The title pretty much contains the question, but here's some elaboration:

The following is one of the first results one encounters while learning about Ultrafilters.

Fact: If $\mathfrak{U}$ is an ultrafilter on an index set $I$, and $X\subset I$, then exactly one of $X$ and $I\backslash X$ is in $\mathfrak{U}$.

Question: Can this be generalized to a collection $X_{1}, ... X_{n}$ of disjoint subsets of $I$ for any $n\geq 1$? That is, if $\mathfrak{U}$ is an ultrafilter on $I$, then there is exactly one value of $j\in\{1,...,n\}$ such that $X_{j}\in \mathfrak{U}$?

I haven't been able to find it in the literature anywhere or prove it myself (though I thought I did briefly) so I'm beginning to suspect it's false in general.

Second Question: If the answer to the first question is false, is it true if $\mathfrak{U}$ is an ultrafilter on $\mathbb{N}$ containing the order filter?


If $X_1, \dots, X_n$ is a collection of disjoint sets with $\bigcup_{i=1}^nX_i = I$, then $X_k \in \mathfrak{U}$ for precisely one $k$.

Note that for each $k$, either $X_k \in \mathfrak{U}$ or $\bigcup_{i \neq k}X_i \in \mathfrak{U}$. If $X_k \not\in \mathfrak{U}$ for all $k$, then $\bigcup_{i\neq k}X_i \in \mathfrak{U}$ for every $k$, and so the intersection of such sets would also belong to $\mathfrak{U}$. This is a contradiction as the intersection is empty, i.e. $\bigcap_{k=1}^n\bigcup_{i\neq k}X_i = \emptyset$. Therefore, there exists $k \in \{1, \dots, n\}$ such that $X_k \in \mathfrak{U}$. If there were $l \in \{1, \dots, n\}$, $l \neq k$, with $X_l \in \mathfrak{U}$, then $X_l\cap X_k \in \mathfrak{U}$, but this intersection is empty, so no such $l$ exits. Therefore, $k$ is unique.

Note, you need the sets $X_i$ to cover $I$. If they don't cover $I$, choose $m \in I\setminus\bigcup_{i=1}^nX_i$ and consider the principal ultrafilter $\mathfrak{U}_m$. As $m \not\in X_i$ for all $i$, $X_i \not\in \mathfrak{U}_m$ for all $i$.


Let $X_1,X_2,\dots, X_n$ be a finite collection of sets (not necessarily pairwise disjoint) such that the union $X_1\cup X_2\cup \cdots\cup X_n$ is in the ultrafilter. Then at least one of the $X_i$ is in the ultrafilter. If the $X_i$ are pairwise disjoint, then exactly one of the $X_i$ is in the ultrafilter. The proof is straightforward, by induction.

Added: From comments, it became clear that the OP knew the standard proof that if $D$ is an ultrafilter on the index set $I$, then $X$ or $X^c$ is in $D$. The fact that if the union $X_1\cup X_2$ is in $D$, then $X_1$ or $X_2$ is in $D$ follows. For let $O$ be the "rest" of $I$. If $X_1$ is not in $D$, then its complement $X_2\cup O$ is in $D$. Also, we know that $X_1\cup X_2$ is in $D$. Now note that $X_2=(X_2\cup O)\cap(X_1\cup X_2)$, so $X_2\in D$.

Remark: I think it helps the intuition to think of an ultrafilter as defining a two-valued "measure" $\mu$ on the collection of subsets of $I$, where $\mu(X)=1$ if $X$ is in the ultrafilter, and $\mu(X)=0$ otherwise. The "measure" is almost always not a real measure, since it is ordinarily not countably additive. But it is finitely additive. The finite additivity makes the answer to your question clear. If the $X_i$ all had measure $0$, their union would have measure $0$.

  • $\begingroup$ It is the intuition you describe that led me to guess that the result might be true. I could not come up with a proof, however. Thanks for your additional information! $\endgroup$ – roo May 26 '15 at 1:12
  • 1
    $\begingroup$ To see how easy the induction is, let's do it for $3$ sets. Let $X_1\cup X_2\cup X_3$ be in the ultrafilter $D$. Then $(X_1\cup (X_2\cup X_3))\in D$. By the $n=2$ case, either $X_1\in D$, and we are finished, or $(X_2\cup X_3)\in D$, in which case again by the $n=2$ case we are finished. $\endgroup$ – André Nicolas May 26 '15 at 1:41
  • $\begingroup$ I agree that the induction part of the proof is indeed easy. But you are using the following lemma, which I think is the tricky part: If $X_{1}\cup X_{2}\in \mathfrak{U}$, with $X_{1}\cap X_{2} = \phi$, then exactly one of $X_{1}$ or $X_{2}$ is in $\mathfrak{U}$. This is slightly stronger than the fact I mentioned, and cannot be proved in quite the same way (using a brief maximality argument). The lemma clearly follows immediately from Michael Albanese's argument above, but I do not see an easy way to get it directly. $\endgroup$ – roo May 26 '15 at 1:55
  • 1
    $\begingroup$ That it cannot be both is obvious, for part of the usual definition of a filter $D$ is that $\emptyset\not\in D$. For the proof that at least one is, it depends on the definition of ultrafilter. If we define it as a maximal filter, one shows that if neither $A$ nor $A^c$ is in $D$, then $A$ can be added to $D$, along with all intersections of elements of $D$ with $A$, and their supersets, to form a larger filter $D'$. $\endgroup$ – André Nicolas May 26 '15 at 2:06
  • 1
    $\begingroup$ Is it the part $O$ "outside" $X_1\cup X_2$ that bothers you? If $X_1$ is not in $D$, then its complement $X_2\cup O$ is in $D$, and by assumption $X_1\cup X_2$ is in $D$, so the intersection of $X_2\cup O$ and $X_1\cup X_2$ is in $D$, that is, $X_2$ is in $D$. $\endgroup$ – André Nicolas May 26 '15 at 3:54

It is true, granted the sets form a partition of $I$. Otherwise it is easy to come by counterexamples, simple consider a few singletons and a free ultrafilter.

The proof is simple by induction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.