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Let $\Omega \subset \mathbb{C}$ be a simply connected set with a smooth boundary (i.e., there is a differentiable function $\gamma : [a, b] \to \mathbb{C}$ so that $\gamma([a, b]) = \partial \Omega$). How can I show that $\overline{\Omega}$ is path-connected? We know that the boundary is path-connected via $\gamma$, and the interior is path-connected because an open, connected set is path-connected. I'm not sure how to connect these two parts via a path though. I did see this question, but it's a bit beyond my current knowledge.

Frank
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It only remains the case of connecting a point $z_1 \in \Omega$ with a point $z_2 \in \partial \Omega$.

$\partial \Omega = \gamma([a, b])$ is compact because $\gamma$ is continuous, therefore $$ \{ t > 0 \mid z_1 + t \in \partial \Omega \} $$ has a minimum $t^*$. Then the segment $[z_1, z_1+t^*]$ connects $z_1$ with $z_3= z_1+t^* \in \partial \Omega$ in $\overline \Omega$. Now connect $z_3$ with $z_2$ with a restriction of $\gamma$.

Martin R
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  • So if the boundary is only continuous and $\Omega$ is merely connected, then this still holds? – Frank Jun 03 '21 at 08:52
  • @Frank: Yes. If the boundary is the image of a continuous function $\gamma$ then it is compact. So you can consider a line from $z_1$ and define $z_3$ as the first intersection of that line with the boundary. – Martin R Jun 03 '21 at 08:55
  • This might be a dumb question, but how do we know there is a point that we can treat as the “first point” on $\partial \Omega$ that the straight line intersects? What if for every point on $\partial \Omega$ we intersect, there is a point not in $\overline{\Omega}$ that we intersect first? – Frank Jun 03 '21 at 09:00
  • @Frank: Is it more clear now? – Martin R Jun 03 '21 at 09:03
  • Yes! That makes sense, thanks. – Frank Jun 03 '21 at 09:07