Rigour behind showing that $f(x) = c$ where $c$ is a constant

Question

Rigour behind showing that $f(x) = c$ where $c$ is a constant.

When being taught that a constant function is continuous usually professors just gloss over that the $\delta$ could be any value. So it got me thinking of how to demonstrate it completely rigorously.

So by definition a function is continuous at a point $a$ if:

$\forall \epsilon > 0, \ \exists \ \delta > 0 \ s.t \ if\ |x-a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon$

Let $f(x) = c$, $f: \mathbb{R} \rightarrow \mathbb{R}$:

So to show this explicitly I would have to "extract" a $|x-a|$ out of my expression and then obtain a $\delta$ by this relationship. I'm curious if I "extracted" the $|x-a|$ in the correct fashion:

$$|c-c| \leq |ac - xc + xc -ac| \leq |x-a||c-c| < \delta |c-c| = \delta \bullet 0 < \epsilon$$

The step in this inequality that is troubling me is whether I can get away with the $|c-c| \leq |ac-ac|$ it feels as if that shouldn't be an issue, but I'm not 100% sure. Thoughts?

Answer 1

A formal proof could be written in the following way, to show that $f(x) = c$ is continuous at some $a \in \mathbb{R}$:

Let $\epsilon > 0$. Pick $\delta = 1$; then for every $x \in \mathbb{R}$ such that $|x-a|<\delta$, we have $|f(x) - f(a)| = |c-c| = 0 < \epsilon$.

Hence $\forall \epsilon > 0$, $\exists \delta > 0$ such that $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. So $f$ is continuous at $a$.

Remark, of course, that the choice of $\delta>0$ was completely arbitrary, since it had no impact on the conclusion.