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 $\star SC$?
  • $\begingroup$ @NoahSchweber The first line says that ⋆-complement is a operator that is not complement that satisfy the conditions $\endgroup$ – Holo Oct 25 '19 at 14:35
  • $\begingroup$ Whoops, I'm good at reading. But that said, I don't immediately see why the existence of a $\star$-complement kills amorphous sets. $\endgroup$ – Noah Schweber Oct 25 '19 at 14:41
  • $\begingroup$ 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 $X^c=X^*$ $\endgroup$ – Holo Oct 25 '19 at 15:08
  • $\begingroup$ 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? $\endgroup$ – Noah Schweber Oct 25 '19 at 15:24
  • 2
    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Oct 30 '19 at 5:54

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