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From: "Jirí Adámek" <j.adamek@tu-bs.de>
To: <ptj@maths.cam.ac.uk>
Cc: Categories mailing list <categories@mta.ca>
Subject: Re: Non-cartesian closedness of Met
Date: Mon, 19 Dec 2022 09:50:30 +0100	[thread overview]
Message-ID: <E1p7SFy-000167-EN@rr.mta.ca> (raw)
In-Reply-To: <E1p6HOO-0005FO-RF@rr.mta.ca>

Dear Peter,

Thanks for pointing out that my argument has been incomplete. Here
is a proof that Met is not a ccc:

We use that limits in Met are the limits in Set with the supremum
metric (coordinate-wise). Let us denote by Met(X,Y) the hom-sets with the
supremum metric d'(f,g) = sup_x d(fx,gx). The addition of real
numbers from RxR to R is not non-expanding, but its curried from
R to Met(R,R) is; thus all we need to do is to show that if Met were a
ccc, one could take [X,Y]=Met(X,Y).

The underlying set of [X,Y] can be taken to be the hom-set, using
adjoint transposes of morphisms from 1 to [X,Y]. And the universal
morphism eval:[X,Y]xX -> Y can be taken to be the evaluation map
(precompose it with morphisms fxX for f:1->[X,Y]). The metric d of
[X,Y] satisfies d \geq d', using adjoint transposes of morphisms
from 2-element spaces to [X,Y]. To prove d \leq d', we first consider
a finite space X. Let id: n->X be the identity map from the discrete
space on the underlying set of X, a coproduct of n copies of 1. Then
[id,Y]: [X,Y]->[n,Y]= Y^n demonstrates that d= d'. For X arbitrary,
express it as a directed colimit X=colim X_i of all finite subspaces X_i.
Then [X,Y] = lim [X_i,Y] carries the supremum metric.

Best regards,
Jiri


[For admin and other information see: http://www.mta.ca/~cat-dist/ ]


  parent reply	other threads:[~2022-12-19  8:50 UTC|newest]

Thread overview: 5+ messages / expand[flat|nested]  mbox.gz  Atom feed  top
2022-12-16 16:41 ptj
2022-12-17  0:27 ` Non-cartesian closedness of Met (ptj@maths.cam.ac.uk) Vaughan Pratt
2022-12-19  8:50 ` Jirí Adámek [this message]
2022-12-17  9:20 Non-cartesian closedness of Met Dirk Hofmann
2022-12-18 13:04 ptj

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