Graph of Nonlinear Additive Function is Dense in the Plane

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Theorem

Let $f: \R \to \R$ be an additive function which is not linear.

Then the graph of $f$ is dense in the real number plane.


Proof

From Additive Function is Linear for Rational Factors:

$\map f q = q \map f 1$

for all $q \in \Q$.

Without loss of generality, let:

$\map f q = q$

for all $q \in \Q$.

Since $f$ is not linear, let $\alpha \in \R \setminus \Q$ be such that:

$\map f \alpha = \alpha + \delta$

for some $\delta \ne 0$.


Consider an arbitrary nonempty circle in the plane.

Let:

its centre be $\tuple {x, y}$ where $x \ne y$ and $x, y \in \Q$
its radius be $r > 0$.

We will show how to find a point of the graph of $f$ inside this circle.

As $x \ne y$ and $r$ can be arbitrarily small, this will prove the theorem.


Since $\delta \ne 0$, let $\beta = \dfrac {y - x} \delta$.

Since $x \ne y$:

$\beta \ne 0$

As Rationals are Everywhere Dense in Topological Space of Reals, there exists a rational number $b \ne 0$ such that:

$\size {\beta - b} < \dfrac r {2 \size \delta}$

As Rationals are Everywhere Dense in Topological Space of Reals, there also exists a rational number $a$ such that:

$\size {\alpha - a} < \dfrac r {2 \size b}$

Now put:

$X = x + b \paren {\alpha - a}$
$Y = \map f X$

Then:

$\size {X - x} = \size {b \paren {\alpha - a} } < \frac r 2$

so $X$ is in the circle.

Then:

\(\ds Y\) \(=\) \(\ds \map f {x + b \paren {\alpha - a} }\) Definition of $Y$ and $X$
\(\ds \) \(=\) \(\ds \map f x + \map f {b \alpha} - \map f {b a}\) Definition of Cauchy Functional Equation
\(\ds \) \(=\) \(\ds x + b \map f \alpha - b \map f a\) Additive Function is Linear for Rational Factors
\(\ds \) \(=\) \(\ds y - \delta \beta + b \map f \alpha - b \map f a\) Definition of $y$
\(\ds \) \(=\) \(\ds y - \delta \beta + b \paren {\alpha + \delta} - b a\) Additive Function is Linear for Rational Factors
\(\ds \) \(=\) \(\ds y + b \paren {\alpha - a} - \delta \paren {\beta - b}\)

Therefore

$\size {Y - y} = \size {b \paren {\alpha - a} - \delta \paren {\beta - b} } \le \size {b \paren {\alpha - a} } + \size {\delta \paren {\beta - b} } \le r$

so $Y$ is in the circle as well.

Hence the point $\tuple {X, Y}$ is inside the circle.

$\blacksquare$


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