Topology is hard

Does anybody want to give me a hint on problem 3?

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].

However, I don't know how to execute this.

Other urls found in this thread:

dropbox.com/s/d0ai3iqzicz4kc9/hyperspace 2-4.pdf?dl=0
en.wikipedia.org/wiki/Hausdorff_distance
twitter.com/SFWRedditGifs

what is the topology on H?

I guess it's the usual topology on R where open sets are open intervals with the Euclidean metric.

? how so? Isn't H a collection of compact sets?

Yeah, non-empty compact subsets of R.

so? what is the topology? show me an open set in H

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.

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

btw, which textbook are you using?

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.

No textbook. We're using some notes our professor compiled.

Here's the chapter this problem is from. dropbox.com/s/d0ai3iqzicz4kc9/hyperspace 2-4.pdf?dl=0

After reading the notes I'm pretty sure the answer is

It's the Vietoris topology.

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.

Continuity is a topologic property, if we don't know the topology, we can't help you. (what is h in problem 2?)

It's the Hausdorff distance
en.wikipedia.org/wiki/Hausdorff_distance

The topology is generated by this metric.

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.

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?).

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 ?

Okay, that makes sense. By ]a-r , a+x+r[ do you mean the complement of [a-r,a+x+r]?

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.

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?

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.

>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.