The Unapologetic Mathematician

Mathematics for the interested outsider

Products of Algebras of Sets

As we deal with algebras of sets, we’ll be wanting to take products of these structures. But it’s not as simple as it might seem at first. We won’t focus, yet, on the categorical perspective, and will return to that somewhat later.

Okay, so what’s the problem? Well, say we have sets X_1 and X_2, and algebras of subsets \mathcal{E}_1\subseteq P(X_1) and \mathcal{E}_2\subseteq P(X_2). We want to take the product set X=X_1\times X_2 and come up with an algebra of sets \mathcal{E}\subseteq P(X). It’s sensible to expect that if we have E_1\in\mathcal{E}_1 and E_2\in\mathcal{E}_2, we should have E_1\times E_2\in\mathcal{E}. Unfortunately, the collection of such products is not, itself, an algebra of sets!

So here’s where our method of generating an algebra of sets comes in. In fact, let’s generalize the setup a bit. Let’s say we’ve got \mathcal{R}_1\subseteq P(X_1) which generates \mathcal{E}_1 as the collection of finite disjoint unions of sets in \mathcal{R}_1, and let \mathcal{R}_2\subseteq P(X_2) be a similar collection. Of course, since the algebras \mathcal{E}_1 and \mathcal{E}_2 are themselves closed under finite disjoint unions, we could just take \mathcal{R}_1=\mathcal{E}_1 and \mathcal{R}_2=\mathcal{E}_2, but we could also have a more general situation.

Now we can define \mathcal{R} to be the collection of products R_1\times R_2 of sets R_1\in\mathcal{R}_1 and R_2\in\mathcal{R}_2, and we define \mathcal{E} as the set of finite disjoint unions of sets in \mathcal{R}. I say that \mathcal{R} satisfies the criteria we set out yesterday, and thus \mathcal{E} is an algebra of subsets of X.

First off, \emptyset is in both \mathcal{R}_1 and \mathcal{R}_2, and so \emptyset\times\emptyset=\emptyset is in \mathcal{R}. On the other hand, X_1\in\mathcal{R}_1 and X_2\in\mathcal{R}_2, so X_1\times X_2=X is in \mathcal{R}. That takes care of the first condition.

Next, is \mathcal{R} closed under pairwise intersections? Let R_1\times R_2 and S_1\times S_2 be sets in \mathcal{R} A point (x_1,x_2) is in the first of these sets if x_1\in R_1 and x_2\in R_2; it’s in the second if x_1\in S_1 and x_2\in S_2. Thus to be in both, we must have x_1\in R_1\cap S_1 and x_2\in R_2\cap S_2. That is,

\displaystyle(R_1\times R_2)\cap(S_1\times S_2)=(R_1\cap S_1)\times(R_2\cap S_2)

Since \mathcal{R}_1 and \mathcal{R}_2 are themselves closed under intersections, this set is in \mathcal{R}.

Finally, can we write (R_1\times R_2)\setminus(S_1\times S_2) as a finite disjoint union of sets in \mathcal{R}? A point (x_1,x_2) is in this set if it misses S_1 in the first coordinate — x_1\in R_1\setminus S_1 and x_2\in R_2 — or if it does hit S_1 but misses S_2 in the second coordinate — x_1\in R_1\cap S_1 and x_2\in R_2\setminus S_2. That is:

\displaystyle(R_1\times R_2)\setminus(S_1\times S_2)=\left((R_1\setminus S_1)\times R_2\right)\cup\left((R_1\cap S_1)\times(R_2\setminus S_2)\right)

Now R_1\setminus S_1\in\mathcal{E}_1, and so it can be written as a finite disjoint union of sets in \mathcal{R}_1; thus (R_1\setminus S_1)\times R_2 can be written as a finite disjoint union of sets in \mathcal{R}. Similarly, we see that (R_1\cap S_1)\times(R_2\setminus S_2) can be written as a finite disjoint union of sets in \mathcal{R}. And no set from the first collection can overlap any set in the second collection, since they’re separated by the first coordinate being contained in S_1 or not. Thus we’ve written the difference as a finite disjoint union of sets in \mathcal{R}, and so (R_1\times R_2)\setminus(S_1\times S_2)\in\mathcal{E}.

Therefore, \mathcal{R} satisfies our conditions, and \mathcal{E} is the algebra of sets it generates.

March 17, 2010 Posted by | Analysis, Measure Theory | 3 Comments

   

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