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Let C be the class of all subsets of $\mathbb{R}$ which are either countable or are the complement of a countable set. Show that C is a $\sigma$ field.

This is what I have right now:

(i) $\varnothing$ $\in$ $C$ since $\emptyset$ is countable .

(ii) A $\in$ C since A is countable since given C is countable.

(iii) $\bigcup^{\infty}_{i=1}A_i$ $\in$ C, is true since $A_1,A_2$, are countable then so is $\cup^{}_{n}A_n$ since a countable union of countable sets is countable.

I need help understanding (ii) and (iii).

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    $\begingroup$ Your reasons for $\varnothing \in C$ are not only false, they do not imply that $\varnothing \in C$ in any way. The second point is also very confusing and probably wrong. The third point is wrong, but it's written in a clear way and it's easy to point where you're going wrong: You don't know that all $A_i$s are countable, you know that each of them is either countable or it's complement in $\mathbb R$ is countable. $\endgroup$
    – Git Gud
    Jan 29, 2017 at 23:24
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    $\begingroup$ @FernandoFernandez One step at a time. You want to prove that $\varnothing \in C$. What requirements does $\varnothing$ need to fulfill in order for it to belong to $C$? $\endgroup$
    – Git Gud
    Jan 29, 2017 at 23:35
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    $\begingroup$ @FernandoFernandez No, it's not true that $\varnothing\in \mathbb R$, as ou know $\varnothing$ is not a real number, so how could it belong to the set of real numbers? You're confused about something, but I can't tell what it is. I repeat it: what requirements does $\varnothing$ need to fulfill in order for it to belong to $C$? $\endgroup$
    – Git Gud
    Jan 30, 2017 at 6:59
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    $\begingroup$ Right. So because $\varnothing$ is countable, it follows that $\varnothing \in C$. Do you understand this? For (ii) you want to prove that $A\in C\implies \mathbb R\setminus A\in C$. Assume $A\in C$, then either $A$ is countable or $\mathbb R\setminus A$ is countable. Your goal is to prove that $\mathbb R\setminus A\in C$. What does requirements must $\mathbb R\setminus A$ fulfill in order for it to belong to $C$? $\endgroup$
    – Git Gud
    Jan 31, 2017 at 15:13
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    $\begingroup$ @FernandoFernandez Given $A\in C$, you want to prove that $\mathbb R\setminus A$ is either countable or the complement of a countable set. Note this very important fact: the complement of $\mathbb R\setminus A$ in $\mathbb R$ is $\mathbb R\setminus (\mathbb R\setminus A)$ and this is equal to $A$. Well, since $A\in C$ one has two cases: $A$ is countable or $\mathbb R\setminus A$ is countable. Try to conclude that in both cases $\mathbb R\setminus A$ is either countable or the complement of a countable set. You'll need to use the very important fact I mentioned. $\endgroup$
    – Git Gud
    Feb 2, 2017 at 20:34

1 Answer 1

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Let me point out a minor nitpick before answering:

"Countable" here means "at most countable", i.e. "finite or countable". If we were to take "countable" as strictly meaning "in bijection with the natural numbers", this would not produce a $\sigma$-algebra, since $\varnothing$ is finite (cardinality 0) end its complement is uncountable, and hence $\varnothing$ would not be in our class.

That said, I take most of what I say below from comments.

$$\text{(i: The empty set)}$$

$\varnothing$ is finite, hence at most countable, hence $\varnothing\in C$. I am assuming $\varnothing\in\emptyset$ is a typo for $\varnothing\in C$, otherwise that "formula" is Gibberish to me.

$$\text{(ii: Closure under taking complements)}$$

This point is a bit unclear, but it seems to be trying to show closure of $C$ under complement. The other interpretation would be that it's trying to show the "universe set" is in $C$, but that "universe" is $\mathbb{R}$ here, and it has never been called $A$ in the question. So I assume $A\subseteq\mathbb{R}$, and in fact $A\in C$.

We then have two cases:

  1. $A$ is at most countable;
  2. $A$ is the complement of an at most countable set.

In the first case, $\mathbb{R}\smallsetminus A$ (alias $A^C$) is the complement of an at most countable set, hence $A^C\in C$. In the second case, $A^C$ must be at most countable, since $A$ is not at most countable and hence must be the complement of an at most countable set, and $A=(A^C)^C$;

$$\text{(iii: Closure under countable unions)}$$

If $A_i\in C$ for all $i$, we have three cases:

a) All $A_i$'s are at most countable;

b) All $A_i$'s are complement of at most countable sets;

c) There are some $A_i$'s that are at most countable, and some that are complements of at most countable sets;

in case a), a countable union of countable sets is countable, as you remarked; so your (iii) works for case a), but neglects the other cases. In case b), the union of the $A_i$'s is:

$$\bigcup_iA_i=\bigcup_i(A_i^C)^C=\left(\bigcap_iA_i^C\right)^C,$$

and that intersection is contained in, e.g., $A_1^C$, which is at most countable, and hence is at most countable. In case c), we remark that the union operator is associative and commutative, so we reduce $\bigcup_iA_i$ to:

$$\bigcup_iA_i=\left(\bigcup_{i:|A_i|\leq|\mathbb N|}A_i\right)\cup\left(\bigcup_{i:|A_i^C|\leq|\mathbb N|}A_i\right).$$

So this union is the union of two sets, the first of which is at most countable as shown in case a), and the second of which is the complement of an at most countable set as shown in case b). Hence, all that is left to prove is that, if $|A|\leq|\mathbb N|$ and $|B^C|\leq|\mathbb N|$ (i.e. $A$ is at most countable as is $B^C$), then $A\cup B$ is either at most countable or the complement of an at most countable set. Now, $B\subseteq A\cup B$, so if $B$ is not at most countable, $A\cup B$ cannot be either. Let us look at the complement:

$$(A\cup B)^C=A^C\cap B^C\subseteq B^C,$$

and $B^C$ is at most countable. Hence, $A\cup B$ is the complement of an at most countable set, which completes the proof that $C$ is a $\sigma$-algebra (or $\sigma$-field, as you call it).

I hope this is clear.

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    $\begingroup$ Thank you so much for such a detailed answer, the explanations and steps are clear for me to understand your point. $\endgroup$ Feb 9, 2017 at 16:06
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    $\begingroup$ Very nice effort. Good job, Mick. $\endgroup$
    – Git Gud
    Feb 11, 2017 at 17:57

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