## 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 and , and algebras of subsets and . We want to take the product set and come up with an algebra of sets . It’s sensible to expect that if we have and , we should have . 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 which generates as the collection of finite disjoint unions of sets in , and let be a similar collection. Of course, since the algebras and are themselves closed under finite disjoint unions, we could just take and , but we could also have a more general situation.

Now we can define to be the collection of products of sets and , and we define as the set of finite disjoint unions of sets in . I say that satisfies the criteria we set out yesterday, and thus is an algebra of subsets of .

First off, is in both and , and so is in . On the other hand, and , so is in . That takes care of the first condition.

Next, is closed under pairwise intersections? Let and be sets in A point is in the first of these sets if and ; it’s in the second if and . Thus to be in both, we must have and . That is,

Since and are themselves closed under intersections, this set is in .

Finally, can we write as a finite disjoint union of sets in ? A point is in this set if it misses in the first coordinate — and — or if it does hit but misses in the second coordinate — and . That is:

Now , and so it can be written as a finite disjoint union of sets in ; thus can be written as a finite disjoint union of sets in . Similarly, we see that can be written as a finite disjoint union of sets in . 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 or not. Thus we’ve written the difference as a finite disjoint union of sets in , and so .

Therefore, satisfies our conditions, and is the algebra of sets it generates.

If this is the product, what are the morphisms?

Comment by Chad | March 18, 2010 |

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

Comment by John Armstrong | March 18, 2010 |

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

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