I am not sure what you are trying to do, but your code probably doesn't even compile, as subset() wants the starting and ending elements of the set, ie. it wants balls.

updated the example, i need a subset depending on the range of a slider.

Your stuff SortedSet losts its type...
There is no get() method on a TreeSet, perhaps you should read the JavaDoc more closely.
You can get an iterator on the TreeSet, do the iteration, maintaining a counter, and get your elements this way, perhaps.
Why are you using a SortedSet, anyway?

that's how far I got

the code runs
the actual sorting works (in balls)

I couldn't figure out the subSet part, mainly my b1 and b2 don't work and there's no get() 

it runs, but it doesn't work

1. // The type SortedSet is not generic; it cannot be parameterized with arguments
2. SortedSet<Ball> balls = new TreeSet<Ball>(new BallComparator());
3. void setup() {
4.   Ball b1= new Ball(random(100));
5.   Ball b2= new Ball(random(100));
6.   for (int i = 0; i < 50; i++) {
7.     Ball item1 = new Ball(random(100));
8.     println( item1.size ) ;
9.     balls.add(item1);
10.     if (i==25) {
11.       b1=balls.last();
12.     }
13.     if (i==48) {
14.       b2=balls.last();
15.     }
16.   }
17.   println("-----------------------------------------");
18.   int count = 0;
19.   for (Ball b : balls) {
20.     println( count + ": " + b.size );
21.     count ++;
22.   }
23.   //  Ball b1=balls.get(25);
24.   //  Ball b2=balls.get(50);
25.   SortedSet<Ball> test = balls.subSet(b1, b2);
26.   // SortedSet<Ball> test = balls.subSet(balls.get(25), b2); 
27.   println("-----------------------------------------");
28.   count = 0;
29.   for (Ball b : test) {
30.     // println(b.size);
31.     println( count + ": " + b.size );
32.     count ++;
33.   }
34. }
35. class Ball {
36.   float size;
37.   Ball(float size) {
38.     this.size = size;
39.   }
40. }
41. public class BallComparator implements Comparator<Ball> {
42.   public BallComparator() {
43.   }
44.   public int compare(Ball b1, Ball b2) {
45.     if (b1.size > b2.size) {
46.       return 1;
47.     }
48.     else if (b1.size < b2.size) {
49.       return -1;
50.     }
51.     return 0;
52.   }
53. }

both thanks. Didn't thought there would be so many downsides to it.