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Feb 9, 2021 at 1:53 history edited Gabe Goldberg CC BY-SA 4.0
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Feb 9, 2021 at 0:36 comment added user141903 Wow, what an amazing answer. Martin's Maximum... was not expecting that.
Feb 8, 2021 at 22:34 comment added zeb This is a remarkably complete answer!
Feb 8, 2021 at 22:34 vote accept zeb
Feb 8, 2021 at 18:49 comment added Gabe Goldberg Should be fixed now
Feb 8, 2021 at 18:45 history edited Gabe Goldberg CC BY-SA 4.0
deleted 68 characters in body
Feb 8, 2021 at 18:36 comment added Gabe Goldberg No, you're right. I confused myself! Let me fix it up
Feb 8, 2021 at 17:29 comment added zeb I don't see how $\mathcal{T}(\alpha)$ is Turing equivalent to the $1$-type of $\alpha$: this doesn't seem to be true for $\alpha = 1$, for instance (the $1$-types of $0$ and $1$ are Turing equivalent, but $\mathcal{T}(1)$ is not computable from $\mathcal{T}(0)$).
Feb 8, 2021 at 16:39 history edited Gabe Goldberg CC BY-SA 4.0
added 1 character in body
Feb 8, 2021 at 7:38 history edited Gabe Goldberg CC BY-SA 4.0
added 143 characters in body
Feb 8, 2021 at 7:29 history edited Gabe Goldberg CC BY-SA 4.0
deleted 38 characters in body
Feb 8, 2021 at 7:19 history edited Gabe Goldberg CC BY-SA 4.0
added 23 characters in body
Feb 8, 2021 at 7:12 history answered Gabe Goldberg CC BY-SA 4.0