Is the Empty Set Axiom Redundant in Ultrafilters?

[alexfloo]

One of the four defining axioms of an ultrafilter is that it doesn't contain the empty set (according to Wikipedia, and a talk I was listening to today). Isn't this implied by the other axioms?

If an ultrafilter U on X contained the empty set, then it also contains every superset, including X. Therefore, it doesn't contain the complement of X, which is the empty set, and therefore we have a contradiction.

I understand that the axiom is necessary if the filter is not ultra to eliminate trivial cases, but am I missing something or is it redundant in the ultra case?

 

[micromass]

I have a hard time following your argument. Could you perhaps tell me what your definition is of a filter and what your definition is of an ultrafilter??

One thing that I don't understand in your reasoning is why you say that an ultrafilter cannot both contain a set and its complement. This fact is true, but it follows from the fact that an ultrafilter cannot contain the empty set. So I don't think you can use the fact in your argument.

 

[alexfloo]

Apologies, I didn't realize that there were a number of different developments of the idea. The one I saw in the talk, which is also the first one listed on Wikipedia, is as follows:

1. \emptyset\notin U
2. U contains any superset (in P(X)) of any set in U
3. U is closed under intersection
4. For any subset A of X, U contains exactly one of A and its complement.

Furthermore, it was noted that a "filter" satisfies only the first 3. That is, it is the 4th property that makes a filter "ultra."

The proof above (I think) uses the last three to establish the first. Since it depends on the fourth, it doesn't hold in the case of general filters, but is my logic sound for ultrafilters?

 

[micromass]

Apologies, I didn't realize that there were a number of different developments of the idea. The one I saw in the talk, which is also the first one listed on Wikipedia, is as follows:

1. \emptyset\notin U
2. U contains any superset (in P(X)) of any set in U
3. U is closed under intersection
4. For any subset A of X, U contains exactly one of A and its complement.

Furthermore, it was noted that a "filter" satisfies only the first 3. That is, it is the 4th property that makes a filter "ultra."

The proof above (I think) uses the last three to establish the first. Since it depends on the fourth, it doesn't hold in the case of general filters, but is my logic sound for ultrafilters?

Aah. In that case you are correct that (1) follows from (2) and (4) (we don't even need (3)).

Some notes:
- Filters are often also required to be nonempty. Indeed, \emptyset trivialy satisfies (1),(2) and (3). This is often undesirable.

- Ultrafilters are often defined in an other manner. A filter is still something that satisfies (1),(2),(3). An ultrafilter \mathcal{U} is defined as a maximal filter: that is, if \mathcal{F} is a filter such that \mathcal{U}\subseteq \mathcal{F}, then \mathcal{U}=\mathcal{F}.

- The biggest problem with ultrafilters is the existence. To prove existence-results, we often need to use Zorns lemma. With your definition (1)-(4) it isn't clear why Zorns lemma should work.

 

[alexfloo]

When you say that existence results depend on Zorn's lemma, do you mean to say the existence of non-principal ultrafilters? My understanding is that ultrafilters containing exactly those subsets with contain some x\in X can be established finitarily, but the existence of other ultrafilter requires AC. Is this correct?

 

[micromass]

When you say that existence results depend on Zorn's lemma, do you mean to say the existence of non-principal ultrafilters? My understanding is that ultrafilters containing exactly those subsets with contain some x\in X can be established finitarily, but the existence of other ultrafilter requires AC. Is this correct?

That is correct. The existence of point-filters is easy. But the existence of free ultrafilters (= non point-filters) requires the AC.

In general, we have the following existence result: for each filter \mathcal{F}, there exists an ultrafilter \mathcal{U} such that \mathcal{F}\subseteq \mathcal{U}. This is called the "ultrafilter lemma". This depends crucially on the Axiom of Choice or one of its related forms.

In general, there are a lot more free ultrafilters than point-filters. And free ultrafilters can not be constructed at all: you can never give an example of a free ultrafilter!
However, note that if X is finite then X has no free ultrafilter.

 

[alexfloo]

That seems reasonable, since ultrafilters on a set form a poset and Zorn's lemma can be used to establish the existence of maximal elements.

Thanks a lot!