Is $\sin(x^2)$ uniformly continuous for all $x \in (0, +∞)$? Where is my error?

Question

The answer is that is not uniformly continuous. But I get the contrary. Here is my solution. Let $f(x) = \sin(x^2)$, if $x ≠ 0$ and 0 if $x = 0$. In this case $f \in \Bbb{R}_{≥0}$ and $f(x)$ is continuous on $\Bbb{R}_{≥0}$. According to Heine–Cantor's theorem, if a function is continuous on $[a, b]$ then it is uniformly continuous. Let $a = 0$, and $b_n \to +\infty$ is the sequence of all numbers that belong to $\Bbb{R}_{≥0}$. Then $f(x)$ is uniformly continuous on $[0, b_n]$. And so it is uniformly continuous on $\Bbb{R}_{≥0}$. Now, I think that we can claim that $\sin(x^2)$ is uniformly continuous for any real $x$. Where is my error?

Answer 1

The formal error in your proof is extrapolating from compact intervals to all of the real line. Other answers and comments have noted this.

The informal/intuitive reason is that uniform continuity is essentially a bound on how steep the function is, since the slope at any point determines (in a way) how the $\delta$ in the definition of uniform continuity depends on the $\epsilon$.

In this case it's easy to see that the derivative $2x\sin(x)$ is unbounded on the real line. (It is bounded on the open unit interval, as in the title of the question.)

Answer 2

$f$ is uniformly continuous on $[0,b_n]$ for each $n$. But this does not imply that $f$ is uniformly continuous on $\Bbb{R}_{≥0}.$