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I recently saw a different definition for simply connected which I had never seen before. A connected subset $\Omega\subset\Bbb C$ is called simply connected if the boundary is the image of a simple closed curve. Is this equivalent to the usual definitions?

Thanks

Mike
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  • This does not coincide with the usual definition, because the boundary of a simply connected open set of $\Bbb R^2$ could be much more complicated than a simple closed curve. – 23rd Jun 10 '13 at 09:14
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    could you provide an example, reference? Thanks – Mike Jun 10 '13 at 09:15
  • Imagine an $8$-shaped area. It's interior is not simply connected, but it's boundary can be an image of a simple closed curve, like a circle, and even a continuous image of it. – SBF Jun 10 '13 at 09:19
  • But this curve would not be simple. It would self intersect. – Mike Jun 10 '13 at 09:22
  • @Mike: You may see the pictures in page 15-1 here. Pictures $(b)$, $(c)$, $(d)$ are counter-examples. There are more pathological counter-examples, but let me leave the answer to others. – 23rd Jun 10 '13 at 09:31
  • @Mike: it isn't a simple curve, but it's the image of a simple curve. Nevertheless, you require the set to be connected - and in my example it's not. – SBF Jun 10 '13 at 11:58

1 Answers1

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image of a simple closed curve

Interpretation 1: the word "image" refers to the fact that a curve is really a map (parametrization), and the geometric shape that we think of as a curve is the image of that map. Some people, including me, would omit the words "image of" in this case.

With this interpretation, we have the definition of a Jordan domain, which is a smaller class of domains than simply connected. For example, the slit disk $\{z:|z|<1\}\setminus [0,1]$ is a simply connected domain but is not a Jordan domain.

image of a simple closed curve

Interpretation 2: the word "image" means continuous image. A set is a continuous image of a simple closed curve if and only if it is compact, connected, and locally connected (this follows from the Hahn-Mazurkiewicz Theorem. Again, we have a more restrictive definition that simple-connectedness:

  • the boundary of a simply connected domain need not be compact (consider the halfplane)
  • the boundary of a simply connected domain need not be locally connected. To construct an example, remove the vertical line segments from $1/n$ to $1/n+i$, $n=1,2,\dots$ from the open quadrant $x>0,y>0$.

Remarks:

  1. I guess the first interpretation was the intended one.

  2. It is true that a connected subset of $\Omega\subset\mathbb C$ is simply connected if and only if the set $\overline{\mathbb C}\setminus \Omega $ is connected in the topology of the Riemann sphere $\overline{\mathbb C}$. In the special case when $\Omega$ is bounded, this amounts to $\partial \Omega$ being connected.

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