According to Heine's Theorem, "If a function is continuous on a closed and bounded interval, then the function is uniformly continuous on that interval."
I understand why Heine says "closed" but I don't understand why he says "bounded."
Can't I just say "If a function is continuous on a closed interval, then the function is uniformly continuous on that interval." ? Can somebody give me a counter example for this?
On
So, on the set $[0,+\infty)$ continuity does not imply uniform continuity. $x \mapsto x^2$ is a simple counterexample. And there's some disagreement as to whether or not $[0,+\infty)$ should be considered a "closed interval".
There are issues with the definition of "interval" - can the upper end be infinite? Also, naively, the fact that it doesn't include "the point at infinity" might make you say that it's not closed. But as you learn more you'll see that it is in fact a closed set in the topological sense. Including the term "bounded" clarifies (or removes) all these issues.
Also, there is a generalization of Heine's theorem to metric spaces, in which this version for $\mathbb R$ becomes just a special case. In the generalized version the requirement is that the domain be "compact", and you'll learn that the compact sets in $\mathbb R$ are the closed and bounded sets, so it's nice to get used to how "boundedness" plays a role.
The function $f:\mathbb{R} \longrightarrow \mathbb{R}$ you mentioned does not violates the statement.
For an example function $f:D \longrightarrow \mathbb{R}$ violates the statement, (which is actually impossible) the following two condition should be satisfied; i. $D$ is closed and bounded interval and $f$ is continuous on $D$ ii. $f$ is not uniformly continuous on the interval.
Since your function $f$ does not satisfies 'i', it is not a counterexample of the statement.