In an countably infinite set $\Omega$ (sample space), is it possible $P(\omega_1) \neq P(\omega_2)$ for $\omega_1, \omega_2 \in \Omega$ Let $\Omega$ be a countably infinite set and let $P:2^{\Omega} \to \mathbb{R}$ be a probability measure. Show that there exist $\omega_1, \omega_2 \in \Omega$ such that $P(\{\omega_1\}) \neq P(\{\omega_2\})$. Can it hold that $P(\omega) > 0$ for all $\omega \in \Omega$.
 A: Hint: Since $P$ is a probability measure it is $\sigma$-additive. In particular, for any enumeration $(\omega_i)_{i\geq 1}$ of the points of $\Omega$,
$$1=P(\Omega)=P(\cup_{i=1}^\infty \omega_i)=\sum_{i=1}^\infty P(\omega_i).$$  What happens if $P(\omega_i)=P(\omega_j), \:\forall  i,j\geq 1$? Can you really rule out $P(\omega)>0\: \forall\omega\in\Omega$?
A: Thanks for the help! I think this is the answer.
Question 1: Show that there is $\omega_1, \omega_2 \in \Omega$ such that $P(\{\omega_1\}) \neq P(\{\omega_2\})$. 
Suppose $P(\{\omega_i\}) = P(\{\omega_j\})$ for all $i,j \in \mathbb{N}$. 
If $P(\{\omega_i\}) = P(\{\omega_j\}) = 0$, then $P(\omega_1) + P(\omega_2) + \ldots = 0$. We also that if $A_1, A_2$ are disjoint events in $\mathcal{F} = 2^\Omega$ (event space), then $$
\sum_{i=1}^{\infty} P(A_i) = P\left( \bigcup_{i=1}^{\infty} A_i\right) 
$$
since $P: 2^{\Omega} \to \mathbb{R}$ is a probability measure. Since $\{\omega_i\} \cap \{\omega_j\} = \emptyset$ for all $i,j$ it means they are all disjoint events in $\mathcal{F}$. So it means that $$
P(\omega_1) + P(\omega_2) + \ldots = \sum_{i=1}^{\infty} P\{\omega_i\} = P\left(\bigcup_{i=1}^{\infty} \{\omega_i\} \right) = P(\Omega) = 0
$$
This leads to a contradiction, since $P(\Omega) = 1$ must hold, since $P$ is a probability measure.
Suppose $P(\{\omega_i\}) = P(\{\omega_j\}) = p, \forall i,j \in \mathbb{N}$ with $p \in [0,1]$. Then the sum$$
P(\{\omega_1\}) + P(\{\omega_2\}) + \ldots = \sum_{i=1}^{\infty} P(\{\omega_i\}) = \sum_{i=1}^{\infty} p = \infty 
$$
would diverge. And this would mean that $$
\sum_{i=1}^{\infty} P(\{\omega_i\}) = P\left(\bigcup_{i=1}^{\infty} \{\omega_i\} \right) = P(\Omega) = \infty
$$
what would be an obvious contradiction with $P$ being a probability measure. In conclusion $\exists i,j$ such that $$
P(\{\omega_i\}) \neq P(\{\omega_j\})
$$
As for the second question: Can it hold that $P(\omega) > 0$ for all $\omega \in \Omega$. If we look at how many times we have to throw a fair coin to throw $k$ consecutive heads, then $\Omega = \mathbb{N}$, $\mathcal{F} = 2^{\Omega}$ and $P(\{n\}) = \left(\frac{1}{2}\right)^n$ for all $n \in \Omega$. So it can hold that $P(\Omega) > 0$ for all $\omega \in \Omega$
