Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

share|cite|improve this answer
  • The empty set belongs to $A$ because it is countable.
  • $X\in A$ because $X^c$ is the empty set, therefore, countable.
  • if $B\in A$ is countable then $B=(B^c)^c$ is countable. Theferefore, $B^c\in A$
  • if $A_n$ is countable for all n, then $\cup{A_n}$ is countable. If $A_n^c$ is countable for some m, then, since $A_m\subset \cup A_n$, we have $(\cup A_n)^c\subset A_m^c$,which is countable. Then $(\cup A_n)^c$ is countable and, by the third item, $\cup A_n$ is countable.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.