1. ## Help with proof

If S is uncountable. Show it is impossible that P({i}) > 0 for every i∈S.

2. The relevant constraint is that $\displaystyle \sum P(i) = 1$ 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
$\displaystyle (P_1 + P_2 + P_3 + ...) \geq (x + x + x + x ....)$

So its sufficient to show that
$\displaystyle (x + x + x + x ....) > 1$

This is true, since you are adding up a strictly positive number an uncountably large number of times (ie, more than 1/x times).

3. Originally Posted by SpringFan25
Equivalent statement: Min (P{i}) > 0
Can we always talk of a minimum when the set under consideration is countable ?

Also, it seems that your proof does not really use uncountability.

4. Can we always talk of a minimum when the set under consideration is countable ?
I think you mean uncountable, and I dont understand why that would be a problem. For example the subset of real numbers $\displaystyle [2,\infty)$ is uncountable but clearly has a minimum of 2.

Also, it seems that your proof does not really use uncountability.
it did use and require uncountability. The final step was to show that (x + x + x + x + x... ) > 1
This is true iff there are more than $\displaystyle \displaystyle \lceil \ 1/x \rceil$ 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 $\displaystyle \displaystyle \lceil \ 1/x \rceil$ elements.

(note that $\displaystyle \displaystyle \lceil \ 1/x \rceil$ is well defined as the proposition was that x>0)

5. Originally Posted by Sneaky
If S is uncountable. Show it is impossible that P({i}) > 0 for every i∈S.
You haven't said that P is supposed to denote a probability on the set S, so that $\displaystyle \sum P(\{i\}) = 1$. 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 $\displaystyle A_n$. If P({i}) > 0 then it must be true that P({i}) > 1/n for some n and so $\displaystyle i\in A_n$, and therefore i is in the union of all the $\displaystyle A_n$. 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.