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  "body": "\n## Proof\n\n1. Let $(\\Omega, \\mathcal{F}, P)$ be a [[probability space|0ZYBZ34RT2]] and let $A \\in \\mathcal{F}$ be an [[event|00SQGJNRTB]].\n2. The family $\\mathcal{F}$ is a [[σ-algebra|04WSW8T3SY]]. So the [[complement|0B1FSS1VZW]] $A^c$ is in $\\mathcal{F}$, and so is $\\varnothing = \\Omega^c$.\n3. We first show $P(\\varnothing) = 0$. Let $c = P(\\varnothing)$, a number in $[0, 1]$.\n4. 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|0VGMEBVW1M]] $P$, we get $c = \\sum_{n=1}^{\\infty} c$.\n5. Suppose that $c \u003e 0$. By the [[Archimedean property|0A3GWZ86AY]], there is $N$ with $N c \u003e 1$. Then the partial sums of $\\sum_n c$ exceed $1$ from the index $N$ on, so the series does not converge to $c \\le 1$.\n6. This contradicts step 4. So $c = 0$, that is, $P(\\varnothing) = 0$.\n7. 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.\n8. Their union is $A \\cup A^c = \\Omega$.\n9. Countable additivity and step 6 give $P(\\Omega) = P(A) + P(A^c) + \\sum_{n \\ge 3} 0 = P(A) + P(A^c)$.\n10. Since $P(\\Omega) = 1$, we get $P(A) + P(A^c) = 1$. Hence $P(A^c) = 1 - P(A)$.\n\n## Idea\n\nAn event and its complement split the sample space into two disjoint pieces. Additivity turns this split into a sum of probabilities, and the whole space has probability $1$. The only care needed is to get finite additivity from countable additivity, which uses $P(\\varnothing) = 0$.\n",
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