ it is equivalent to $h(k) \leq \omega_0$.

$(v)$: It is easy to define inductively a ruler of size $cf(k)$. (and there is a cofinal one if $cf(k) > \omega_0$)

$(vi)$: A bit long, but one can define inductively on $\alpha \times cf(k)$ with the lexicographic order a sequence of bounded rulers $r_{(\beta,\lambda)}$ of length ${\alpha}^\lambda.\beta$ whose reunion is a ruler of size ${\alpha}^{cf(k)}$ (not necessarly bounded)

$(vii)$: Using $(vi)$, and classical inequalities of ordinal arithmetics, one can prove that $\forall cf(k) \leq \alpha < h(k), \alpha.cf(k) < h(k)$. This is enough to prove that $\sup( \{{cf(k)}^{\gamma} \ | \ cf(k)^{\gamma} \leq h(k)\}) = h(k)$; then it is not difficult to prove that $\lambda$ is $1$ or a limit ordinal.

$(viii)$ If $h(k) > |h(k)|$, then given a bounded ruler $r$ of length $|h(k)|$ over $k$ and an uper bound $M$ of $r$, because $cf(k) > \omega_0$, $\{M^n \ | \ n \in \mathbb{N}\}$ has an upper bound $M'$. Using $r$, you can define inductively on any ordinal $\alpha \in |h(k)|^+$ an embedding $\alpha \rightarrow k \cap [0;\frac{1}{2}]$ whose range is $\frac{1}{M'}$-separated, for its distinct elements are distant of some power of $\frac{1}{M}$. The proof is similar to this proof that every countable ordinal embeds into $\mathbb{Q}$, where in the limit case, instead of the ruler of length $\omega_0$, you use $r$. This embedding can be used to define a ruler $r'$ of length $\alpha$ bounded by $\frac{M'}{2}$. Since $h(k) \in |h(k)|^+$, you get $h(k) \in h(k)$, which is impossible. So $h(k) = |h(k)|$.

*$k_1$ is $\mathbb{Q}$. $k_2$ is $\mathbb{Q}(X)$ where $X > \mathbb{Q}$. $h(k_1) = \omega_0$ and $h(k_2) = \omega_1$ for reasons similar to those invoked in the proof of $(viii)$.


1 Answer 1


After some time, I finally proved that $h(k)$ is a cardinal, and that it can take any cardinal value (including singular cardinals).

The proof is a bit long so I'll only post a scheme if asked.


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