Suppose a variety V is defined by an infinite minimal set of identities. Show that V is a subvariety of at least continuum many varieties.
1 Answers
Let $V = Mod(\Sigma)$, where $\Sigma$ is the infinite minimal set of identities.
$Mod(\cdot)$ operator is antitone and so for any subset $\Sigma' \subseteq \Sigma$ of identities we get the following inclusion: $V = Mod(\Sigma) \subseteq Mod(\Sigma')$. Hence we have the correspondence between the varieties containing $V$ and the subsets of $\Sigma$ which is given by $Mod(\cdot)$ operator.
This correspondence is injective because of the minimality of $\Sigma$.
Suppose that $Mod(\Sigma_1) = Mod(\Sigma_2)$ while $\Sigma_1 \neq \Sigma_2$, where $\Sigma_1, \Sigma_2 \subseteq \Sigma$. W.l.o.g. $\Sigma_2 \setminus \Sigma_1 \neq \varnothing$.
It implies that $Mod(\Sigma_1 \cup \Sigma_2) = Mod(\Sigma_1) \cap Mod(\Sigma_2) = Mod(\Sigma_1)$, showing that the set of identities $\Sigma_1 \cup \Sigma_2$ can be reduced to $\Sigma_1$. Hence we can remove $\Sigma_2 \setminus \Sigma_1$ identities from $\Sigma$, which contradicts its minimality.
Since $|\Sigma| \geqslant \aleph_0 \Rightarrow 2^{|\Sigma|} \geqslant 2^{\aleph_0} = \mathfrak{c}$, we have at least continuum many varieties containing $V$ which correspond to the subsets of $\Sigma$.
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