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The Parity Principle states that

if $X\neq \emptyset$ is a set, then there is $\mathcal B\subseteq \mathcal P(X)$ such that whenever $a,b\in \mathcal P(X)$ with $a\mathbin\Delta b = \{x\}$ for some $x\in X$, then $\mathcal B$ contains exactly one of $a$, $b$.

The Axiom of Choice for $2$-element sets $\mathbf{C}_2$ is strictly weaker than the original Axiom of Choice, and $\mathbf{C}_2$ implies the Parity Principle.

Does the Parity Principle imply $\mathbf{C}_2$?

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    $\begingroup$ Have you asked the question you intended? It seems we know that it does by the linked question, as you say. $\endgroup$ Commented Sep 13, 2023 at 10:53
  • $\begingroup$ Thanks @JoelDavidHamkins for spotting my mistake, I intended to ask about the other implication and have corrected my error. $\endgroup$ Commented Sep 14, 2023 at 11:16
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    $\begingroup$ @AndreasLietz I don't follow. The two different possibilities will have the same parity. $\endgroup$ Commented Sep 15, 2023 at 16:29
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    $\begingroup$ @AndreasLietz: I don’t follow either, but for a different reason than Elliot Glazer. In trying to proving $\mathbf{C}_2$ (or even full choice) over a well-ordered base by induction, the difficulty at limit steps isn’t in choosing the “next point”, it’s in picking earlier choices for all previous steps together. $\endgroup$ Commented Sep 15, 2023 at 16:33
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    $\begingroup$ Youre right, I was talking nonsense! $\endgroup$ Commented Sep 15, 2023 at 17:36

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Over ZFA, the Parity Principle is strictly weaker. We'll show it follows from Multiple Choice, the assertion that for any family of nonempty sets $\mathcal{F},$ there is $g: \mathcal{F} \rightarrow [\bigcup \mathcal{F}]^{<\omega} \setminus \{\emptyset\}$ such that $g(S) \subseteq S$ for all $S \in \mathcal{F}.$ Multiple Choice holds in the second Fraenkel Model ($\mathcal{N}_2$ in Consequences of the Axiom of Choice), in which $\mathbf{C}_2$ fails badly (in particular, its atoms are a family of Russell socks).

Fix $X.$ Let $\sim$ be the equivalence relation on $\mathcal{P}(X)$ defined by $S \sim T$ if $S \triangle T \in [X]^{<\omega}.$ Let $\mathcal{F}=\mathcal{P}(X) / \sim$ and let $g: \mathcal{F} \rightarrow [\bigcup \mathcal{F}]^{<\omega} \setminus \{\emptyset\}$ be such that $g(S) \subset S$ for all $S \in \mathcal{F}.$ Define a choice function $h: \mathcal{F} \rightarrow \bigcup \mathcal{F}$ by $h(S)=\bigcup g(S).$ Then $\mathcal{B}:=\{S \subseteq X: |S \triangle h([S]_{\sim} )| \in 2\mathbb{Z}\}$ is as desired.

Note that over ZF, Multiple Choice is equivalent to full Choice, so this approach doesn't immediately separate Parity Principle from $\mathbf{C}_2$ over ZF.

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    $\begingroup$ In human language the Axiom of Multiple Choice says you can assign to each nonempty set a finite nonempty subset. Since the connected components of the Hamming graph on $2^X$ are sublattices, if you choose a nonempty finite subset of each component, you can choose an element of each component by taking the supremum. As the graph has no odd cycles, once you establish a base point in each component, there is a unique proper $2$-coloring with all the base points colored blue. I post this redundant comment for the benefit of those like me who get a headache from deciphering a lot of notation. $\endgroup$
    – bof
    Commented Sep 16, 2023 at 3:28
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    $\begingroup$ @bof Thank you! $\endgroup$ Commented Sep 16, 2023 at 5:53

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