Moving an Ultrafilter Along a Function
A function sends points of one set to points of another, but it does not generally send a filter to a filter by simply sending each member set to its image. The reliable construction tests subsets of the target by asking whether their inverse images belong to the original filter. This construction preserves the defining decision property of ultrafilters, which is why it applies in particular to the coordinate projections from a product.
Filters and ultrafilters depend only on the underlying sets in this discussion. When the sets are topological spaces, their topologies do not enter the definition below. For a product of topological spaces, we use the product topology.
A proper filter can exist only when \(X\ne\varnothing\), since otherwise \(X=\varnothing\) would have to belong to the filter. For a point \(x\in X\), the principal ultrafilter at \(x\) is \(\mathcal{U}_x=\{A\subseteq X:x\in A\}\).
This definition applies even if \(f\) is not injective or surjective. Its use of inverse images is essential: inverse images preserve inclusions and finite intersections, and the inverse image of a complement is the complement of the inverse image.
A Decision Property of Ultrafilters
Proof. Since \(A\cap(X\setminus A)=\varnothing\) and \(\varnothing\notin\mathcal{U}\), the two sets cannot both belong to \(\mathcal{U}\). To prove that at least one belongs, suppose \(A\notin\mathcal{U}\). If every \(C\in\mathcal{U}\) met \(A\), then the collection of sets containing some \(C\cap A\), with \(C\in\mathcal{U}\), would be a proper filter containing both \(\mathcal{U}\) and \(A\). It is a filter because intersections of two such generating sets have the form \((C_1\cap C_2)\cap A\), with \(C_1\cap C_2\in\mathcal{U}\); it is proper because the supposition makes each \(C\cap A\) nonempty. This would strictly enlarge \(\mathcal{U}\), contradicting maximality. Therefore some \(C\in\mathcal{U}\) satisfies \(C\cap A=\varnothing\). Then \(C\subseteq X\setminus A\), so upward closure gives \(X\setminus A\in\mathcal{U}\). Thus at least one of the two sets belongs, and the first part shows exactly one belongs. \(\square\)
The proposition is the key step in proving that image filters remain ultrafilters. We first verify the filter conditions directly, then use the decision property for each subset of the target.
Proof. The inverse image of \(Y\) is \(X\), so \(Y\in f_*(\mathcal{U})\). The inverse image of \(\varnothing\) is \(\varnothing\), so \(\varnothing\notin f_*(\mathcal{U})\). If \(B\subseteq C\subseteq Y\) and \(B\in f_*(\mathcal{U})\), then \(f^{-1}(B)\subseteq f^{-1}(C)\). Upward closure of \(\mathcal{U}\) gives \(f^{-1}(C)\in\mathcal{U}\), and hence \(C\in f_*(\mathcal{U})\). If \(B,C\in f_*(\mathcal{U})\), then \(f^{-1}(B),f^{-1}(C)\in\mathcal{U}\), so their intersection belongs to \(\mathcal{U}\). Since \(f^{-1}(B\cap C)=f^{-1}(B)\cap f^{-1}(C)\), this gives \(B\cap C\in f_*(\mathcal{U})\). Thus \(f_*(\mathcal{U})\) is a proper filter. Finally, for any \(B\subseteq Y\), the proposition applied to \(f^{-1}(B)\subseteq X\) says that either \(f^{-1}(B)\) or \(X\setminus f^{-1}(B)\) belongs to \(\mathcal{U}\). The latter set is \(f^{-1}(Y\setminus B)\). Therefore either \(B\) or \(Y\setminus B\) belongs to \(f_*(\mathcal{U})\). A proper filter with this decision property is maximal: any proper filter containing it cannot contain a set \(B\) that it does not already contain, because it already contains \(Y\setminus B\), and then it would contain the empty intersection. Hence \(f_*(\mathcal{U})\) is an ultrafilter. \(\square\)
Coordinate Projections
Let \(\{X_j\}_{j\in J}\) be a family of topological spaces, and give \(X=\prod_{j\in J}X_j\) the product topology. For each \(j\in J\), the coordinate projection is the function \(\pi_j:X\to X_j\) defined by \(\pi_j((x_i)_{i\in J})=x_j\).
Proof. Each \(\pi_j\) is a function from \(X\) to \(X_j\). Applying the theorem with \(f=\pi_j\) shows that the filter \(\{B\subseteq X_j:\pi_j^{-1}(B)\in\mathcal{U}\}\) is an ultrafilter on \(X_j\). The product topology is compatible with the usual meaning of the coordinate projections, but the filter conclusion uses only the underlying sets and the fact that \(\pi_j\) is a function. If \(J=\varnothing\), there are no indices \(j\), so the assertion is vacuous. \(\square\)
The explicit membership test is often the most useful form of the result: \(B\subseteq X_j\) belongs to the image ultrafilter precisely when the set of product points whose \(j\)-th coordinate lies in \(B\) belongs to \(\mathcal{U}\).
Worked Example: A Principal Ultrafilter on a Product
Give \(\mathbb{R}\) its standard topology and \(\mathbb{R}\times\mathbb{R}\) the product topology. Let \(p=(2,-1)\), and let \(\mathcal{U}_p\) be the principal ultrafilter at \(p\). Compute the image ultrafilters under \(\pi_1\) and \(\pi_2\).
For \(A\subseteq\mathbb{R}\), the membership test gives \(\pi_1^{-1}(A)=A\times\mathbb{R}\). This set belongs to \(\mathcal{U}_p\) exactly when it contains \(p=(2,-1)\), which holds exactly when \(2\in A\). Therefore \((\pi_1)_*(\mathcal{U}_p)=\mathcal{U}_2\), the principal ultrafilter at \(2\).
For \(B\subseteq\mathbb{R}\), \(\pi_2^{-1}(B)=\mathbb{R}\times B\). It belongs to \(\mathcal{U}_p\) exactly when \(-1\in B\), so \((\pi_2)_*(\mathcal{U}_p)=\mathcal{U}_{-1}\).
Check: for \(A=\{2\}\), the first inverse image is \(\{2\}\times\mathbb{R}\), which contains \(p\), so \(\{2\}\) belongs to the first image ultrafilter. For \(A=\{3\}\), the inverse image \(\{3\}\times\mathbb{R}\) misses \(p\), so \(\{3\}\) does not belong. These are exactly the membership decisions of the principal ultrafilter at \(2\). The corresponding second-coordinate test gives membership for \(\{-1\}\) and nonmembership for \(\{0\}\).
Worked Example: A Nonprincipal Ultrafilter and Parity
Give \(\mathbb{N}\) and \(\{0,1\}\) their discrete topologies. We construct a nonprincipal ultrafilter on \(\mathbb{N}\) containing the even numbers, then compute its image under the parity function \(f:\mathbb{N}\to\{0,1\}\), where \(f(n)=0\) for even \(n\) and \(f(n)=1\) for odd \(n\).
Let \(E=\{2n:n\in\mathbb{N}\}\). Start with the proper filter \(\mathcal{F}=\{A\subseteq\mathbb{N}: E\setminus F\subseteq A\text{ for some finite }F\subseteq\mathbb{N}\}\). It contains \(\mathbb{N}\), excludes \(\varnothing\), and is upward closed. If \(A,B\in\mathcal{F}\), choose finite \(F,G\subseteq\mathbb{N}\) with \(E\setminus F\subseteq A\) and \(E\setminus G\subseteq B\). Then \(E\setminus(F\cup G)\subseteq A\cap B\). This set is nonempty: \(F\cup G\) is finite, while \(E\) contains infinitely many distinct even integers. Thus \(A\cap B\in\mathcal{F}\) and \(\varnothing\notin\mathcal{F}\), as required.
By the Ultrafilter Extension Principle, extend \(\mathcal{F}\) to an ultrafilter \(\mathcal{U}\). This extension contains \(E\) and every cofinite subset of \(\mathbb{N}\). It contains no finite set \(F\): it also contains the cofinite complement \(\mathbb{N}\setminus F\), and membership of both would force their empty intersection into the proper filter. In particular, it contains no singleton, so it is not principal. This construction ensures nonprincipality; merely extending the filter generated by \(E\) would not rule out a principal extension at an even integer.
Now \(f^{-1}(\{0\})=E\), so \(\{0\}\in f_*(\mathcal{U})\). Also \(f^{-1}(\{1\})=\mathbb{N}\setminus E\), which cannot belong because \(E\in\mathcal{U}\). The target has exactly four subsets: \(\varnothing,\{0\},\{1\},\{0,1\}\). The empty set is excluded, the whole target belongs, and the two singleton decisions have just been computed. Therefore \(f_*(\mathcal{U})\) is the principal ultrafilter at \(0\).
Check: the inverse image of \(\{0,1\}\) is \(\mathbb{N}\), which belongs to \(\mathcal{U}\), while the inverse image of \(\varnothing\) is \(\varnothing\), which does not. The computed image filter therefore has the required whole-set and empty-set membership decisions.
Worked Example: A Nonprincipal Ultrafilter on the Diagonal
Give \(\mathbb{N}\) the discrete topology, and give \(\mathbb{N}\times\mathbb{N}\) the product topology. Let \(\mathcal{V}\) be a nonprincipal ultrafilter on \(\mathbb{N}\), whose existence follows from the cofinite-filter extension construction in the preceding example. Define \(g:\mathbb{N}\to\mathbb{N}\times\mathbb{N}\) by \(g(n)=(n,n)\), and set \(\mathcal{U}=g_*(\mathcal{V})\). By the theorem, \(\mathcal{U}\) is an ultrafilter on the product.
For \(C\subseteq\mathbb{N}\), the inverse image under the first projection is \(\pi_1^{-1}(C)=C\times\mathbb{N}\). By the definition of \(\mathcal{U}\), \[ \pi_1^{-1}(C)\in\mathcal{U} \quad\Longleftrightarrow\quad g^{-1}(C\times\mathbb{N})\in\mathcal{V}. \] But \(g^{-1}(C\times\mathbb{N})=C\), since \(g(n)=(n,n)\) has first coordinate in \(C\) exactly when \(n\in C\). Thus \(C\in(\pi_1)_*(\mathcal{U})\) exactly when \(C\in\mathcal{V}\), proving \((\pi_1)_*(\mathcal{U})=\mathcal{V}\).
For the second projection, \(\pi_2^{-1}(C)=\mathbb{N}\times C\), and \(g^{-1}(\mathbb{N}\times C)=C\), since the second coordinate of \(g(n)\) is also \(n\). Hence \((\pi_2)_*(\mathcal{U})=\mathcal{V}\) as well.
Check: for \(C=\{1,2\}\), each projection's inverse image pulls back under \(g\) to \(\{1,2\}\). This finite set is not in the nonprincipal ultrafilter \(\mathcal{V}\), so it belongs to neither coordinate image. For \(C=\mathbb{N}\), both pullbacks are \(\mathbb{N}\), which is in \(\mathcal{V}\), so both coordinate images contain \(\mathbb{N}\), as required.
A Common Mistake: Taking Direct Images
The notation \(f_*(\mathcal{U})\) can suggest applying \(f\) to each member of \(\mathcal{U}\), but the family \(\{f(A):A\in\mathcal{U}\}\) need not even be a filter on the target. For a concrete check, give \(X=\{a\}\) and \(Y=\{0,1\}\) the discrete topologies, let \(f(a)=0\), and let \(\mathcal{U}\) be the principal ultrafilter at \(a\). Its members are \(\{a\}\) and \(X\). Their direct images are both \(\{0\}\), so the direct-image family is \(\{\{0\}\}\). It does not contain \(Y\), and therefore fails a filter axiom. In contrast, the inverse-image definition gives \(f_*(\mathcal{U})=\{\{0\},Y\}\): a subset of \(Y\) belongs exactly when it contains \(0\). This is the principal ultrafilter at \(0\).
The coordinate-projection result is useful because it lets one transfer an ultrafilter on a product to each factor without needing the projection to be one-to-one or the original ultrafilter to be principal. The proof relies only on inverse images and the ultrafilter decision property; it does not assert that an arbitrary family of ultrafilters on the factors comes from one ultrafilter on the product.
Check Your Understanding
Use the inverse-image definition in each question, and state which subsets belong to the relevant image ultrafilter.
- Let \(\mathcal{U}_x\) be principal at \(x\in X\). Prove directly that \(f_*(\mathcal{U}_x)\) is principal at \(f(x)\).
- For the parity map in the worked example, determine whether \(\{1\}\) belongs to the image ultrafilter and justify the decision using its inverse image.
- Let \(X=\mathbb{R}^2\) with the product of the standard topologies. Describe the inverse image under \(\pi_1\) of an interval \((a,b)\subseteq\mathbb{R}\).
- Explain why the proof that \(f_*(\mathcal{U})\) is a filter uses preservation of finite intersections by inverse images.
- Give a reason the direct-image family in the common-mistake example fails to be a filter on \(Y\).