I've noticed that semi continuity is a 'sort of' continuity with respect to one sided limits:
We say that $f:X\to \mathbb{R}$ is upper (or respc. lower) semi-continuous at $x_0\in X$ if for all $\epsilon>0$ there exists a neighbourhood $U_\epsilon(x_0)$ such that $$f(x)\leq f(x_0)+\epsilon \qquad \text{or respc.} \qquad f(x)\geq f(x_0)-\epsilon $$ for all $x\in U_\epsilon(x_0)$. This seems to indicate to me that $f$ is upper semi-continuous at $x_0$ if $f^{-1}[V]$ is a neighbourhood of $x_0$, while $f(x_0) \in V$ and $V$ is open in the left order topology. Similarly, it looks like $f$ is lower semi-continuous at $x_0$ if $f^{-1}[V]$ is a neighbourhood of $x_0$, while $f(x_0) \in V$ and $V$ is open in the right order topology.
i.e., semi-continuity is just continuity of $f:X\to \mathbb{R}$ with respect to the appropriate one sided order topology on $\mathbb{R}$. I have not seen this stated explicitly anywhere though, which makes me think that perhaps I am overlooking something.
Is this indeed the case, or can someone point out why this is not true?