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I'm proving the following proposition taken from Royden 4th Edition,

Let $\mathcal{F}$ be a collection of subsets of a set $X$. Then the intersection $\mathcal{A}$ of all $\sigma$-algebras of subsets of $X$ that contain $\mathcal{F}$ is a $\sigma$-algebra that contains $\mathcal{F}$.

I didn't have much trouble proving this proposition, but I am having some issues with the notation and the word containment. The way I interpreted the above proposition in symbols is,

Let $\mathcal{F}$ be a collection of subsets of a set $X$. Furthermore, let $\mathcal{A}$ be any $\sigma$-algebra such that $\mathcal{F}\in\mathcal{A}$ and let $\Sigma_\mathcal{F}$ denote the collection of all such $\sigma$-algebras. We will show that $\mathcal{C}=\bigcap\limits_{\mathcal{A}\in\Sigma_\mathcal{F}}\mathcal{A}$ is a $\sigma$-algebra, $\mathcal{F}\in\mathcal{C}$ and that $\mathcal{C}$ is the smallest $\sigma$-algebra that contains $\mathcal{F}$.

My concern is whether $\mathcal{F}$ is an element of $\mathcal{C}$ and we write $\mathcal{F}\in\mathcal{C}$ or is the elements of $\mathcal{F}$ in $\mathcal{C}$ and so $\mathcal{F}\subset\mathcal{C}$.

I do know that $\sigma$-algebras are collection of subsets, but when it says containment, I'm wondering if it's the elements of the subset or the subset itself.

Thanks for any help or feedback!

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First of all, I object to this language:

let $\mathcal{A}$ be any $\sigma$-algebra such that $\mathcal{F}\in\mathcal{A}$ and let $\Sigma_\mathcal{F}$ denote the collection of all such $\sigma$-algebras.

It is confusing for the reader, and risky for the mathematician, to create a variable and then suddenly universalize it. You should say "Let $\mathbb{N}$ be the set of all integers," rather than "Let $n$ be an integer, and let $\mathbb{N}$ be the set of all $n$."

I am sorry if this sounds pedantic, but it is important to get these basic things straight before worrying about more complicated matters.

Okay, on to the heart of the matter. We have $\Sigma_\mathcal{F} = \{\mathcal{A} \bigm| \text{$\mathcal{A}$ is a $\sigma$-algebra, and } \mathcal{F} \subset \mathcal{A} \}$, and $\mathcal{C}=\bigcap\limits_{\mathcal{A}\in\Sigma_\mathcal{F}}\mathcal{A}$.

You ask: do we have $\mathcal{F} \in \mathcal{C}$, or $\mathcal{F} \subset \mathcal{C}$? Well, $\mathcal{F} \subset \mathcal{A}$ for each $\mathcal{A}$ in the intersection, so $\mathcal{F} \subset \mathcal{C}$ certainly looks like the correct thing to write. And this is what we want: a $\sigma$-algebra which, at the very least, contains all the same sets that $\mathcal{F}$ does. So $U\in \mathcal{F} \implies U\in \mathcal{C}$, which is exactly what we mean by $\mathcal{F} \subset \mathcal{C}$.

$\mathcal{F} \in\mathcal{C}$, on the other hand, would mean that, for any $\mathcal{A}\in\Sigma_F$, $\mathcal{F} \in \mathcal{A}$ and $\mathcal{F} \subset \mathcal{A}$. Hmm. This certainly looks suspicious. In short, this is not the right thing to think about, for the following reason: every element of every $\mathcal{A}\in\Sigma_\mathcal{F}$ is a subset of $X$. But is $\mathcal{F} \subset X$? No, $\mathcal{F} \subset \mathcal{P}(X)$.

This is why the notion is what it is. All the standard capital letters, $U,X$, etc., represent the sets that we are chopping up, and the curvy ones, $\mathcal{A}, \mathcal{C}, \mathcal{F}$, and so on, represent collections of chopped up pieces of things in the first category. So if I give you $T$ and $\mathcal{W}$, already you should know that $T\in \mathcal{W}$ could happen, but $T\subset \mathcal{W}$ is probably unreasonable. Think of capital letters like $T$ as representing puzzle pieces, and curvy letters like $\mathcal{W}$ as representing boxes of puzzle pieces.

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  • $\begingroup$ Thanks for the response and the clarification on my wording. What your saying is $\mathcal{W}$ contains only sets (boxes) and not the individual elements(puzzle pieces)? And so, when it says $\mathcal{A}$ contains $\mathcal{F}$, since their both collections of sets, its refering to the elements of $\mathcal{F}$ which are sets. $\endgroup$ – Shant Danielian Oct 14 '13 at 3:24
  • $\begingroup$ @ShantDanielian No, I mean to say that $\mathcal{W}$ contains puzzle pieces, which contain, well, bits of wood pulp. It's not a perfect analogy, since puzzle pieces are disjoint, but the point is that if $X$ were a picture of a sailboat, then the subsets of $X$ are different chunks of that picture, and a collection $\mathcal{F}\subset\mathcal{P}(X)$ is like a box full of a lot of different chunks that you can assemble however you like. When working with $\sigma$-algebras, you often don't need to think about the actual elements of $X$, so I find "chunks" a better mental picture than "sets". $\endgroup$ – Slade Oct 14 '13 at 4:15
  • $\begingroup$ After thinking about it this, I think I understand what you were trying to say. What I came to was that $\sigma$-algebras are sets whose elements are sets. So when we compare two $\sigma$-algebras they could potentially be subsets. Also, when we look at the individual sets in a $\sigma$-algebra, there just elements. I think this is what you meant with given $T$ and $\mathcal{W}$, $T\in\mathcal{W}$ since $T$ is an element(set), whereas if we have another $\sigma$-algebra $\mathcal{A}$(a collection of sets or set of sets), then $\mathcal{A}\subset\mathcal{W}$ could happen. $\endgroup$ – Shant Danielian Oct 16 '13 at 6:10
  • $\begingroup$ @ShantDanielian Glad that things are getting more clear. $\endgroup$ – Slade Oct 16 '13 at 6:16
  • $\begingroup$ Me too, thanks for the help! $\endgroup$ – Shant Danielian Oct 16 '13 at 6:18

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