# Irrationals are Everywhere Dense in Reals

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## Theorem

Let $T = \left({\R, \tau}\right)$ be the Euclidean space of real numbers.

Let $\R \setminus \Q$ be the set of irrational numbers.

Then $\R \setminus \Q$ is everywhere dense in $T$.

## Proof

Let $x \in \R$.

Let $U \subseteq \R$ be an open set of $T$ such that $x \in U$.

From Basis for Euclidean Topology on Real Number Line, there exists an open interval $V_0 = \left({x - \epsilon \,.\,.\, x + \epsilon}\right) \subseteq U$ for some $\epsilon > 0$ such that $x \in V_0$.

$\exists p \in \Q: p \in \left({x - \epsilon \,.\,.\, x + \epsilon}\right)$

Thus, we can define the open interval $V_1 = \left({x \,.\,.\, p}\right) \subseteq V_0$.

Similarly:

$\exists q \in \Q: q \in \left({x \,.\,.\, p}\right)$

We can then define an open interval $V_2 = \left({q \,.\,.\, p}\right) \subseteq V_1$.

We have $V_2 \subseteq V_1$, $V_1 \subseteq V_0$ and $V_0 \subseteq U$.

By successively applying Subset Relation is Transitive, it follows that $V_2 \subseteq U$.

Note that $x \notin V_2$, since $x < q < p < x + \epsilon$.

From Between two Rational Numbers exists Irrational Number, there exists $y \in \R \setminus \Q: y \in \left({p \,.\,.\, q}\right) = V_2$.

As $x \notin V_2$, it must be the case that $x \ne y$.

Since $V_2 \subseteq U$, $U$ is an open set of $T$ containing $x$ which also contains an element of $\R \setminus \Q$ other than $x$.

As $U$ is arbitrary, it follows that every open set of $T$ containing $x$ also contains an element of $\R \setminus \Q$ other than $x$.

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

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

The result follows from the definition of everywhere dense.

$\blacksquare$