I'm confused about the following exercise in the textbook Vector Calculus:
Let $f:(0,1) \rightarrow \mathbb{R}$ be s.t. $t \mapsto \frac{1}{t}$.
Part a) Prove that $f$ is continuous at every point $p \in (0,1)$
Part b) Prove that $f$ is not uniformly continuous
For part a, the solution given is as follows:
Let $\delta = \min( \frac{1}{2}p, \frac{1}{2}\epsilon p^2)$. Then from $|t-p| < \delta$ it follows that $t>\frac{1}{2}p$ and also $$|f(t)-f(p)| = \frac{|t-p|}{|tp|} < \frac{ \frac{1}{2}\epsilon p^2} { \frac{1}{2}p^2} = \epsilon$$
For part b, this proof is given:
Take $\epsilon = \frac{1}{2}$. Let $\delta$ be arbitrary. Then choose $n$ s.t. $\delta > \frac{1}{n}$, and let $p = \frac{1}{n}$ and $t = \frac{1}{n+1}$. We get that $|t-p| < \delta$ but $f(t) - f(p) = 1 > \epsilon$.
Textbook defines only uniform continuity explicitly:
A function $f:D \rightarrow \mathbb{R}$ is uniformly continuous on $D$ if to each $\epsilon>0$ there corresponds $\delta>0$ such that $$|f(t) - f(s)| < \epsilon \text{ whenever } |t-s| < \delta$$ for $t,s \in D$.
Part a is asking us to prove that given a point, the function is continuous at that point. I.e. $\forall p \forall \epsilon \exists \delta$. But part b is asking us to prove that $\forall \epsilon \exists \delta \forall p$.
While I understand the logical difference, I am extremely confused as to the interpretation of these definitions. How can a function be continuous at every point, but not uniformly continuous? Does this have to do with uniformity of slopes?
Uniform continuity, as defined here, seems to require that all points are some function of $epsilon$ away from each other, regardless of the choice of point. In other words, uniform continuity is independent of the point.
How should I interpret this notion intuitively?
The geometric interpretation of uniform continuity, for real-valued functions on intervals of $\mathbb R$, if you like thinking about the graph of a function is:
For your given $\epsilon$ value, draw a "band" around your graph by plotting $f(x) + \epsilon$ and $f(x) - \epsilon$. You can also think of them as lower and upper guardrails, limiting the motion of your graph.
Now try to wiggle the original graph back and forth horizontally by $\delta$, which corresponds to graphs of $f(x-\delta_1)$ for $| \delta_1 | \lt \delta$.
If you can do the back-and-forth wiggling and keep the wiggled graph within the band you created, congratulations, you have found a $\delta$ that works for that $\epsilon$ in the definition of uniform continuity.
Unfortunately, for the standard non-uniformly continuous example, $1/x$ on $(0,1)$, we are so bad at visualizing what's happening near $0$ (we really want to draw the $\epsilon$-band transverse to the graph, not purely in the $y$-axis direction) that we can't see how this fails. So I almost end up falling back on the "bounded derivative" version, even though it's less general.