Ultrafilters and Convergence

Ultrafilters are maximal filters -- they represent "points at infinity" or "ideal limit processes." Their existence relies on the axiom of choice (via Zorn's lemma), and they are the key ingredient in the proof of Tychonoff's theorem.

Definition

Definition8.6Ultrafilter

A filter $\mathcal{U}$ on a set $X$ is an ultrafilter if it is maximal with respect to inclusion among all filters on $X$ . That is, if $\mathcal{F}$ is a filter with $\mathcal{U} \subseteq \mathcal{F}$ , then $\mathcal{F} = \mathcal{U}$ .

Theorem8.2Characterization of Ultrafilters

A filter $\mathcal{U}$ on $X$ is an ultrafilter if and only if for every $A \subseteq X$ , either $A \in \mathcal{U}$ or $X \setminus A \in \mathcal{U}$ (but not both).

Proof

( $\Rightarrow$ ): Suppose $\mathcal{U}$ is an ultrafilter and $A \notin \mathcal{U}$ . We show $X \setminus A \in \mathcal{U}$ . If $X \setminus A \notin \mathcal{U}$ , then $A \cap U \neq \emptyset$ for all $U \in \mathcal{U}$ (otherwise, $U \subseteq X \setminus A$ , and $X \setminus A \in \mathcal{U}$ by upward closure). So $\mathcal{B} = \{A \cap U : U \in \mathcal{U}\}$ is a filter base generating a filter $\mathcal{F}$ containing both $\mathcal{U}$ and $A$ . By maximality, $\mathcal{F} = \mathcal{U}$ , so $A \in \mathcal{U}$ , contradicting our assumption.

( $\Leftarrow$ ): If $\mathcal{F} \supseteq \mathcal{U}$ is a filter and $A \in \mathcal{F}$ , then $A \in \mathcal{U}$ or $X \setminus A \in \mathcal{U} \subseteq \mathcal{F}$ . The latter gives $\emptyset = A \cap (X \setminus A) \in \mathcal{F}$ , impossible. So $A \in \mathcal{U}$ , hence $\mathcal{F} = \mathcal{U}$ .

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Existence of Ultrafilters

Theorem8.3Ultrafilter Lemma

Every filter on a set $X$ is contained in an ultrafilter. In particular, ultrafilters exist on every nonempty set.

Proof

Let $\mathcal{F}$ be a filter on $X$ . Consider the partially ordered set of all filters on $X$ containing $\mathcal{F}$ , ordered by inclusion. This is nonempty ( $\mathcal{F}$ itself) and every chain has an upper bound (the union of a chain of filters is a filter). By Zorn's lemma, there exists a maximal element, which is an ultrafilter containing $\mathcal{F}$ .

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RemarkAxiom of Choice

The ultrafilter lemma is a consequence of Zorn's lemma (which is equivalent to the axiom of choice). It is strictly weaker than AC but independent of ZF. Many important results in topology (Tychonoff's theorem, existence of non-principal ultrafilters) require the ultrafilter lemma.

Types of Ultrafilters

Definition8.7Principal and Non-Principal Ultrafilters

An ultrafilter $\mathcal{U}$ on $X$ is:

Principal (or fixed) if $\bigcap \mathcal{U} \neq \emptyset$ . In this case, $\bigcap \mathcal{U} = \{x\}$ for some $x$ , and $\mathcal{U} = \{A \subseteq X : x \in A\}$ .
Non-principal (or free) if $\bigcap \mathcal{U} = \emptyset$ .

ExampleUltrafilter Examples

For any $x \in X$ , the principal ultrafilter $\mathcal{U}_x = \{A \subseteq X : x \in A\}$ is the unique ultrafilter converging to $x$ in the discrete topology.
On $\mathbb{N}$ , the Fr'echet filter (cofinite sets) is not an ultrafilter (e.g., neither the even nor odd integers are cofinite). But it extends to a non-principal ultrafilter (by the ultrafilter lemma). Such ultrafilters are non-constructive.
On a finite set, every ultrafilter is principal.

Ultrafilter Convergence and Compactness

Theorem8.4Compactness via Ultrafilters

A topological space $X$ is compact if and only if every ultrafilter on $X$ converges.

Proof

( $\Rightarrow$ ): Let $X$ be compact and $\mathcal{U}$ an ultrafilter. Suppose $\mathcal{U}$ does not converge. Then for each $x \in X$ , there exists an open $U_x \ni x$ with $U_x \notin \mathcal{U}$ . By the ultrafilter characterization, $X \setminus U_x \in \mathcal{U}$ for each $x$ .

$\{U_x\}$ is an open cover; extract a finite subcover $U_{x_1}, \ldots, U_{x_n}$ . Then $X \setminus U_{x_i} \in \mathcal{U}$ for each $i$ , so: $\bigcap_{i=1}^n (X \setminus U_{x_i}) = X \setminus \bigcup_{i=1}^n U_{x_i} = X \setminus X = \emptyset \in \mathcal{U},$ contradicting the filter axiom.

( $\Leftarrow$ ): Suppose every ultrafilter converges. Let $\{C_\alpha\}$ be a collection of closed sets with the FIP. The set $\{C_\alpha\}$ generates a filter base (by FIP, finite intersections are nonempty). Extend to a filter, then to an ultrafilter $\mathcal{U}$ . By hypothesis, $\mathcal{U} \to x$ for some $x$ . Since $C_\alpha \in \mathcal{U}$ and $\mathcal{U} \to x$ , every neighborhood of $x$ meets every $C_\alpha$ , so $x \in \overline{C_\alpha} = C_\alpha$ . Thus $x \in \bigcap C_\alpha \neq \emptyset$ .

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Ultrafilters on Products

Theorem8.5Image Ultrafilter

If $f: X \to Y$ is a function and $\mathcal{U}$ is an ultrafilter on $X$ , then the pushforward $f_*(\mathcal{U}) = \{B \subseteq Y : f^{-1}(B) \in \mathcal{U}\}$ is an ultrafilter on $Y$ .

If $f$ is continuous and $\mathcal{U} \to x$ in $X$ , then $f_*(\mathcal{U}) \to f(x)$ in $Y$ .

Proof

$f_*(\mathcal{U})$ is a filter: $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2) \in \mathcal{U}$ if both $B_i$ are in the pushforward. It is an ultrafilter: for $B \subseteq Y$ , either $f^{-1}(B) \in \mathcal{U}$ or $X \setminus f^{-1}(B) = f^{-1}(Y \setminus B) \in \mathcal{U}$ .

If $\mathcal{U} \to x$ and $V$ is a neighborhood of $f(x)$ , then $f^{-1}(V) \in \mathcal{N}(x) \subseteq \mathcal{U}$ , so $V \in f_*(\mathcal{U})$ .

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RemarkUltrafilters in Mathematics

Ultrafilters appear throughout mathematics:

Nonstandard analysis: Ultrafilters on $\mathbb{N}$ give rise to ultrapowers, leading to hyperreal numbers.
Model theory: Ultrapowers and ultraproducts are fundamental constructions.
Combinatorics: Ramsey-type results (e.g., the Hales-Jewett theorem) use ultrafilter arguments.
Topology: Compactness, Stone-Cech compactification, and Stone duality.