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    Convex and star-shaped open sets are simply connected

    E-SS3.EX-21

    Area
    Complex Analysis
    Topics
    Argument Principle, Meromorphic Functions, Residue Theorem
    Status
    draft
    1. Certain sets have geometric properties that guarantee they are simply connected.

    1. An open set \(\Omega \subset \mathbb { C }\) is convex if for any two points in Ω, the straight line segment between them is contained in Ω. Prove that a convex open set is simply connected.

    2. More generally, an open set \(\Omega \subset \mathbb { C }\) is star-shaped if there exists a point \(z _ { 0 } \in \Omega\) such that for any \(z \in \Omega ,\) the straight line segment between \(z_0\) and \(z\) is contained in Ω. Prove that a star-shaped open set is simply connected. Conclude that the slit plane \(\mathbb { C } - \{ ( - \infty , 0 ] \}\) (and more generally any sector, convex or not) is simply connected.

    3. What are other examples of open sets that are simply connected?

    Solution

    1. Let \(\Omega\) be convex and fix \(z_0\in\Omega\). For any closed curve \(\gamma:[0,1]\to\Omega\), define \[ H(s,t)=(1-s)\gamma(t)+sz_0, \qquad 0\le s,t\le1. \] By convexity, \(H(s,t)\in\Omega\) for every \((s,t)\). Thus \(H\) is a homotopy in \(\Omega\) from \(\gamma\) to the constant loop at \(z_0\). Hence every closed curve is null-homotopic, so \(\Omega\) is simply connected.

    2. The same argument works for a star-shaped open set: by definition there is \(z_0\in\Omega\) such that every segment from \(z_0\) to \(z\in\Omega\) lies in \(\Omega\). Therefore the same formula \[ H(s,t)=(1-s)\gamma(t)+sz_0 \] contracts every loop to \(z_0\).

    The slit plane \[ \mathbb C\setminus(-\infty,0] \] is star-shaped with respect to any positive real point, for example \(1\): if \(z\) is not on the nonpositive real axis, then the segment from \(1\) to \(z\) cannot cross that ray except possibly at its endpoint, which is excluded. Hence it is simply connected. Likewise a sector \(\{z\ne0:\alpha<\arg z<\beta\}\) with \(0<\beta-\alpha\le2\pi\) is star-shaped with respect to any point \(c\) of its bisecting ray: every \(z\) in the sector has argument within \((\beta-\alpha)/2\le\pi\) of \(\arg c\), strictly less than \(\pi\), so along the segment from \(c\) to \(z\) the argument moves monotonically between \(\arg c\) and \(\arg z\) and the segment avoids \(0\). Hence every such sector is simply connected.

    1. Other examples include discs, half-planes, strips, convex polygons with interior, and the whole plane. Each is convex (hence star-shaped), so part (a) applies.

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