Sigma-Algebra is Delta-Algebra
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Theorem
A $\sigma$-algebra is also a $\delta$-algebra.
Proof
Let $\SS$ be a $\sigma$-algebra whose unit is $\mathbb U$.
Let $A_1, A_2, \ldots$ be a countably infinite collection of elements of $\SS$.
Then:
\(\ds \forall i: \, \) | \(\ds \mathbb U \setminus A_i\) | \(\in\) | \(\ds \SS\) | $\SS$ is closed under relative complement with $\mathbb U$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \bigcup_{i \mathop = 1}^\infty \paren {\mathbb U \setminus A_i}\) | \(\in\) | \(\ds \SS\) | $\SS$ is closed under countable unions | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbb U \setminus \bigcap_{i \mathop = 1}^\infty A_i\) | \(\in\) | \(\ds \SS\) | De Morgan's Laws | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \bigcap_{i \mathop = 1}^\infty A_i\) | \(\in\) | \(\ds \SS\) | $\SS$ is closed under relative complement with $\mathbb U$ |
Thus $\SS$ is a $\delta$-algebra.
$\blacksquare$