Rationals are Everywhere Dense in Reals

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Theorem

Let $\left({\R, \tau_d}\right)$ be the real number line under the usual (Euclidean) topology.

Let $\Q$ be the set of rational numbers.


Then $\Q$ is everywhere dense in $\left({\R, \tau_d}\right)$.


Proof

Let $x \in \R$.

Let $U \subseteq \R$ be an open set of $\left({\R, \tau_d}\right)$ such that $x \in U$.


From Basis for Euclidean Topology on Real Number Line, there exists an open interval $V = \left({x - \epsilon \,.\,.\, x + \epsilon}\right) \subseteq U$ for some $\epsilon > 0$ such that $x \in V$.

Now consider the open interval $\left({x \,.\,.\, x + \epsilon}\right) \subseteq V$.

By Subset Relation is Transitive it follows that $\left({x \,.\,.\, x + \epsilon}\right) \subseteq U$.


Note that $x \notin \left({x \,.\,.\, x + \epsilon}\right)$.

From Between Every Two Reals Exists a Rational, there exists $y \in \Q: y \in \left({x \,.\,.\, x + \epsilon}\right)$.

As $x \notin \left({x \,.\,.\, x + \epsilon}\right)$, it must be the case that $x \ne y$.

That is, $V$ is an open set of $\left({\R, \tau_d}\right)$ containing $x$ which also contains an element of $\Q$ other than $x$.

As $V$ is arbitrary, it follows that every open set of $\left({\R, \tau_d}\right)$ containing $x$ also contains an element of $\Q$ other than $x$.

That is, $x$ is by definition a limit point of $\Q$.

As $x$ is arbitrary, it follows that all elements of $\R$ are limit points of $\Q$.

The result follows from the definition of everywhere dense.

$\blacksquare$