Absolute Value Function is Continuous
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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 $x$.
Then $f$ is a continuous real function.
Proof
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 $f$ | ||||||||||
\(\ds \) | \(\le\) | \(\ds \size {x - a}\) | Reverse Triangle Inequality on Real Numbers | |||||||||||
\(\ds \) | \(<\) | \(\ds \delta\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \epsilon\) |
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.
$\blacksquare$