Absolute Value Function is Continuous

Theorem

Let $f$ be the real function defined as:

$\forall x \in \R: \map f x = \size x$

where $\size x$ denotes the absolute value of $x$.

Let $a \in \R$.

Let $\epsilon \in \R_{\mathop > 0}$.

Let $\delta \le \epsilon$

We have:

$\ds \forall x \in \R : \size{x - a} < \delta: \, $

$\ds \size{\map f x - \map f a}$

$=$

$\ds \bigsize {\size x - \size a}$

Definition of $f$

$\ds $

$\le$

$\ds \size {x - a}$

$\ds $

$<$

$\ds \delta$

$\ds $

$\le$

$\ds \epsilon$

Since $\epsilon$ was arbitrary, it follows that $f$ is continuous at $a$ by definition.

Since $a$ was arbitrary, it follows that $f$ is continuous everywhere by definition.

$\blacksquare$