Number of possible unions of a countable number of sets If $\{ A_{n} \}_{n=1}^{\infty}$ is a countable sequence of distinct sets, then is the number of possible distinct unions between any two or more of the sets in the sequence uncountable? I would like a proof.
By possible distinct unions I mean unions such as $A_{1} \cup A_{2}$, $A_{1} \cup A_{2} \cup A_{3}$, $A_{2} \cup A_{3}$, etc.
Background
I am doing self study on measure theory, and am trying to prove that a $\sigma$-algebra that contains a countable number of distinct sets is uncountable.
 A: Yes it can be uncountable, let ${A_n}=\{n\}$, so the sets are $\{1\},\{2\},\{3\}\dots$
Then it is clear that the sets that are the union of two or more of these sets are the subsets of $\mathbb N$ with at least two elements.
This subset is not countable, we can find a bijection with the set of infinite binary sequences with at least two ones (which is uncountable).
The bijection is as follows: Send a set $A$ to the sequence where term $i$ is one if $i\in A$ and the term $i$ is zero if $i\not\in A$

The reason why the set of binary sequences with at least two ones is countable is that the set of binary sequences with $1$ one or no ones is countable.

Added: The set can also be countable. Consider the following sets: $\mathbb N-\{1\},\mathbb N-\{2\},\mathbb N-\{3\},\mathbb N-\{4\}\dots$
The union of at least two sets must be $\mathbb N$. So the set of possible unions has exactly $1$ element and is countable.
A: Similar to another answer, but to further address your main question: If you have countably infinite number of DISJOINT sets in your $\sigma$-algebra, then you can map every possible union to a subset of natural numbers simply by enumerating your disjoint sets as $A_1,A_2,A_3, \ldots$ and then for a given union, just form the set of indices $i$ of sets $A_i$ that are included in the union. It is a well known fact that the power set of natural numbers, i.e. the set of all subsets of natural numbers, is uncountable, and one way to see it is by the famous so-called diagonalization argument, where you suppose there are countably many natural number subsets, call them $X_i$ and then you construct a new subset $Y$ where $i \in Y$ if and only if $i \notin X_i$. Then $Y$ is not any of your enumerated subsets $X_i$.
