Proof of the probability of the complement

Proof

1. Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $A \in \mathcal{F}$ be an event.
2. The family $\mathcal{F}$ is a σ-algebra. So the complement $A^c$ is in $\mathcal{F}$, and so is $\varnothing = \Omega^c$.
3. We first show $P(\varnothing) = 0$. Let $c = P(\varnothing)$, a number in $[0, 1]$.
4. The sets $E_n = \varnothing$ for $n \ge 1$ are pairwise disjoint and their union is $\varnothing$. By the countable additivity of the probability measure $P$, we get $c = \sum_{n=1}^{\infty} c$.
5. Suppose that $c > 0$. By the Archimedean property, there is $N$ with $N c > 1$. Then the partial sums of $\sum_n c$ exceed $1$ from the index $N$ on, so the series does not converge to $c \le 1$.
6. This contradicts step 4. So $c = 0$, that is, $P(\varnothing) = 0$.
7. Now let $E_1 = A$, $E_2 = A^c$ and $E_n = \varnothing$ for $n \ge 3$. These sets are pairwise disjoint, because $A \cap A^c = \varnothing$ and $\varnothing$ meets no set.
8. Their union is $A \cup A^c = \Omega$.
9. Countable additivity and step 6 give $P(\Omega) = P(A) + P(A^c) + \sum_{n \ge 3} 0 = P(A) + P(A^c)$.
10. Since $P(\Omega) = 1$, we get $P(A) + P(A^c) = 1$. Hence $P(A^c) = 1 - P(A)$.