# Non-Principal Ultrafilters Confused!!

I've just started learning about filters and non-principal ultrafilters. I'm getting confused on the requirement: $U$ contains no finite subsets of $J$; where $U$ is the ultrafilter and $J$ is a set.

I believe what this means is that $U$ can only contain sets that are infinite? (if it is non-principal).

Furthermore I'm getting somewhat confused by the proof that non-principal ultrafilters can exist. By the following " Take the filter of all cofinite sets and extend to an ultrafilter". I can't seem to find a definition of a cofinte set?

Sorry if this is incoherent, getting frustrated by this.

• Principal, not principle. – Mariano Suárez-Álvarez Jul 11 '15 at 19:12
• A set is cofinite if its complement is finite. – Mariano Suárez-Álvarez Jul 11 '15 at 19:12
• And yes, «$U$ contains no finite subsets of $J$» means that all the elements of $U$ are infinite sets (since its elements are subsets of $J$) – Mariano Suárez-Álvarez Jul 11 '15 at 19:13
• Mariano thanks for replying (and the typo spot). So it is the complement of the confinite set that is finite, so then the actual finite set is infinite? – user253919 Jul 11 '15 at 19:16
• Do you know of any simple examples of cofinite sets, I'm relatively new to set theory. – user253919 Jul 11 '15 at 19:16

Yes, an ultrafilter which contain a finite set is principal. To see why, note that if $A$ is this finite set, then either some $a\in A$ satisfies $\{a\}$ is in the ultrafilter, in which case it is principal; or else $X\setminus\{a\}$ is in the ultrafilter for all $a\in A$, so the finite intersection $A\cap\bigcap_{a\in A}(X\setminus\{a\})$ is also in the ultrafilter.

So a non-principal ultrafilter must contain only infinite sets. In particular, if $X$ is finite, then every ultrafilter on $X$ is principal.

The definition of a cofinite set is relative to $X$, and it simply means that $X\setminus A$ is finite. Co being short for "complement [of]". And if $X$ is an infinite set, then the collection of cofinite subsets of $X$ makes a filter. Moreover if you extend this filter in any way, you will never add a finite set to it. Therefore an ultrafilter extending the cofinite filter is necessarily non-principal.

• Asaf, thanks for this I'm beginning to understand a bit better. When you put {a} are you simply meaning just the single element a in A? And then is X\{a} the complement of the {a}? – user253919 Jul 11 '15 at 19:22
• Also just to double check the reason that we cannot construct non-principal ultrafiliters explicitly is because of the fact that they do contain no finite sets? – user253919 Jul 11 '15 at 19:27
• Yes on the first comment. For the second comment, the answer is no, we can't explicitly construct one because we know that it is consistent that there are no non-principal ultrafilters in some universes of set theory where Zorn's lemma fails (in those universes it fails pretty badly, as you can imagine). – Asaf Karagila Jul 11 '15 at 19:32
• Thank you. So just to double check my understanding of the first comment, {a} are all the single elements in A. X is some subset of the ultrafilter, so that X\{a} is the complement of {a} i.e. everything in X which aren't single elements. Thus the interesction of the two will be finite. Hence non-principal ultrafilters contain infinite sets... (PS you're very good at explaining complex ideas). – user253919 Jul 11 '15 at 19:48
• No, $X$ is the set on which we have the filter. $A$ is a finite subset of $X$. $a$ is an element of $A$, so $\{a\}$ is a subset of $A$. And $X\setminus\{a\}$ is the complement of $\{a\}$ (which is a cofinite subset of $X$). – Asaf Karagila Jul 11 '15 at 19:55

You can also use the following construction: Take X to be any infinite set, choose N a countable subset of X and enumerate N as a sequence, $N={\{x_n\}}_{n\in\mathcal{N}}$. Set $B_1=N$ and $B_n=N/\{x_1,\ldots,x_{n-1}\},n\in\mathcal{N}$

Then $\mathcal{B}=\{B_n : n\in\mathcal{N}\}$ is a filter base that can be extended to a non principal ultrafilter on X.