Complement of a set.

Question

I am learning about σ-algebra. I am looking at an example of a set F ⊂ Ω and the authors are proving that F is a σ-algebra.

Let |A| denote the number of elements of a set A ⊂ Ω. Then, F is defined as F = {A ⊂ Ω: either |A| is finite or |A^c| is finite. So, among other things, they trying to prove that in line with the standard definition of σ-algebra that if a subset B ⊂ F ⇒ B^c ⊂ F.

For THIS particular example, they are defining the complement of |A| as |A^c|^c = |A|. I am unable to understand this particular equality. Can someone please explain how they came up with this equality?

Answer 1

I had a similar question in an exercise I'm about to post (replace the statement "$|A|$ is finite" with "$A$ is countable").

We're trying to show that if $A$ is in this algebra, then so is $A^C$. So suppose $A$ is in the algebra, and, WLOG, suppose that A is a finite set. Consider the set $A^C$. To be in the algebra, either this set, or its complement, $A$, must be in it. We already know $A$ is in, so $A^C$ is in.

Remember that complement has an involution property: the complement of the complement is the original set.

Answer 2

First, let us correct several mistakes in your question. Suppose we have a set $\Omega$ which we will call our universal set. Suppose we have a collection of subsets of $\Omega$ which we will call $\mathcal{F}$. In this case we have $\mathcal{F}\subseteq 2^{\Omega}$ where $2^{\Omega}$ represents the power set of $\Omega$, the set of all subsets of $\Omega$. (You wrote that $\mathcal{F}\subset \Omega$. That is incorrect, that would have $\mathcal{F}$ being a subset of $\Omega$, not $\mathcal{F}$ being a collection of subsets of $\Omega$). Now, we say that $\mathcal{F}$ is an algebra if it satisfies certain properties.

In your example, we should probably have $\mathcal{F} = \{B\subset \Omega~:~|B|~\text{is finite or }|B^c|~\text{is finite}\}$. In the process of proving that $\mathcal{F}$ is an algebra we would want to prove that if $B\in \mathcal{F}$ then $B^c\in\mathcal{F}$. You wrote $B\subseteq F$ which is not at all what you want. Be more aware of the distinction between being an element of something and being a subset of something.

On to your actual question:

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The base of your question seems to be "How is a complement of a set defined?"

Given a universal set $\Omega$ and a set $A$, we define the relative complement of the set $A$ as:

$$A^c = \{x\in \Omega~:~x\notin A\} = \Omega\setminus A$$

You might continue to ask "How is the cardinality of a complement of a set defined?"

It is defined in exactly the same way as it is if the set were not a complement. Notice that if $A$ is a set, then $A^c$ is also a set.

You say "For this example they are defining the complement of $|A|$ as $|A^c|^c=|A|$"

This makes no sense on several fronts. We talk about complements of sets... not complements of cardinalities. $A$ is a set and has a complement, $A^c$. $|A|$ however is not a set anymore (unless being pedantic) and the complement of $|A|$ is undefined in this context. You grossly misunderstood what was being said.

Now... rather than talking about complements of cardinalities you /can/ talk about cardinalities of complements. $|A^c|$ makes sense to talk about. $|A|^c$ does not. As mentioned, $|A^c|$ is defined already using the same usual definition of cardinality.

A simple result you should be familiar with that should hopefully build some intuition for you: If $\Omega$ is finite then $|A^c|=|\Omega|-|A|$. Note however that "$\infty - \infty$" is undefined and so you cannot claim that just because $|A|$ is infinite that it should imply $|A^c|$ is finite. Another simple result you should be familiar with is that $(A^c)^c = A$. (Note, the use of parentheses here, not the use of vertical lines which usually denote cardinality)

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A set $A$ which has finitely many elements is called "finite" and we also say that $|A|$ is finite.

A set $A$ in the context of a universal set $\Omega$ such that $A^c$ is finite is called "cofinite." If the universal set is finite, it follows that every cofinite set would be finite as well. If the universal set is infinite however, it follows that every cofinite set would be infinite as well.

As an example, consider our universal set as the set of integers, $\Bbb Z$. The set $\{1,2,3\}$ is a finite set. The set $\Bbb Z\setminus \{-1,0,1\} = \{\dots,-5,-4,-3,-2,2,3,4,5,\dots\}$ is a cofinite set.

On the other hand, $\{2,4,6,8,10,12,\dots\}$ is neither finite nor cofinite but is infinite.