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/ ]