I'm studying for my probability exam and I'm dealing with this proof:
Consider the metric space $(\mathtt{N}, 2^\mathtt{N})$. For each $A \epsilon 2^\mathtt{N}$ define $$P\{A\} = \sum_{n \epsilon A}\frac{2}{3^n}$$ Show that P is a probability function.
I know that I have to show that $P\{\Omega\} = 1$; $P\{A\} \ge 0$ and that $P\{\bigcup_{i = 1}^\infty A_i \} = \sum_{i = 1}^\infty P\{A_i\}$, the part that I don't understand is how is $A\epsilon2^\mathtt{N}$. Is the event $\{1\}$ has the same probability as $\{2\}$ or $\{3\}$? How do I compute the probability of a set such as $\{1, 2, 3\}$ or $\{4, 6, 8\}$? Would be a great help if any of you could explain to me.
To prove that $\sum P(A_k)=P(\cup_n A_k)$ when $A_k$'s are disjoint write $P(A_k)$ as $\sum\limits_{n=1}^{\infty} I_{A_k}(n) \frac 2 {3^{n}}$ (and a similar expression for $P(\cup_n A_k)$. We can always interchange two sums when the terms are non-negative. Hence $\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} I_{A_k}(n) \frac 2 {3^{n}}=\sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} I_{A_k}(n) \frac 2 {3^{n}}$. From this conclude that $\sum P(A_k)=P(\cup_n A_k)$. [ Note that if $A=\cup_k A_k$ then $I_A=\sum\limits_{k=1}^{\infty} I_{A_k}$ because $A_k$'s are disjoint].