Open Balls on Rational Centers form Basis for Usual Topology on Plane

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Theorem

Let $\R^2$ be the real number plane with the usual (Euclidean) topology.

Let $S$ be the set defined as:

$S = \set {\tuple {x, y} \in \R^2: x, y \in \Q}$

That is, let $S$ be the set of all points in $\R^2$ whose coordinates are rational numbers.


Let $\BB$ denote the set defined as:

$\BB = \set {\map {B_q} s: s \in S, q \in \Q}$

That is, let $\BB$ be the set of all open $q$-balls with centers in $S$ and rational radius.


Then $\BB$ forms a basis for $\R^2$.


Proof

Let $d: \R^2 \times \R^2 \to \R$ be the usual (Euclidean) metric on $\R^2$.

Let $U$ be an open set of $\R^2$.

Let $z = \tuple {x, y} \in U$.

Then by definition of open set:

$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} z \subseteq U$

By Rationals are Everywhere Dense in Reals:

$\exists \tuple {x', y'}$ such that:
$\size {x - x'} < \dfrac \epsilon 5$
$\size {y - y'} < \dfrac \epsilon 5$

Let $s = \tuple {x', y'}$.

Then:

$\map d {z, s}< \dfrac {e \sqrt 2} 5 < \dfrac \epsilon 3$

Let $q \in \Q$ such that $\dfrac \epsilon 3 < q \le \dfrac {2 \epsilon} 3$.

Then:

$z \in \map {B_q} s \subseteq \map {B_\epsilon} z \subseteq U$


As $z$ is arbitrary, it follows that $U$ is the union of open balls with centers in $S$ and rational radius.

Hence the result.

$\blacksquare$


Sources

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