Example

Let $X$ be a topological space, $A$ a subset of $X$, $x$ a point of $X$. In order that the trace on $A$ of the neighbourhood filter $\mathfrak{B}$ of $x$ should be a filter on $A$, it is necessary and sufficient that every neighbourhood of $x$ meets $A$, i.e. that $x$ lies in the closure of $A$ (§ 1, no. 6, Definition 10).

This example of an induced filter is of interest for two reasons: first because it plays an important role in the theory of limits (§ 7, no. 5) and secondly because every filter can be defined in this way. Indeed, let $\mathfrak{F}$ be a filter on a set $X$ and let $X'$ be the set obtained by adjoining a new element $\omega$ to $X$, $X$ being identified with the complement of $\{ \omega \}$ in $X'$ (Set Theory, R, § 4, no. 5); let $\mathfrak{F}'$ be the filter on $X'$ consisting of the sets $M \cup \{ \omega \}$ where $M$ runs through $\mathfrak{F}$. For each point $x \neq \omega$ of $X'$, let $\mathfrak{B}(x)$ be the set of all subsets of $X'$ which contain $x$, and let $\mathfrak{B}(\omega)$ be $\mathfrak{F}'$; then the $\mathfrak{B}(x)$ for $x \in X'$ obviously satisfy axioms $(\mathrm{V}_1), (\mathrm{V}_{\mathrm{II}}), (\mathrm{V}_{\mathrm{III}})$ and $(\mathrm{V}_{\mathrm{IV}})$ and therefore define a topology on $X'$ for which they are the neighbourhood filters of points. Finally $\omega$ lies in the closure of $X$ in this topology, and $\mathfrak{F}$ is induced by $\mathfrak{F}' = \mathfrak{B}(\omega)$ on $X$. The topology thus defined on $X'$ (resp. the set $X'$ with this topology) is called the topology (resp. the topological space) associated with $\mathfrak{F}$.