Let be such that for every distinct , .
Show that is countable.
Assume the contrary, i.e. is uncountable.
Consider the sets for . The union of these sets is , so one of them, say , must contain infinitely many points; otherwise could be written as the union of countably many finite sets and would be countable, contrary to our assumption.
Now what does that say about the sum , where ?
The point was to show that there exists an such that is infinite (no matter whether countably infinite or uncountably so). Therefore, by secting finitely many from , you can get arbitrarly large sums , which contradicts . Thus, the assumption that for uncountably many must be rejected.
even if each is finite, i don know how could be always true. say . If has finitely many elements in it and if you pick 3 elements from this set, then the sum is already greater than 1. or i m not supposed to pick points like that? i guess i m not understanding this funtion.
Well, that is a given fact about the particular function f in question. There is no need to prove that it holds. You just assume it holds and show, what you are asked to show.
Well, yes, if it is true that has at least 3 members, then you get the contradiction to the assumed property of f that for any n different elements of .say . If has finitely many elements in it and if you pick 3 elements from this set, then the sum is already greater than 1. or i m not supposed to pick points like that?
Similarly, if we know that , has at least m members (infinitely many, say), then you get the contradiction , for different elements from .
You need not understand this function: you only have to assume that it has the properties that you are told it has and move on to show what you are told to show.i guess i m not understanding this function.
But if you want an example of a function that has this property you may consider