How to prove $P(\emptyset)=0$ from the axioms of probability? A text states that you can prove that a probability of a null set is 0 through one of the axioms of probability. I know the three axioms, but I fail to employ these axioms to prove the above. I understand it only intuitively but would like a more formal explanation. 
 A: What are the axioms of probability  according to you or your textbook or instructor?
If your third axiom merely says

If $A$ and $B$ are mutually exclusive events, then $P(A\cup B) = P(A) + P(B)$

then choosing $A = \Omega$ and $B = \emptyset$ suffices as in @BruceET's comment on your question. A more formal (pedantic?) approach might slip in the assertion that we know that $\Omega$ is an event, and so we have that $\Omega^c = \emptyset$ is also an event and so everything is kosher when we choose $\Omega$ and $\emptyset$ as our two events.
On the other hand, if your third axiom says

If $A_1, A_2, \ldots$ is a countably infinite sequence of mutually exclusive events, then $$P\left(\bigcup_{i=1}^\infty A_i\right) = \sum_{i=1}^\infty P(A_i)$$ 

then you have to work a little harder. We will borrow the previous pedantry to prove that $\emptyset$ is an event and then take the countably infinite sequence of mutually exclusive events as $\Omega, \emptyset, \emptyset, \ldots$. This gives
$$P\left(\bigcup_{i=1}^\infty A_i\right) = P\left(A_1 \cup \bigcup_{i=2}^\infty A_i\right) = P\left(\Omega \cup \bigcup_{i=2}^\infty \emptyset\right) = 1 + \sum_{i=2}^\infty P(\emptyset).$$
But $\displaystyle \Omega \cup \bigcup_{i=2}^\infty \emptyset = \Omega$ and so we have shown that
$$1 = P(\Omega) = 1 + \sum_{i=2}^\infty P(\emptyset).$$
Can you take it from here to complete the proof that it must be the case that $P(\emptyset)=0$?
