Is open always Zariski open in the context of Algebraic geometry, unless otherwise mentioned?

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Going through Smith et al.'s Invitation to Algebraic Geometry I sometimes find myself wondering whenever they use the word open, do they mean open in the usual sense on $\mathbb{A}^1$ for example, or Zariski open? Many times, it is clear from the context. But I am unsure about the context of Section 4.3 on Regular functions.

Especially this definition of to be regular on an affine variety $V$:

Let $U$ be any open set of an affine variety $V$. A complex-valued function $f : U \to \mathbb{C}$ is regular at a point $p \in U$ if there exist functions $g$ and $h$ in (the coordinate ring) $\mathbb{C}[V]$ such that $h(p) \neq 0$ and $f$ agrees with the function $g/h$ in some neighborhood of $p$. The function $f$ is regular on $U$ if it is regular at every point of $U$. The set of all regular functions on $U$ is denoted by $O_V(U)$.

Is $U$ allowed to be ordinary open, or always Zariski open? And does $f$ have to agree with $g/h$ in some Zariski neighbourhood, or can it be an ordinary neighbourhood (which are usually allowed to be much smaller). In the second case, if we allow for the standard opens in $\mathbb{A}^n$, then it seems possible from $h(p) \neq 0$ to find an open around $p$ such that $h \neq 0$ on that open but that is also not that clear to me why it has to be the case for Zariski opens.