Linear Function is Continuous/Proof 1

Theorem

Let $\alpha, \beta \in \R$ be real numbers.

Let $f : \R \to \R$ be a linear real function:

$\map f x = \alpha x + \beta$

for all $x \in \R$.

Then $f$ is continuous at every real number $c \in \R$.

First assume $\alpha \ne 0$.

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

Let $\delta = \dfrac \epsilon {\size \alpha}$.

Then, provided that $\size {x - c} < \delta$:

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

$=$

$\ds \size {\alpha \paren {x - c} }$

$\ds $

$=$

$\ds \size {\alpha} \cdot \size {x - c}$

$\ds $

$<$

$\ds \size {\alpha} \delta$

$\ds $

$=$

$\ds \epsilon$

So, we have found a $\delta$ for a given $\epsilon$ so as to make $\size {\map f x - \map f c} < \epsilon$ provided $\size {x - c} < \delta$.

So $\ds \lim_{x \mathop \to c} \map f x = \map f c$ and so $f$ is continuous at $c$, whatever $c$ happens to be.

Now suppose $\alpha = 0$.

Then $\forall x \in \R: \map f x - \map f c = 0$.

So whatever $\epsilon \in \R_{>0}$ we care to choose, $\size {\map f x - \map f c} < \epsilon$, and whatever $\delta$ may happen to be is irrelevant.

Continuity follows for all $c \in \R$, as above.

$\blacksquare$