Let be such that for every distinct , .
Show that is countable.
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 ?
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, 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