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Feb 5, 2020 at 17:30 comment added YCor [Of course another counterexample for the first fact is for degenerate varieties, namely consisting only of sets of cardinal $\le 1$, for which all free algebras on nonempty sets are singletons. A nondegenerate variety has models of arbitrary large infinite cardinals (consider a fixed model of cardinal $\ge 2$ and its powers). So $\mathcal{V}_2$ is interesting because it's also nondegenerate.]
Jan 31, 2020 at 16:06 history answered YCor CC BY-SA 4.0