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Aug 7, 2019 at 17:52 comment added Alex Kruckman @MarkSmith Right. "Infinite" here refers to the cardinality of the resulting structure, not to the cardinality of the set of factors in the ultraproduct.
Aug 7, 2019 at 14:07 comment added Mark Smith Thanks! So if $\mathcal{C}=\{A_i\}_{i\in I}$, "an infinite ultraproduct of members of $\mathcal{C}$" does Not mean only a member of the set $\left\{\prod_{i\in I} A_i /\mathcal{U} : \mathcal{U} \text{ is an ultrafilter on } I \right\}$ ?
Aug 7, 2019 at 13:47 vote accept Mark Smith
Aug 7, 2019 at 3:27 history answered Alex Kruckman CC BY-SA 4.0