\begin{definition}[Definition:Irrational Slope Topology]
Let $S = \set {\tuple {x, y}: y \ge 0, x, y \in \Q}$.
Let $\theta \in \R \setminus \Q$ be some fixed irrational number.
:500pxrightthumb$\epsilon$-neighborhood $\map {N_\epsilon} {x, y}$: intervals are of [[Definition:Length of Real Intervallength $\epsilon$]]
Let $\BB$ be the set of $\epsilon$-neighborhoods of the form:
:$\map {N_\epsilon} {x, y} = \set {\tuple {x, y} } \cup \map {B_\epsilon} {x + \dfrac y \theta} \cup \map {B_\epsilon} {x - \dfrac y \theta}$
where:
:$\map {B_\epsilon} \zeta := \set {r \in \Q: \size {r - \zeta} < \epsilon}$
that is, the open $\epsilon$-ball at $\zeta$ in $\Q^2$.
Hence each $\map {N_\epsilon} {x, y}$ consists of:
:the singleton $\set {\tuple {x, y} }$
together with:
:two open intervals of length $\epsilon$ in the rational numbers whose midpoints are at the irrational points $x \pm \dfrac y \theta$
The lines joining $\tuple {x, y}$ to $\tuple {x \pm \dfrac y \theta, 0}$ have slope $\pm \theta$ and are not part of $\map {N_\epsilon} {x, y}$.
Let $\tau$ be the topology generated from $\BB$.
$\tau$ is referred to as the '''irrational slope topology'''.
\end{definition}