Motivation
$$ \gdef\vois#1#2{\mathcal{V}_{#1}(#2)} $$
Nets and filters are used for describing convergence in a non-metric space $X$.
Denote the collection of (open) neighbourhoods of $x \in X$ by $$\vois{X}{x}$$.
Definitions and examples
- Directed set
- A partially ordered set $I$ such that $$\forall i, j \in I: i \le j, \exists k \in I: k \ge j.$$
- Net
- A function in $X^I$, where $I$ is a directed set.
- example: any sequence in $X^\N$
- Convergence of nets to a point
- $x_i \to x$ if
$$\forall A \in \vois{X}{x}, \exists j \in I: \forall k \ge j, x_k \in A.$$
- example: absolute convergence of series ($I$ is the collection of finite subsets of $\N$, finite sum $\Sigma \in \R^I$.)
- example: Riemann integral ($I$ is the collection of tagged partitions, the partial order doesn’t depend on tags, $\int \in \R^I$.)
- Filter base
- A nonempty collection $\mathcal{F} \subseteq \mathcal{P}(X) \setminus
{\varnothing}$ such that
$$\forall F,G \in \mathcal{F},\exists H \in \mathcal{F}: H \subseteq F \cap G.$$
(contains nonempty part of intersection)
Difference with topological basis: sets have to be nonempty here - Filter
- A filter base $\mathcal{F}$ so that
- contains supersets: $\forall F \in \mathcal{F}, \forall G \supseteq F, G \in \mathcal{F}$
- contains intersection: $\forall F, G \in \mathcal {F}, F \cap G \in \mathcal{F}$
The image of a filter $\mathcal{F}$ under a function $f$ is also a filter, denoted by $f[[\mathcal{F}]]$.
Filters can be made from a filter base.
Let $\mathcal{F}$ be a filter base. Then $\mathcal{G} = \lbrace G \subseteq X \mid \exists F \in \mathcal{F}: F \subseteq G\rbrace$ is a filter. (We say $\mathcal{F}$ is a filter of $\mathcal{G}$.)
A picture guides through the direct verification. Note that all intersections should be nonempty in this context.
- Convergence of a filter base to a point
- $\mathcal{F} \to x$ if $$\vois{X}{x} \subseteq \mathcal{G}$$, where $\mathcal{G}$ is the filter generated by $\mathcal{F}$.
Equivalent ways to define continuity
- Open preimage (usual way)
- $x_i \to x \Longrightarrow f(x_i) \to x$
- $\mathcal{F} \to x \Longrightarrow f[[\mathcal{F}]] \to f(x)$
The axiom of choice is needed to some stages.