# Complement-like operator and the axiom of choice

We say that an operator $$^*$$ on $${\cal P}(A)$$ is $$\star$$-complement if $$^*$$ is not the complement operator and for all $$X⊆A$$ we have:

• $$X^*∪X=A$$
• $$X^{**}=X$$

We say that $$^*$$ is $$\star$$-strong complement if it is $$\star$$-complement and for all $$X\subseteq A$$ we also have:

• $$X^*=X^c⇔$$ $$X$$ is finite or cofinite in $$A$$

We can prove that these three propositions are equivalent:

1. There exists $$\star$$-complement operator on $${\cal P}(A)$$
2. There exists a set $$Z⊆{\cal P}(A)$$ such that $$(Z,\supsetneq)≅(\Bbb Z,>)$$
3. $$A$$ is countable union of infinite disjoint sets

We can also prove that these two propositions are equivalent:

1. There exists $$\star$$-strong complement operator on $${\cal P}(A)$$
2. The set of infinite co-infinite subsets of $$A$$ can be partitioned such that for every $$Z$$ in the partition we have $$(Z,\supsetneq)≅(\Bbb Z,>)$$.

Sketch of the proof of 1 ⇔ 2 ⇔ 3:

(1) $$\implies$$ (2): Let $$^*$$ be such operator, take $$X⊆A$$ such that $$X^*≠X^c$$, we have $$X^*\supsetneq X^c$$ and $$X^*$$ is bijective. It is injective because $$x^*=y^*⇒x=x^{**}=y^{**}=y$$. It is surjective because $$(x^*)^*=x$$. Hence we can create the set: $$X, (X^*)^c,(((X^*)^c)^*)^c,\ldots, X^{(*c)^n},\ldots$$, which when ordered by $$\supsetneq$$ is isomorphic to $$(\Bbb N, >)$$). Because $$^*$$ and $$^c$$ are both bijections, we can also continue this sequence backwards.

(2) $$\implies$$ (1): Let $$Z⊆{\cal P}(A)$$ be that set, then define $$Z_n^*=Z_{n+1}^c$$.

(2) $$\implies$$ (3): We set $$C_i=Z_i\setminus Z_{i+1}$$, then by letting $$f:\Bbb N^2→\Bbb Z$$ be bijection, we have $$\bigcup_{j∈ω}(\bigcup_{k∈ω} C_{f(j,k)})$$ is a subset of $$A$$ that is countable union of infinite disjoint sets.

(3) $$\implies$$ (2): Let $$(A_i)_{i∈\mathbb Z}$$ be the sequence of dijoint sets such that $$\bigcup_{i∈\mathbb Z}A_i=A$$, then let $$Z_i=\bigcup_{k>i} A_i$$.

We can prove in $$ZF$$ that there exists a set with $$\star$$-complement.

Consider the axioms: $$\star C: \text{for every infinite }A, \text{ there is a } \star \text{-complement operator on } {\cal P}(A)$$ $$\star SC: \text{for every infinite }A, \text{ there is a strong } \star \text{-complement operator on } {\cal P}(A)$$

Clearly $$\star C$$ is not provable in $$ZF$$, since it proves that there are no amorphous sets.

My questions are:

• What can we say about the existence of $$\star$$-strong complement operator in $$ZF$$?
• Can we prove in $$ZF$$ that $${\cal P}(\Bbb N)$$ has $$\star$$-strong complement?
• How strong are the axioms $$\star C$$ and $$\star SC$$?
• @NoahSchweber The first line says that ⋆-complement is a operator that is not complement that satisfy the conditions – Holo Oct 25 '19 at 14:35
• Whoops, I'm good at reading. But that said, I don't immediately see why the existence of a $\star$-complement kills amorphous sets. – Noah Schweber Oct 25 '19 at 14:41
• It is immediate from the 3rd equivalence, if $A$ is amorphous, and ${\cal P}(A)$ has ⋆ -complement, then $A$ is countable union of infinite sets, hence it is not amorphous. You can also prove by induction that if $X⊆A$ is finite or cofinite, then $X^c=X^*$ – Holo Oct 25 '19 at 15:08
• Maybe I'm being thick, but I don't see either the argument(s) for that equivalence or the induction you mention at the moment; can you sketch one of them? – Noah Schweber Oct 25 '19 at 15:24
• Well, this certainly follows from $a+a=a$ for all infinite cardinals, since that implies $a=a\cdot\aleph_0$, and therefore every set is equipotent with the disjoint union of countably many infinite sets. This axiom, however, is not terribly strong. – Asaf Karagila Oct 30 '19 at 5:54