Peter Krautzberger · on the web

Eternal preliminaries part 3, the problem of extending partial semigroups

In the previous post I have totally ignored my focus from the first post, namely that I want to focus on partial semigroups, not full semigroups. As mentioned in the first part this is not really a problem as any partial semigroup operation is in essence a restriction of a full semigroup operation. So the full semigroup is always a neat thing to fall back to when you’re in doubt. Nevertheless, if, as I claimed, it is an advantage to work with partial semigroups, can we not get around this problem?

Extending a partial operation to .

Luckily, it is very much possible to do extend a partial operation. (Un)fortunately, nobody’s perfect that is I don’t know how to get all the theory to run smoothly so in what follows I might have to fall back for some things. We’ll see.

So how do you extend the operation? Easy! Use the brute force method!

Extending the operation to For , consider those with . These sets may or may not yield an ultrafilter. If they do, we call it .

Ok, that’s a bit weird. The way I wrote it there is no reason to believe that such will ever form a filter, let alone an ultrafilter! Now, since the operation is partial, we had to expect a little bit of weirdness. But accept (or ignore) the unreasonable definition at this point and I’ll try to explain why this could work.

First to note is: watch out! Maybe I’m trying to trick you with the definition, hiding behind the weirdness. As I wrote in the last post, the -notation is a “general nonsense” notation (even worse, I came up with it myself…), a simplified notation that hides complicated structures (though, hopefully, to a later advantage). So we better check it in detail.

But, assuming you’re a forgiving reader, I think it still makes sense: just make to mean “ and it is defined”. In other words, the original definition should really read

This is fine from the point of view of a full semigroup (since always ) and it really captures what we want to capture: those that maps to . Of course, the convention that entails is really standard ever since the paper by Bergelson, Blass and Hindman. So I’m luckily in good company. To finish the introduction, we get the same phenomenon as we did before.

if and only if there exists such that .

Again, with the analogous convention that .

The most important observation

So after this short delay, we can say we have defined a partial operation on where the product is as defined in the previous part but only if the definition yields an ultrafilter. One question is rather immediate.

Luckily, this question can be answered rather easily. But let me begin with the most crucial observation and the goal of the rest of this post.

The semigroup Given an (adequate) partial semigroup and the above extension of the operation to , it turns out that is a full semigroup.

Kaboom! This is extremely nice. Even though our operation is partial, we get a relatively good piece of that is a full, compact, and (we’ll see later) right-topological semigroup — the three key properties for the next part of these preliminaries. This will be our motivation for the rest of this post. Luckily, the proof is essentially done by answering the above questions.

The partial semigroup .

As the heading says, the nicest thing about this extension is that it yields a partial semigroup. This is positively surprising because strong associativity seems difficult to conserve. The key observation is the following.

Proposition For , the product is defined if and only if .

This is a pretty natural observation. You’d expect the product to work out on the ultrafilters if they contain sets where the multiplication behaves nicely. This is, of course, a common phenomenon with ultrafilters: properties of an ultrafilter are often reflected by its elements and vice versa.


And once again, I have taken the liberty of skipping a little bit ahead. In fact, I will spend the bigger part of the next post talking about other tricks like that the notation has to offer.

But with this we can actually claim.

Theorem is a partial semigroup.

The proof is long and rather boring. Using the previous proposition you can compare what it means for say and to be defined. Unsurprisingly, this comparison boils down to a comparison of elements in — where, of course, we have strong associatitivity. I’ll gladly update this post if there’s interest and you can find it as Proposition B.3 in my thesis.

It’s almost midnight and I’m getting tired so let’s wrap up the proof of the supposedly important observation.

Proof that is a semigroup.

Ok, after this small detour, we’re ready for some ancient and brilliant theorems. Ellis’s Lemma and Hindman’s Theorem coming right up.