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Let $X$ be an infinite set and let $\mathcal F$ be a family of subsets of $X$ such that

$\bullet$ $X \in \mathcal F$ and $\mathcal F$ is closed under taking complements;

$\bullet$ $\mathcal F$ is closed under the symmetric difference $\Delta;$

$\bullet$ the group $\langle \mathcal F; \Delta\rangle$ is of index two in the group $\langle 2^X; \Delta\rangle.$

How a typical $\mathcal F$ like that can be constructed? How many are there the families $\mathcal F$ with the above properties up to the action of the symmetric group $\mathrm{Sym}(X)?$

P.H.
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    Requiring $\mathcal F$ to be closed under complementation is redundant, as the complement of any $Y \in \mathcal F$ is just the symmetric difference $X \Delta Y$. – Geoffrey Trang Jan 01 '25 at 20:11
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    Sure, but it is helpful to stress something sometimes. – P.H. Jan 01 '25 at 20:19

1 Answers1

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You are asking about $1$-codimensional subspaces in $\mathbb{F}_2^X$ that contain $1$, where $\mathbb{F}_2$ is the field with two elements. They are the kernels of non-zero linear maps $\mathbb{F}_2^X \to \mathbb{F}_2$ that map $1 \mapsto 0$.

The dimension of $\mathbb{F}_2^X$ is $2^d$ when $d := \mathrm{card}(X)$ by MO/49551 or MSE/58548. Hence, there are $2^{2^d}$ linear maps $\mathbb{F}_2^X \to \mathbb{F}_2$. For the cardinality, mapping $1 \mapsto 0$ and being non-zero doesn't matter. The action of $\Sigma_X$ won't change it either because it just has cardinality $2^d$ by MO/27785.

Martin Brandenburg
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