Topology

Ultra filter

A filer $\mathcal{F}$ containing either $Y$ or $Y^\complement$ for any $Y \subseteq X$.

Two days ago, I spent an afternoon to understand Dudley’s proof of this little result.

A filter is contained in some ultrafilter. A filter is an ultrafilter iff it’s maximal.

At the first glance, I didn’t even understand the organisation of the proof! I’m going to rephrase it for future reference.

• only if: let $\mathcal{F}$ be an ultrafilter contained in another filter $\mathcal{G}$. If $\mathcal{F}$ isn’t maximal, let $Y \in \mathcal{G} \setminus \mathcal{F}$. Since $\mathcal{F}$ is an ultrafilter, either $Y \in \mathcal{F}$ or $Y^\complement \in \mathcal{F}$. By construction of $Y$, only the later option is possible, so $Y^\complement \in \mathcal{G}$ by hypothesis, but this contradicts our assumption $Y \in \mathcal{G}$: $\varnothing = Y \cap Y^\complement \in \mathcal{G}$, which is false since $\mathcal{G}$ is a filter.

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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 $f[[\mathcal{F}]]$.