Can there exist a sigma-algebra that is infinite and countable?

Question

Not sure if such sigma-algebra exists

Answer 1

No. Let us suppose, for the sake of contradiction, that there was such a sigma algebra enumerated using the bijection h: n -> h(n)

Since sigma algebras, by definition, are closed under complementation and the taking of countable unions, and hence the taking of countable intersections, each of the sets j(s),

where : s is a unending sequence of 0s and 1s
and j(s) is the intersection of the sets h(s,n)
for n = 1 to infinity, where

h(s,n) = h(n) if the n-th element of the
sequence s is 1, and the complement
~h(n) otherwise

will be a member of our sigma algebra. By construction, the j(s)'s will be mutually disjoint, and every element of the sigma algebra will be a union of j(s)'s.

We know that infinitely many j(s)'s have to be non-null, because otherwise our sigma algebra, being the collection of the unions taken from a finite collection of sets will, itself, be finite, contrary to assumption. But if this is the case, since every infinite set includes a countably infinite subset, we can find a sequence

k1, k2, k3 , ...

where kn is the n-th j(s) enumerated in that sequence. Thus, if for any series of 0s and zeroes t, we define k(t) to be the union of the sets k(t,n), where

k(t,n) = kn if t(n) = 1
k(t,n)= the null set if t(n) = 0

We'll get a one-to-one mapping of those series into the sigma algebra. But this is impossible, because the set of all such sequences, courtesy of Cantor's diagonal argument, can be seen to be noncountable.

Answer 2

"It is easy to construct a sequence {A_n} of members of the sigma-algebra such that each A_n contains, but is not equal to, A_{n+1}. The key point is that, if A_n is a member of the sigma-algebra that contains infinitely many other members of the sigma-algebra, and B is any member (other than A_n or the empty set) contained in A, then B or A_n \ B (or both) contains
infinitely many members of the sigma-algebra. So you take A_{n+1} to be either B or A_n \ B.

Now B_n = A_n \ A_{n+1} form a sequence of mutually disjoint members of the sigma-algebra. For each set of positive integers, the union of B_n for n in the set is a member of the sigma-algebra, and these are all distinct."