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.
Brody Evans
No textbook. We're using some notes our professor compiled.
After reading the notes I'm pretty sure the answer is
Andrew James
It's the Vietoris topology.
Jackson James
For every [math]a \in \mathbb{R} [/math] and [math] x \in [0,1] [/math] we have [eqn] \lim_{y \to 0} h(f_a(x+y) , f_a(x)) = \lim_{y \to 0} y = 0 [/eqn] Therefore [math]f_a [/math] is continuous.
Connor Robinson
Continuity is a topologic property, if we don't know the topology, we can't help you. (what is h in problem 2?)
Alternatively in this metric every open set of the functions range takes the form of all intervals [a,a+x] where x is in some (a,b) contained in [0,1], hence every open set of the functions range has an open inverse image.
Alexander Murphy
Thanks guys, you've helped out a lot.
If anyone would like provide another hint for a different problem it would be much appreciated.
I'm trying to prove that the collection of sets in Definition 1 is a basis for some topology (Vietoris?).
Asher Miller
Ok so an open ball of center fa(x) and radius r would be the set of all compacts of R contained in ]a-r , a+x+r[ . Are you ok with that, because you seem confused about what an open set of H looks like ?
Jack Bailey
Okay, that makes sense. By ]a-r , a+x+r[ do you mean the complement of [a-r,a+x+r]?
Matthew Martin
Thats not quite correct. An open ball of center fx(x) and radius r would be the set of all sets of the form [a, a + x + h], where |h| < r. The left endpoint must be "a" for interval to be in the range of f.
John Bailey
Not OP but can someone clear up the confusion of a topology noob
How is [0, 1] a topological space when paired with the usual Euclidean topology? The book hasn't mentioned subset topologies at all. Does the restriction of the Euclidean metric to [0, 1] somehow induce a subset topology? Otherwise how can there be an open interval around 0 or 1?
Camden Myers
He means (a-r, a+x+r).
He's right, the metric is the Hausdorff metric.
An open set in the subspace doesn't need to be open in the entire space. [0, 1/2) is open in [0, 1], for example.
Lucas Carter
>An open set in the subspace doesn't need to be open in the entire space. [0, 1/2) is open in [0, 1], for example. Alright so I'm guessing that the open ball function in (R, d) is different from the one in [math](I, d\vert_I)[/math]? That would make sense.