by 0000005
Last Updated August 14, 2019 15:20 PM

For example: How can we tell that for $$\vec{F} = \bigg(\dfrac{x}{x^2+y^2},\dfrac{y}{x^2+y^2},z^2\bigg),$$ $D$ is **connected** but **not simply connected**? Where $D$ is the domain of $\vec{F}$.

Is this true? We can tell it is connected because $\vec{F}$ is conservative since we can test that the $(\nabla\times\vec{F})=0$. If this is true, is there an *easier* way to see that this is connected?

In essence, $D$ is connected but not simply connected because there are a series of points which are undefined in the domain.

Consider the origin, $(0,0,0)$; well this point is undefined because we have $0$ in the denominator of the $x$ and $y$ coordinates.

Additionally, there are **singularities** $\forall (0,0,z)$ points, i.e. all points along the $z$-axis.

As you have asserted, we know that $\vec{F}$ is conservative because $\nabla\times\vec{F}=0$. These aforementioned singularities are **removable discontinuities.** If you put these ingredients together we have that $D$ is connected.

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