# The Unapologetic Mathematician

## 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.

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Posted by John Armstrong | Analysis, Measure Theory

## 3 Comments	»

1. If this is the product, what are the morphisms?

Comment by Chad | March 18, 2010 | Reply

2. “We won’t focus, yet, on the categorical perspective, and will return to that somewhat later.”

Comment by | March 18, 2010 | Reply

3. [...] fact, we’ve seen that given rings and we can define the product as the collection of finite disjoint unions of [...]

Pingback by | July 15, 2010 | Reply

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