Let $f(x) = x^3 - 3x$. Then we have $f'(x) = 3(x^2 - 1)$, $f'(x) > 0$ if and only if $x \in (-\infty, -1) \cup (1, \infty)$. Then we can deduce that $f(x)$ is increasing over two intervals $(-\infty, -1)$ and $(1, \infty)$.
However, I have seen other people write "$f(x)$ is increasing over $(-\infty, -1) \cup (1, \infty)$". I believe that this notation is incorrect. $f(x)$ is (strictly) increasing over a set $S \subset \mathbb{R}$ if and only if $f(u) > f(v)$ for any $x, y \in S$ and $x > y$. We have $f(-1.5) = 1.125$ and $f(1.5) = -1.125$. Thus $f(-1.5) > f(1.5)$ while $-1.5 < 1.5$.
What are your thoughts on this?