User:RaisinBread

RaisinBread's User Page.

Work In Progress
Adapting the proof of Rationals Dense in Reals to create the page Irrationals Dense in Reals.

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 $x \in V_0$.

From Between Every Two Reals Exists a Rational, $\exists p\in\Q:p\in (x . . x+\epsilon)$

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

Similarly, we see $\exists q\in\Q:q\in (x . . p)$.

We can then define an open interval $V_2=(q. . p)\subseteq V_1$.

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

By successively applying Subsets Transitive, it follows that $V_2 \subseteq U$.

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

From Between Every Two Rationals Exists an Irrational, there exists $y \in \R\setminus\Q: y \in (p. . q)=V_2$.

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 $y\in\R\setminus\Q$ 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.