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Oct 15, 2023 at 11:11 comment added Asaf Karagila Indeed, that is one way of finding such $X$.
Oct 15, 2023 at 4:39 comment added bof You left as an exercise the construction of $\mathfrak m$ such that $1+\mathfrak m\le^*\mathfrak m\lt1+\mathfrak m$. I guess one way to do this is to take a Dedekind-finite infinite set $X$ and let $\mathfrak m$ be the cardinality of the set of all non-null finite sequences of distinct elements of $X$.
Oct 15, 2023 at 4:26 history edited bof CC BY-SA 4.0
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Oct 14, 2023 at 19:53 history answered Asaf Karagila CC BY-SA 4.0