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Jan 24, 2017 at 14:22 comment added Dominic van der Zypen OK - thanks Tomas, maybe a variant of your argument can be used to show there is no isomorphism in general? (Although my intuition was, the two posets are isomorphic, but intuitions in mathematics can of course be misleading.)
Jan 23, 2017 at 15:57 comment added Tomáš Jakl I see, your question is more general. My example explains why the embedding $\text{PropCov} \hookrightarrow \text{Cov}$, given by $A \mapsto [A]_\simeq$, is not an isomorphism.
Jan 23, 2017 at 13:25 comment added Joel David Hamkins Could you explain how you intend to use this to show that the partial orders are not isomorphic?
Jan 23, 2017 at 10:47 history answered Tomáš Jakl CC BY-SA 3.0