Given a vector field, $\vec{F}$, and let $D$ be the domain of $\vec{F}$, how can we tell if $D$ is connected and/or simply connected?

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?



Answers 1


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.

JohnColtraneisJC
JohnColtraneisJC
August 14, 2019 14:59 PM

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