# How to prove this collection is a sigma algebra

This is one of the previous comp question. I would appreciate if somebody can give me a proof.

Let $\mathbb A= \{E \subset X: E$ is a countable or $E^c$ is countable $\}$. To prove this collection is a sigma algebra.

What I think is I need to prove there is a empty set in the collection which is vacuously true because empty set has finite number of elements, namely $0$ elements. To prove complement is there I think I need to show for the two cases. That's where I confused. Would somebody explain that to me? I also want to see how the infinite union of the elements of the collection belong to it?

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The complement should be immediate. $E \in \mathbb A$ means $E$ is countable or $E^c$ is countable. If $E \in X$ is countable, then $(E^c)^c = E$ is countable. If $E^c$ is countable, then $E^c$ is countable. (I'm repeating myself here.) In both cases, we get either $(E^c)$ is countable or $(E^c)^c$ is countable, meaning $E^c \in \mathbb A$. –  Tunococ Jan 19 '13 at 1:03
@Tunococ, how about the infinite collection? –  Ganga Jan 19 '13 at 1:04
Consider $A \in \mathbb{A}$. By definition $A$ is countable or $A^c$ is countable. If $A$ is countable, then $(A^c)^c = A$ is countable and hence $A^c \in \mathbb{A}$. If $A^c$ is countable, then since $A^c$ is countable, $A^c \in \mathbb{A}$. –  user17762 Jan 19 '13 at 1:04
A countable union of countable sets is countable. And the countable union of sets whose complement is countable should make you reach for de Morgan's laws and think for a bit. –  user108903 Jan 19 '13 at 1:06
For countable union, suppose $E = \bigcup_n E_n$. If all $E_n$ are countable, then it's obvious that $E$ is countable. Otherwise, one of $E_n$ must have a countable complement, let's say it's $E_1$. So $E_1^c$ is countable. We see that $E^c = \bigcap_n E_n^c = E_1^c \cap \bigcap_{n>1} E_n^c \subset E_1^c$, so $E^c$ must also be countable. –  Tunococ Jan 19 '13 at 1:06

Take $E\in \mathbb A$. Then $E$ ou $E^c$ is countable. So $E^c\in\mathbb A$. Now take a infinite enumerable sets $E_n$ of $\mathbb A$. If all $E_n$ is enumerable than $\cup E_n$ is also enumerable. Suppose at least one of $E_n$ is not countable. Then $(\cup E_n)^c$ is countable because $(\cup E_n)^c=\cap E_n^c$ and at least one of $E_n^c$ is enumerable