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Jan 9, 2023 at 13:43 vote accept Zuhair Al-Johar
Jan 9, 2023 at 12:04 history edited Joel David Hamkins CC BY-SA 4.0
added 339 characters in body
Jan 9, 2023 at 11:38 history edited Joel David Hamkins CC BY-SA 4.0
added 75 characters in body
Jan 9, 2023 at 11:36 comment added Joel David Hamkins Yes. I had said "definable", which to me means by default without parameters, but I have edited to clarify.
Jan 9, 2023 at 11:35 comment added Emil Jeřábek You need $\alpha\mapsto U_\alpha$ to be definable without parameters.
Jan 9, 2023 at 11:28 history answered Joel David Hamkins CC BY-SA 4.0