Question about conditions for conservative field In common textbooks' discussions about conservative vector field. There is always two assumptions about the region concerned, namely the region is simply connected and open.
Usually in textbooks there is not much explanations on why these assumptions are necessary, no proof is given on why conservative field is not possible if the region is not simply connected or not open.
I wonder whether these two assumptions are just for computational convenience or it is really logically not possible to have a conservative field in region that is not simply connected or is not open?
Simple-connectedness
It is certainly possible to have a conservative vector field on a set that is not simply connected: Pick any non-simply connected set $U$ and any differentiable function $f: U \subset \Bbb R^n \to \Bbb R$; then, by definition, the vector field $$\nabla f := \left\langle \frac{\partial f}{\partial x^1}, \ldots, \frac{\partial f}{\partial x^n} \right\rangle$$ is conservative. (By definition, all examples arise this way.)
An key result involving conservative vector fields that relies on simple-connectedness is the following:
The idea here is that one can recover a potential function $f$ for $\bf F$ by choosing a base point $p \in U$, choosing for each $q \in U$ a path $\gamma$ from $p$ to $q$ and defining $$f(q) := \int_{\gamma} {\bf F} \cdot d{\bf s}.$$ Then, since $\bf F$ satisfies $(\ast)$, Green's Theorem guarantees that this definition of $f(q)$ is independent of the only choice we made, namely the path $\gamma$; doing this for every point $q \in U$ defines a function that, by construction, is a potential for $\bf F$.
This theorem fails, however, when we omit the hypothesis of simple connectedness: For example, the so-called vortex vector field $${\bf V} := \frac{1}{x^2 + y^2} \langle -y, x\rangle$$ defined on $\Bbb R^2 - \{ 0 \}$ (which is not simply connected) satisfies $(\ast)$ but it is not conservative: For any circular path $C$ oriented anticlockwise and enclosing the origin, we have $$\int_C {\bf V} \cdot d{\bf s} = 2 \pi.$$ (Try showing this yourself; evaluating this integral is easiest when $C$ is centered at the origin.) So, we can see that simple-connectedness really is essential for this result and some that depend on it. (Note, by the way, that Green's Theorem implies that the restriction of $\bf V$ to an simply connected subset is conservative; up to a suitable constant, any potential is a so-called branch of the argument (function).)
Remark For vector fields $\bf F$ on subsets of $\Bbb R^3$, the appropriate replacement for $(\ast)$ is that $\text{curl } {\bf F} = 0$ (the scalar components of this equation are each similar to $(\ast)$); there are similar conditions on higher-dimensional Euclidean spaces.
Openness
The requirement of openness is more technical. In short, it ensures that one can always (attempt to) take derivatives of functions: In short, if we want to differentiate a function (or vector field) at a point, as we need to do in many cases, including the above theorem, we must be able to compare the its value at a point with nearby points, and openness is simply the requirement that all sufficiently nearby points are already in the domain. This requirement is also less essential than simple-connectedness in that in many cases we can still produce reasonable results with weaker hypotheses on the domain at the cost of some technicality, namely, of making sense of derivatives when, a priori, one cannot compare values of the object being differentiated in all directions. For a particularly simple case of this, recall that we can define derivatives at the endpoints $a, b$ of functions $[a, b] \to \Bbb R$ simply by replacing the limit in the usual difference quotient with (respectively) the right-handed or left-handed limit.