Q:

The infinite countable space

I saw a question asking to prove that
"there is no probability defined on a countable infinite space $\Omega$ assigning the same probability to each $ \omega \in \Omega$."
The proof contains the following equalities
$$1= p(\Omega)= p\left(\bigcup\limits_{i=1}^{\infty} \{w_i \}\right) = \sum_{i=1}^{\infty}p(w_i)= \sum_{i=1}^{\infty}p$$
I wonder based on what it assumes that $ p\left(\bigcup\limits_{i=1}^{\infty} \{w_i \}\right) = \sum_{i=1}^{\infty}p(w_i)$ without considering that the events are disjoint?
Can we consider that events in an infinite countable space are always disjoint? 
For interested people in the reference, you can see pages 14 and 15 in this file

A:

Can we consider that events in an infinite countable space are always disjoint?

No.  This would only be true if the set of events is $\{\emptyset, \Omega\},$ which while allowed is completely uninteresting. 
As written, these events are the singleton subsets of $\Omega$, and any two distinct singletons are disjoint.

