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$X$ is an uncountable set.

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

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

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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 ??

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  • 3
    $\begingroup$ The only thing really needed is that the union of countably many countable sets is countable. $\endgroup$ – André Nicolas Sep 24 '14 at 20:39
  • $\begingroup$ Is this not the power set of $X$? $\endgroup$ – T.J. Gaffney Sep 24 '14 at 20:40
  • $\begingroup$ @Gaffney take $X = [0,1]$, the subset $[0,1/2]$ is not in this $\sigma$-algebra. $\endgroup$ – Xiao Sep 24 '14 at 20:47
  • $\begingroup$ @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 $\endgroup$ – Kamster Sep 24 '14 at 20:49
  • $\begingroup$ To show that a union that involves a cocountable $K$ is cocountable, all we need to observe is that the union contains $K$. $\endgroup$ – André Nicolas Sep 24 '14 at 20:59
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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.
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  • $\begingroup$ (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 $\mathcal{A}$?? $\endgroup$ – Mary Star Sep 27 '14 at 16:06
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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.

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  • $\begingroup$ (1) $\varnothing$ is countable $\Rightarrow \varnothing \in \mathcal{A}$ $\checkmark$ $$$$ (2) $A \in \mathcal{A} \Rightarrow A$ is countable or $A^c$ is countable. So that $A^c \in \mathcal{A}$ it should be $A^c$ is countable or $A$ is countable, which stands, right?? $$$$ (3) I haven`t understood the case where $A_{n_0}$ is uncountable. Why does the relation $$(\cup_{n=1}^\infty A_n)^c =\cap_{n=1}^\infty A_n^c \subset A_{n_0}^c$$ stand?? $\endgroup$ – Mary Star Sep 27 '14 at 15:59
  • $\begingroup$ For two, given $A\in \mathcal{A}$, then either $A = {A^c}^c$ is countable or $A^c$ is countable, we can conclude that $A^c\in \mathcal{A}$. $\endgroup$ – Xiao Sep 27 '14 at 18:17
  • $\begingroup$ For three, the first equality sign $$(A\cup B)^c = A^c \cap B^c.$$ The second inclusion sign $$A\cap B \subset A.$$ $\endgroup$ – Xiao Sep 27 '14 at 18:18

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