Boundary of compact and star domain spaces I wonder if the boundary of a compact and star domain space is connected, or in general if the boundary of a compact and simply connected space is connected?
Thanks in advance. 
 A: Negative for compact and simply connected. Let $D=\{(x,y,z) \in \mathbb{R}^3 ~|~1\leq x^2+y^2+z^2 \leq 2 \}$, i.e., the region between two spheres. Then $\partial D$ is two disjoint copies of $S^2$. 
A: Consider $[0, 1] ⊆ ℝ$. It is compact, simply connected, even contractible, it is a star domain, but $∂[0, 1] = \{0, 1\}$.
Added: Maybe every compact star domain $D$ in $ℂ$ has connected boundary. Here is an idea. Suppose that $0$ is the center of $D$. For every $φ$ and every $0 ≤ r_1 ≤ r_2$ we denote the segment $\{r e^{iφ}: r ∈ [r_1, r_2]\} ⊆ ℂ$ by $[r_1, r_2]_φ$ and similarly for open and half open intervals. Since $D$ is a star domain for every $φ$ there is a point $x(φ) = r(φ) e^{iφ}$ such that $[0, r(φ)]_φ ⊆ D$ and $(r(φ), ∞)_φ ∩ D = ∅$.
It seems that for every $φ$
*

*$r(φ) ≥ \limsup_{φ' \to φ} r(φ')$,

*$S(φ) := [\liminf_{φ' \to φ} r(φ'),\ r(φ)]_φ ⊆ ∂D$,

*$∂D = ⋃_φ S(φ)$,

*for every $r > r(φ)$ there is $U$ a neighborhood $φ$ such that for every $φ' ∈ U$ we have $V(U, r) ∩ S(φ') ≠ ∅$ where $V(U, r) := \{r' e^{iφ'}: 0 ≤ r' < r,\ φ' ∈ U\}$.

From this the claim should follow.
Note that the boundary of a compact start domain in $ℂ$ does not have to be path connected or locally connected. Consider $D = ⋃_{0 ≠ φ ∈ [-π, π]} [0,\ 2 + \cos(1/φ)]_φ ∪ [0, 2]_0$. The boundary is a variant of the Warsaw circle (http://www.wolframalpha.com/input/?i=polar+plot+r%3D+2+%2B+cos(1%2F%CE%B8),+-%CF%80+%3C+%CE%B8+%3C+%CF%80). See Star-shaped domain whose closure is not homeomorphic to $B^n$.
A: In any (real or complex) normed vector space, other than a one-dimensional real vector space, it is true that the boundary of a bounded (not necessarily compact) star domain is connected. This can be seen from two important facts.


*

*Any normed vector space $V$ is unicoherent, i.e. $A\cap B$ is connected
for any two closed connected sets $A, B$ such that $A\cup B = V$.
Since  the closure of a connected set is connected and for any subset $S$ 
we have $\operatorname{Fr} S = \overline{S} \cap \overline{V\setminus S}$, unicoherence implies that a connected set with a connected complement
has a connected boundary.

*In any normed vector space other than a one-dimensional real space, 
spheres are connected. This implies that the complement of a bounded
star domain $S$ is connected, because we can find a sphere about the
centre $c$ of $S$ that lies entirely in $V\setminus S$ and any point $p 
\in V\setminus S$ can be connected to this sphere by a segment, lying in $V\setminus S$, of the line through $c$ and $p$.


Combining these with the obvious path-connectedness of a star domain immediately gives the desired result.
