H(R) is the set of all non-empty compact subsets of the reals (i.e. closed and bounded).
To show that the function is continuous I'm trying to use the fact that it would be equivalent to showing the pre-image of a closed subset of H(R) is closed in [0,1].
I guess it's the usual topology on R where open sets are open intervals with the Euclidean metric.
Alexander Davis
? how so? Isn't H a collection of compact sets?
Elijah Phillips
Yeah, non-empty compact subsets of R.
Kevin Myers
so? what is the topology? show me an open set in H
Elijah Nelson
My guess is that you get an open set in H by taking an open set in the reals, and then defining the set of all compact sets inside this open set to be open.
Ryan Brown
Don't know if this helps, if d([a,a+x],[a,a+y]) = d(x,y) then f_a is an isometry then it's continous
Oliver Green
btw, which textbook are you using?
Nathaniel Walker
I'm not quite sure what an open set would look like but a closed set is something like [a,a]={a} or [a,a+x] for some x in (0,1].
Sorry for the confusion. I don't understand the question very well.