Let $S$ be a bounded and closed star domain of $\mathbb{R}^n$, $n>1$ with not empty interior and with an inner a "good" star-center $x_0$.

I define "good" star-center as a star-center $x_0$ such that for every $x \in S$, the straight line $\overline{x_0 x}$ that crosses $x_0$ and $x$, also intersects $\partial S$ only one point.

I have a sketch of a proof that if the boundary of $S$, $\partial S$, is locally path-connected then $\partial S$ is path-connected.

Proof: Let $B(x_0,r) \subset S^\circ$ be a close ball centered in $x_0$ in the interior of $S$. Let define a bijective map $f:\partial B \rightarrow \partial S$ defined by $f(x) = \partial S \cap \overline{x_0 x}$. As $S$ is bounded, $f(x)$ is defined for all $x \in \partial B$.

Now consider two points $x,y \in \partial S$. As $\partial B$ is path-connected, then there exists a path $\gamma$ on $\partial B$ that connects $f^{-1}(x)$ and $f^{-1}(y)$. Since $\partial S$ is locally path-connected, we can divide $\gamma$ into smaller segments $\{\gamma_i\}_{i=1}^m$ such that $f(\gamma_i)$ is a continuous path in $\partial S$. Hence the union $\cup_i f(\gamma_i)$ is a path that connects $x$ and $y$.

Is correct this proof?

  • $\begingroup$ Your $f$ function simply maps $\partial B$ onto $\partial S$ if well defined. Since $\partial B$ is (path) connected then so is $\partial S$. Is $f$ always well defined? Anyway I don't see how local path-connectness is important here. Actually is it possible for $\partial S$ to not be locally path connected when the star center is in the interior of $S$? $\endgroup$ – freakish Mar 15 '17 at 15:46
  • $\begingroup$ $\partial B$ is a $n-1$ dimensional sphere, and it is path-connected. $\endgroup$ – jaogye Mar 15 '17 at 15:49
  • $\begingroup$ Yes, yes, I've meant $\partial S$. $\endgroup$ – freakish Mar 15 '17 at 15:54
  • $\begingroup$ $\partial S$ is locally path connected by assumption. $\endgroup$ – jaogye Mar 15 '17 at 15:55
  • $\begingroup$ Yes, I know. But I'm asking how is that important? I've already shown you the proof that doesn't depend on that assumption. Plus is there even an example of $S$ with non locally path connected boundary when the star center is in the interior of $S$? $\endgroup$ – freakish Mar 15 '17 at 16:02

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