If S is uncountable. Show it is impossible that P({i}) > 0 for every i∈S.
The relevant constraint is that cannot hold under these conditions.
proposition: If S is uncountable. Show it is impossible that P({i}) > 0 for every i∈S.
Equivalent statement: Min (P{i}) > 0
Denote this minimum = x
Now note that
So its sufficient to show that
This is true, since you are adding up a strictly positive number an uncountably large number of times (ie, more than 1/x times).
I think you mean uncountable, and I dont understand why that would be a problem. For example the subset of real numbers is uncountable but clearly has a minimum of 2.Can we always talk of a minimum when the set under consideration is countable ?
it did use and require uncountability. The final step was to show that (x + x + x + x + x... ) > 1Also, it seems that your proof does not really use uncountability.
This is true iff there are more than terms in the summation. Every element of the set S creates 1 term in the summation. Since S is uncountable, it must have more than elements.
(note that is well defined as the proposition was that x>0)
You haven't said that P is supposed to denote a probability on the set S, so that . As SpringFan25 pointed out, this is crucial to the proof.
For each value of n=1,2,3,..., there are at most n elements i of S with P({i}) > 1/n (otherwise the sum of their probabilities would be greater than 1). Call the (finite) set of all such elements . If P({i}) > 0 then it must be true that P({i}) > 1/n for some n and so , and therefore i is in the union of all the . Thus the set of all elements i such that P({i}) > 0 is a countable union of finite sets and therefore countable.
So if S is uncountable it cannot be true that P({i}) > 0 for all i in S.