General Topology/Filters

Note that the set of open neighbourhoods of a point does not in general form a filter.

Note in particular that the bases are required to be contained within the filter.

Note in particular that the subbases are required to be contained within the filter. Note also that we use the terms base and basis interchangeably; both versions are in use. Note finally that a filter base of a filter $$\mathcal F$$ is a filter subbase of $$\mathcal F$$.

{{proof|Suppose first that whenever $$A_1, \ldots, A_n$$ are elements of $$\mathcal B$$, then $$A_1 \cap \cdots \cap A_n \neq \emptyset$$. Then define
 * $$\mathcal F := \{B \subseteq X| \exists A_1, \ldots, A_n \in \mathcal B: A_1 \cap \cdots \cap A_n \subseteq B$$.

We claim that $$\mathcal F$$ is a filter. Indeed, the empty set can't be in $$\mathcal F$$, since every intersection of elements of $$\mathcal B$$ is, by assumption, nonempty. Further, whenever $$B \in \mathcal F$$ and $$C \supseteq B$$, then $$C \in \mathcal F$$. Finally, suppose that $$B, C \in \mathcal F$$, and choose $$A_1, \ldots, A_n \in \mathcal B$$ so that $$A_1 \cap \cdots \cap A_n \subseteq B$$ and $$D_1, \ldots, D_m \in \mathcal B$$ so that $$D_1 \cap \cdots \cap D_m \subseteq C$$, then
 * $$A_1 \cap \cdots \cap A_n \cap D_1 \cap \cdots \cap D_m \subseteq B \cap C$$,

so that $$B \cap C \in \mathcal F$$. Now suppose that $$\mathcal B$$ is the subbasis of some filter $$\mathcal F$$, and let $$A_1, \ldots, A_n \in \mathcal B$$. Since $$\mathcal B \subseteq \mathcal F$$ and $$\mathcal F$$ is a filter, $$A_1 \cap \cdots \cap A_n \in \mathcal F$$, so that $$A_1 \cap \cdots \cap A_n \neq \emptyset$$.

Let now $$\mathcal F'$$ be an arbitrary filter that has $$\mathcal B$$ as a subbasis. Then $$\mathcal F' \subseteq \mathcal F$$ by the definition of a subbasis. Now since $$\mathcal B \subseteq \mathcal F'$$, we also have $$\mathcal F \subseteq \mathcal F'$$, so that $$\mathcal F = \mathcal F'$$.}}

Analogous to real analysis, we can rephrase continuity in terms of filter convergence. In general, filters are supposed to play the role for topological spaces that sequences play for finite-dimensional real normed spaces; we will see many theorems that are analogous to those on $$\mathbb R^n$$, with sequences replaced by filters.

Note: It has been proven that one can't prove the ultrafilter lemma from Zermelo–Fraenkel axioms alone, but some form of the axiom of choice is needed. Still, the ultrafilter lemma does not imply the axiom of choice in ZF, that is, it is strictly weaker than the axiom of choice.

In this way, we may reformulate the criterion for the existence of a filter containing another filter and a given set: If $$\phi$$ is a filter and $$A$$ a set, then there exists a filter $$\psi$$ with $$A \in \psi$$ and $$\phi \subset \psi$$ iff $$\phi$$ clusters at $$A$$.

Note that principal ultrafilters are ultrafilters, because either $x \in A$ or $x \notin A$ for all $A \subseteq X$.

Exercises

 * 1) Let $$X$$ be a topological space which is not compact. Prove that the set $$\mathcal F$$ of all complements of compact subsets of $$X$$ is a filter. Prove that if $$X$$ is compact, then $$\mathcal F$$ is not a filter.
 * 2) Suppose that $$\phi$$ is an ultrafilter on a set $$X$$, and let $$A_1, \ldots, A_n \subseteq X$$ be finitely many subsets of $$X$$. Prove that if $$\phi$$ does not contain any of the $$A_j$$, it does not contain $$A_1 \cup \cdots \cup A_n$$.
 * 3) Let $$X, Y$$ be sets and $$f:X \to Y$$ a function. Prove that $$f$$ is injective iff for all filters $$\phi$$ in $$Y$$, $$f^{-1}(\phi) = \{f^{-1}(A)|A \in \phi\}$$ is a filter.