# Why is this set a $\sigma$-algebra??

$X$ is an uncountable set.

Why is $\mathcal{A}=\{A \subset X: A \text{ or } X \setminus A \text{ is countable } \}$ a $\sigma$-algebra ??



A $\sigma$-algebra $\mathcal{A}$ on a set $X$ is a collection of subsets of $X$ such that :

(1) $\varnothing \in \mathcal{A}$

(2) $A \in \mathcal{A} \Rightarrow X \setminus A \in \mathcal{A}$

(3) $A_n \in \mathcal{A} \Rightarrow \cup_{n=1}^{\infty} A_n \in \mathcal{A}$



Could you give me a hint how to show that $\mathcal{A}=\{A \subset X: A \text{ or } X \setminus A \text{ is countable } \}$ is a $\sigma$-algebra ??

• The only thing really needed is that the union of countably many countable sets is countable. – André Nicolas Sep 24 '14 at 20:39
• Is this not the power set of $X$? – T.J. Gaffney Sep 24 '14 at 20:40
• @Gaffney take $X = [0,1]$, the subset $[0,1/2]$ is not in this $\sigma$-algebra. – Xiao Sep 24 '14 at 20:47
• @AndréNicolas is it possible though for (3) is a union of uncountable and/or countable sets since its possible that (X\A could be the one thats countable) I feel like you need to know something about intersection also but I could be wrong – Kamster Sep 24 '14 at 20:49
• To show that a union that involves a cocountable $K$ is cocountable, all we need to observe is that the union contains $K$. – André Nicolas Sep 24 '14 at 20:59

## 2 Answers

Well, (1) and (2) are obvious.

Hint for (3): let $A_1,A_2,\dots\in \mathcal A$. Consider two cases:

1. Either all $A_n$'s contain only countable points.
2. Or at least one of them (w.l.o.g., say $A_1$) contains all but countable points.
• (1) $\varnothing$ is countable $\Rightarrow \varnothing \in \mathcal{A}$ $\checkmark$  (2) $A \in \mathcal{A} \Rightarrow A$ is countable or $A^c$ is countable. So that $A^c \in \mathcal{A}$ it should be $A^c$ is countable or $A$ is countable, which stands, right??  (3) If all $A_n$s are countable then their union is also countable. $\checkmark$ If at least one of them is uncountable, how can I conclude that the union is in $\mathcal{A}$?? – Mary Star Sep 27 '14 at 16:06

1 and 2 should be easy, for 3.

Given $A_n \in \mathcal{A}$, if each $A_n$ is countable, then we know $\cup_{n=1}^\infty A_n$ is countable by Andre's comment.

Now suppose one of the $A_n$, say $A_{n_0}$ is uncountable, then $X\setminus A_{n_0}$ is countable, observe that $$(\cup_{n=1}^\infty A_n)^c =\cap_{n=1}^\infty A_n^c \subset A_{n_0}^c .$$ Any subset of a countable set is also countable.

• (1) $\varnothing$ is countable $\Rightarrow \varnothing \in \mathcal{A}$ $\checkmark$  (2) $A \in \mathcal{A} \Rightarrow A$ is countable or $A^c$ is countable. So that $A^c \in \mathcal{A}$ it should be $A^c$ is countable or $A$ is countable, which stands, right??  (3) I havent understood the case where $A_{n_0}$ is uncountable. Why does the relation $$(\cup_{n=1}^\infty A_n)^c =\cap_{n=1}^\infty A_n^c \subset A_{n_0}^c$$ stand?? – Mary Star Sep 27 '14 at 15:59
• For two, given $A\in \mathcal{A}$, then either $A = {A^c}^c$ is countable or $A^c$ is countable, we can conclude that $A^c\in \mathcal{A}$. – Xiao Sep 27 '14 at 18:17
• For three, the first equality sign $$(A\cup B)^c = A^c \cap B^c.$$ The second inclusion sign $$A\cap B \subset A.$$ – Xiao Sep 27 '14 at 18:18