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    • CommentRowNumber1.
    • CommentAuthorTobyBartels
    • CommentTimeApr 5th 2017
    • (edited Apr 6th 2017)

    This theorem, with a constructive proof, is now at convergence space. (The usual proof in undergraduate metric-space theory uses both excluded middle and countable choice1, so I wrote this mostly to verify that it is actually perfectly constructive in the general setting.)


    1. ETA: And the straightforward generalization to nonsequential spaces would use choice of arbitrarily high cardinality. 

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeApr 5th 2017

    Where? Ah, here.

    • CommentRowNumber3.
    • CommentAuthorTobyBartels
    • CommentTimeApr 6th 2017

    Yes, I forgot that I could link directly to it.

    • CommentRowNumber4.
    • CommentAuthorTobyBartels
    • CommentTimeApr 6th 2017

    The proof is pretty sketchy, especially considering that the theorem as stated actually has 4 parts. But I do think that all of the steps are pretty obvious nose-following except for the bit that I wrote down. (Even that is arguably nose-following; I mean, what could you possibly look at besides the neighbourhood filter? But I was worried for a little while that it might not actually be constructive, so it's helpful to me to have it.)

    • CommentRowNumber5.
    • CommentAuthorTobyBartels
    • CommentTimeApr 6th 2017

    One might also write out how they're all reflective subcategories. That's probably also pretty obvious, but I haven't thought the whole thing through.