Theorem: The probability of the empty set is zero:

$\label{eq:prob-emp} P(\emptyset) = 0 \; .$

Proof: Assume that the probability of the empty set is not zero, i.e. $P(\emptyset) > 0$. Then, in an equation derived when proving that probability is monotonic,

$\label{eq:pB} P(B) = P(A) + P(B \setminus A) + \sum_{i=3}^\infty P(E_i) \quad \text{where} \quad E_i = \emptyset \quad \text{for} \quad i \geq 3 \; ,$

the right-hand side would be infinite. However, by the first axiom of probability, the left-hand side must be finite. This is a contradiction. Therefore, $P(\emptyset) = 0$.