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Let $A = \{(x,y) \in\mathbb{R}^2: a \leq (x-c)^2+(y-d)^2 \leq b\}$ for given $a,b,c, d$ real numbers. I want to show that $A$ is path-connected.

How can I do that?

I know that every open subset of $\mathbb R^2$ that is connected is path connected. But this is obviously not open so I cannot use that. Then I thought of multiple cases. If we take arbitrary $x$ and $y$ and draw the line between them and they do not intersect with the circle centred at $(c,d)$ then we can obviously draw a line between the points which is still in the set, so we can then define the function. I am stuck on the other case.

Diego Silvera
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elaRosca
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    What did you try ? – Amr Feb 17 '13 at 20:13
  • I know that every open subset of $\mathbb{R}^2$ that is connected is path connected. But this is obviously not open so I cannot use that. Then I thought of multiple cases. If we take arbitrary x and y and draw the line between them and they do not intersect with the circle centred at (c,d) then we can obviously draw a line between the points which is still in the set, so we can then define the function. I am stuck on the other case. – elaRosca Feb 17 '13 at 20:17
  • Don’t get so tied up in theory that you forget easy, elementary approaches: it’s straightforward to write down an explicit path from one point to another that goes radially to the outer boundary, say, then around the boundary, then radially inward (if necessary). It’s very straightforward if you follow the suggestion in @Seirios’s answer. – Brian M. Scott Feb 18 '13 at 03:26

3 Answers3

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The set $S:=[0,2\pi]\times[a,b]$ is path-connected, being a product of path-connected sets. The annulus $A$ is the image of $S$ under the continuous map $(x,y)\mapsto y\cdot e^{xi}+(c+di)$, where you consider $\mathbb R^2$ as the complex plane.

Stefan Hamcke
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Hint: Translate $A$ so that you can suppose $c=d=0$ and use polar coordinates.

Seirios
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Consider the following set:

$A^\prime=\{(x,y) \in\mathbb{R}^2: a< (x-c)^2+(y-d)^2< b\}$

i.e. exactly your set but with $<$ instead of $\le$. This set is open and connected, so by your comments it's path connected.

Show that the places in your set but not in mine can easily reach mine by a path.

Ben Millwood
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