When can you subtract cardinals? Does $|A|+|C|=|B|+|C|>|C|$ imply $|A|=|B|$ in ZF?

Question

Suppose that $|A|+|C|=|B|+|C|$, and that both sides are strictly larger than $|C|$. Does this imply $|A|=|B|$ in the absence of the axiom of choice?

With choice, cardinals are totally ordered, and $\kappa_1+\kappa_2=\max\{\kappa_1,\kappa_2\}$ for infinite cardinals, so this is trivial. But I worry that things might go wrong when choice is not present.

Answer 1

No, of course not.

Suppose that $A$ is an infinite Dedekind-finite set and let $B=A\cup\{A\}$, then $|A|<|B|$. Now, let $C=\omega$, then $|A|+\aleph_0=|A|+(1+\aleph_0)=(|A|+1|)+\aleph_0=|B|+\aleph_0$.

Since $A\cup\omega$ contains a Dedekind-finite subset, it must be greater in cardinality than $\aleph_0$, so your condition is satisfied.

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More generally, this would be equivalent to the axiom of choice. To see that, let $U$ be any set which cannot be well-ordered and let $\kappa$ be $\aleph(U)$, the Hartogs number of $U$.

Then consider $A=U+\kappa$ and $B=U+\kappa^+$, and let $C=\kappa^{++}$, the argument goes as in the above case. Then the only way that we fail the required condition is that $|C|\nless|A|+|C|=|B|+|C|$, in which case $A$ can be well-ordered.

Answer 2

To complement Asaf's answer, let me mention that there are some interesting $\mathsf{ZF}$ theorems showing that in various cases (not just the well-ordered context) we have a well defined subtraction operation. (The subtraction $\frak x-\frak y$ of cardinals is defined if and only if there is a unique cardinal $\frak z$ such that $\frak y+\frak z=\frak x$, in which case $\frak x-\frak y$ is defined as $\frak z$.)

The best known result in this regard is due to Lindenbaum and Tarski. They claimed in 1926 that (provably without choice) $2^{\frak a} - {\frak a} = 2^{\frak a}$ for all cardinals ${\frak a}\ge\aleph_0$. The first published proof is due to W. Sierpiński in 1947 ("Démonstration de l'égalité $2^{\frak m}-{\frak m}=2^{\frak m}$ pour les nombres cardinaux transfinis", Fund. Math. 34 (1947), 113–118; MR0023315; "Sur la différence de deux nombres cardinaux", Fund. Math. 34 (1947), 119–126; MR0023316]).

The recent paper by Guozhen Shen "Remarks on infinite factorials and cardinal subtraction in $\mathsf{ZF}$", MLQ Math. Log. Q. 68 (2022), no. 1, 67–73, MR4413645, shows that, provably in $\mathsf{ZF}$, for all cardinals $\mathfrak a$, if $\aleph_0\le 2^{\mathfrak a}$ and there is a permutation without fixed points on a set of cardinality $\mathfrak a$, then ${\mathfrak a}!-{\mathfrak a} = {\mathfrak a}!$. (As Shen points out in the comments) the paper also proves that for all cardinals $\mathfrak a,\mathfrak b$, if $\aleph_0\le2^{\mathfrak a}$ and $\mathfrak b\le^* \mathfrak a$ (meaning that there is a surjection from a subset of $\mathfrak a$ onto $\mathfrak b$), then $2^{\mathfrak a}-\mathfrak b=2^{\mathfrak a}$.