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Nov 26, 2016 at 11:23 comment added Franka Waaldijk Oops, I hit enter when I was still editing my comment. Because I wanted to start with saying that I really think that Matt's second proof is very beautiful! I have to get the hang of MO, sorry.
Nov 26, 2016 at 11:21 comment added Franka Waaldijk I still don't have enough reputation to comment on any other answer except my own... So I'll be a little creative.By the way, I said that in Matt's first proof there is a hidden use of countable choice, but it really is a hidden use of dependent choice. This is the "only" failure that I see in that answer (well, it ceases to be an answer then I suppose), and I am surprised that no one noticed the use of DC in that proof.
Nov 26, 2016 at 10:38 comment added Franka Waaldijk Like I said, in CLASS, INT and RUSS we can prove that "an arbitrary pointwise continuous function" coincides with "a #-morphism". That is why I consider my answer to be an answer. For a much more detailed answer see my reply to your same question on constructive-news.
Nov 25, 2016 at 19:27 comment added Mike Shulman This may be interesting, but it does not answer the question, which was specifically about "an arbitrary pointwise continuous function".
Nov 25, 2016 at 10:37 history edited Franka Waaldijk CC BY-SA 3.0
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Nov 25, 2016 at 10:08 review Late answers
Nov 25, 2016 at 10:29
Nov 25, 2016 at 9:58 history edited Franka Waaldijk CC BY-SA 3.0
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Nov 25, 2016 at 9:55 review First posts
Nov 25, 2016 at 12:09
Nov 25, 2016 at 9:53 history answered Franka Waaldijk CC BY-SA 3.0