# Counter example to: If $F$ is a conservative field in region $A$ and $B$, so $F$ is conservative in $A \cup B$

I am looking for a counter example and am explanation for why the following statement is false:

If $$F$$ is a conservative field in region $$A$$ and $$B$$, so $$F$$ is conservative in $$A \cup B$$.

And what about $$A \cap B$$ (is it true in that case)?

The usual example is $$v = \left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\right)$$ defined on the punctured plane $$\mathbb R^2\setminus\{0\}$$. This is not a conservative field, even though we have $$\text{curl}(v(x,y)) = 0$$ for all $$(x,y)\in\mathbb R\setminus \{0\}$$.
However, we may write $$\mathbb R^2\setminus\{0\}$$ as the union of two simply-connected regions. $$v$$ is necessarily conservative on each of these regions, since any vector field on a simply-connected set whose curl is $$0$$ is conservative.
As for the intersection, if any integral of the field over a closed loop in $$A$$ gives zero, then clearly any integral over a closed loop in $$A\cap B$$ must give zero.