No. The basic idea of proving why not is this:
1) The underlying space X is at least countably infinite (of course).
2) Use the properties of a sigma field (aka sigma algebra) to find a countable partition of the space, X = disjiont-union( X_i ).
3) Notice that the union(X_i, s in S) is in the sigma algebra for any subset S of natural numbers.
4) Notice that any union(X_i, s in S) is distinct.
5) Conclude, since the set of subsets of natural numbers is uncountable, so too is your sigma algebra.
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No, it is countably infinite.
There are finite sets, countably infinite sets and uncountably infinite sets.
Closed sets and open sets, or finite and infinite sets.
An infinite set whose elements can be put into a one-to-one correspondence with the set of integers is said to be countably infinite; otherwise, it is called uncountably infinite.
No, it is not.