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- From: Thorsten Altenkirch <Thorsten.Altenkirch AT nottingham.ac.uk>
- To: Ettore Aldrovandi <ealdrov AT math.fsu.edu>
- Cc: "coq-club AT inria.fr" <coq-club AT inria.fr>, agda-list <agda AT lists.chalmers.se>, "coq+miscellaneous AT discoursemail.com" <coq+miscellaneous AT discoursemail.com>, lean-user <lean-user AT googlegroups.com>
- Subject: Re: [Coq-Club] [Agda] Why dependent type theory?
- Date: Thu, 5 Mar 2020 15:39:13 +0000
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Hi Ettore,
In set theory the union of two sets (written \cup) is the set which contains the elements which are in one set or the other. Hence I cannot see what is “mathematically incorrect” in my example
Cheers, Thorsten
From: Ettore Aldrovandi <ealdrov AT math.fsu.edu>
Hi,
I think the example below is not mathematically correct. The problem is that \cup is not the same as \sqcup. The latter is of course a coproduct in the category of sets, whereas the former is a push-out, so a colimit of a more complicated diagram. In the line
of course the two sets {0,1} and {0,1,2,3} are not disjoint, whereas in the line
the union is actually disjoint, i.e. a coproduct. In the example with the sum,
in the first line {0,1} is actually made disjoint from {0,1,2,3}. To turn this around, suppose you do a push-out
{true, false} \coprod_{0,1} {0,1,2,3}
where you use the maps f : {0,1} -> {true, false} and i : {0,1} ->{0,1,2,3} . Then, since f is an isomorphism, you get something isomorphic to the union.
So, this example doesn’t really show that \cup exposes the implementation. But part of this example becomes possible because in sets we have naively “disembodied” elements leading to constructions of this sort…
…I guess, I’m just learning this stuff myself. (First post here, actually!)
Best,
—Ettore
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- Re: [Coq-Club] [Agda] Why dependent type theory?, Thorsten Altenkirch, 03/05/2020
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