we were given this definition:
Let $M \in \mathbb{R}$, $f: M \to \mathbb{R}$ be a function. $f$ has uniform continuity when $\forall \varepsilon > 0 \; \exists \delta >0 \;:\; |f(x) - f(x_0)| < \varepsilon \text{ for all } x, x_0 \in M \text{ with } |x-x_0| < \delta $
In addition to this definiton we have received an example:
Let $a$ be $> 0$. Then $f: [a, \infty) \ni x \mapsto \frac{1}{x} \in \mathbb{R}$ is uniformly continuous. Proof: For $x, x_0 \in [a, \infty)$ is: $|\frac{1}{x} - \frac{1}{x_0}| = \frac{|x-x_0|}{|x|\cdot |x_0|}| \leq \frac{1}{a^2} \cdot |x-x_0| < \varepsilon$ when $|x-x_0| < \delta := a^2 \cdot \varepsilon$
My issue lies in this part: $|\frac{|x-x_0|}{|x|\cdot |x_0|}| \leq \frac{1}{a^2} \cdot |x-x_0| < \varepsilon$. This is an upwards estimate. Because the chosen interval starts with $a$, we can be sure that $a \leq x$ and $a \leq x_0$, right? And that means we at best make the whole equation bigger, hence the $\leq$ in it.
But, wasn't $|\frac{1}{x} - \frac{1}{x_0}|$ meant to be smaller than $\varepsilon$? How can I still guarantee this after this estimation? This doesn't seem to be affected by the choice of $\delta$. Or am I missing something?
Okay, I figured out why I was so confused about this. The definition from the question could be rewritten as this:
$$ ∀ε>0\;\forall x_0, x \in M\;∃δ>0\;:\;|x−x_0|<δ\implies |f(x)−f(x_0)|<ε $$
Writing it like this makes it obvious to me why $\delta$ has an influence on $|f(x)−f(x_0)|$ and why I can just upwards estimate $|f(x)−f(x_0)|$. Suddenly the whole lecture makes sense.