A $\sigma$ algebra of subsets of a set X is defined as a collection of subsets of X which is invariant by taking complements and denumerable unions. And which contains the empty set.

But this last condition is (almost) superfluous. If there exists an element, say $A$, in the $\sigma$-algebra, then it must contain $(A\cup A^c)^c$, which is the empty set. Hence the sole purpose of requiring the empty set to be in the $\sigma$-algebra, is to deny the empty set the right to be a $\sigma$-algebra itself.

I don't really know why the empty set should not be a $\sigma$-algebra, and I don't see any result that would suddenly fail badly if we give the empty set this promotion.

A $\sigma$ algebra of subsets of a set X is defined as a collection of subsets of X which is invariant by taking complements and denumerable unions. And which contains the empty set.

But this last condition is (almost) superfluous. If there exists an element, say $A$, in the $\sigma$-algebra, then it must contain $(A\cup A^c)^c$, which is the empty set. Hence the sole purpose of requiring the empty set to be in the $\sigma$-algebra, is to deny the empty set the right to be a $\sigma$-algebra itself.

I don't really know why the empty set should not be a $\sigma$-algebra, and I don't see any result that would suddenly fail badly if we give the empty set this promotion.

EDIT: Bourbaki, Topology, ch5, section 6 no 3 (TG IX.60) does not require the empty set to be in the $\sigma$-algebra. So I think this is really superfluous.