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For a linear order $(X,\le)$, define a cut to be any downwards closed subset of $X$. In the case $X = \mathbb Q$, there are $2^{\aleph_0}$ cuts, which is more than there are elements in $\mathbb Q$.

My question is: for which cardinals $\kappa$ besides $\aleph_0$ is there a linear order on $\kappa$ with more than $\kappa$ cuts?

Partial answer: Assume GCH and let $\kappa$ be an uncountable cardinal. Let $\le_{\text {Lex}}$ be the lexicographical order on ${\mathbb Q}^\kappa$, the set of functions from $\kappa$ to $\mathbb Q$, which has cardinality $2^\kappa$. Let $A \subseteq {\mathbb Q}^\kappa$ be the set of sequences that are eventually $0$. Notice that $A$ is dense in ${\mathbb Q}^\kappa$: if $a\le_{\text {Lex}}b$ and $\alpha$ is minimal such that $a_\alpha \ne b_\alpha$, define $c_\beta = a_\beta$ for $\beta < \alpha$, $c_\alpha = \frac{a_\alpha + b_\alpha}{2}$ and $c_\gamma = 0$ for $\gamma > \alpha$. Then $a \le_{\text {Lex}} c \le_{\text {Lex}} b$

Because $A$ is dense in ${\mathbb Q}^\kappa$, $(A,\le_{\text {Lex}}\restriction_A)$ has $2^\kappa$ cuts. If $\kappa$ is a successor cardinal, then each element of $A$ corresponds to a function $f:\kappa^- \to \mathbb Q$ and an ordinal $\alpha < \kappa$ (because $|\kappa \cap \alpha | \le \kappa^-$ for all $\alpha < \kappa$), so $|A| = \aleph_0^{\kappa^-} \cdot \kappa\stackrel{GCH}{=} \kappa$.

This shows that if $\kappa$ is a successor cardinal, it is consistent that the statement holds for $\kappa$. In particular there can be an order on $\mathbb R$ with more than $\mathfrak c$ cuts. I'm not sure what can happen to the cardinality of $A$ when $\kappa$ is a limit cardinal, or if it is consistent that the statement fails. Also, while my main question is in ZFC, I would also be interested in results in the absence of choice.

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2 Answers 2

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This exists for any $\kappa$ in ZFC.

Let $\lambda \leq \kappa$ be least so that $2^{\lambda} > \kappa$. Then the lexicographical order on the set $S$ of binary sequences on $\lambda$ which are eventually constant has size $\leq \kappa$, but it has $2^\lambda > \kappa$ many cuts. If $\vert S \vert < \kappa$ just add some dummy elements on the side (for example a copy of the well-order $\kappa$). Then we are guaranteed an order of size exactly $\kappa$ with more than $\kappa$ many cuts.

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  • $\begingroup$ How do you know $|S| \le \kappa$? $\endgroup$
    – Zoe Allen
    Commented Jul 23, 2023 at 10:50
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    $\begingroup$ Ah, I think I've worked it out: by construction, for any cardinal $\alpha < \lambda$, we have $2^\alpha \le \kappa$, and we have that $$|S| = \bigcup_{\alpha < \lambda} 2^\alpha$$ $$\le \lambda \cdot \kappa = \kappa$$ $\endgroup$
    – Zoe Allen
    Commented Jul 23, 2023 at 11:00
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If $$\kappa=\omega_\beta$$ where $\beta$ is a limit ordinal, then your argument about what $A$ looks like still holds. An element of $A$ can be viewed as an ordinal $\gamma < \omega_\alpha$ together with a function $\gamma \to \aleph_0$. So $A$ can be viewed as a union of elements of this form for all ordinals below $\omega_\beta$.

$$|A|=\bigcup_{\alpha<\beta}\aleph^{\omega_\alpha}$$ $$=\bigcup_{\alpha<\beta}\omega_{\alpha+1}$$ Under GCH.

Here it doesn't matter if there's a bit of overlap in the union because the size you get if they're all disjoint works out the same as if there's complete overlap (which there is). The union is clearly bounded below the disjoint union, which is $\omega_\beta$ and $\omega_\beta$ is also the supremum of elements in the union. So $$|A|=\omega_\beta=\kappa$$ So your example works for limit cardinals too, showing that it is consistent with ZFC that all infinite cardinals have this property.

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