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Let $A$, $B$ be sets and denote by $A\leq^* B$ an assertion that if $A$ is non-empty, then there's a surjection from $B$ onto $A$. This might be understood as a dual definition of cardinality. (Typically we use injections to capture the notion of A not being bigger than B. The $\leq^*$ notation is to be read as B not being smaller than A.)

Is this relation trichotomous without using the axiom of choice? Meaning for every two sets $A$ and $B$, it holds $A\leq^* B$ or $B\leq^* A$. (I would be surprised). Would trichotomy imply some interesting extra-ZF axiom?

More broadly, is there anything to be said about $\leq^*$? For example how does it relate to the usual comparisons of cardinality by injections?

I suspect there's no clear answer, since it is somehow reminiscent of Partition Principle ($A\leq^* B \implies A\leq B$) and Weak Partition Principle ($A\leq^* B \implies B\not< A $) which both are independent of ZF.

theDumbGeometer
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  • I read below that you find an almost exact duplicate. It can be closed as such without it being deleted, and it would fulfill the purposes you state in your comment (that's how duplicates are useful). – Pedro Sánchez Terraf Feb 14 '17 at 19:48

1 Answers1

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Not even a little bit.

If $\leq^*$ satisfies the trichotomy, then the axiom of choice holds.

Theorem. (Lindenbaum) If $X$ is a set, then there is an ordinal $\alpha$ such that $\alpha\nleq^*X$.

(Proof. If $\alpha\leq^*X$, then $\alpha$ injects into $\mathcal P(X)$; apply Hartogs' theorem.)

Let us denote the least such (non-zero) ordinal as $\aleph^*(X)$ and name it the Lindenbaum number of $X$.

Theorem. If $X\leq^*\alpha$ for an ordinal $\alpha$, then $X$ can be well-ordered.

(Proof. If $X$ is empty, we're done; otherwise fix a surjection from $\alpha$ onto $X$, and for every $x\in X$ let $\beta_x$ be the least $\beta<\alpha$ mapped to $x$. Then $x\prec y\iff\beta_x<\beta_y$ is a well-ordering of $X$.)

Theorem. If $\leq^*$ satisfies trichotomy, then the axiom of choice holds.

Proof. Compare $X$ and $\aleph^*(X)$, it has to be the case that $X<^*\aleph^*(X)$, and therefore $X$ can be well-ordered.

Now, what can we say about $\leq^*$ in general? Not too much.

It does not have to be anti-symmetric; but if it is anti-symmetric then there are no Dedekind-finite sets.

Asaf Karagila
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