Segment condition: Approximation of sobolev functions by functions which are smooth up to the boundary

Question

The idea behind domains fulfilling the segment seems to exclude the situation that "the domain does not lie on both sides of the boundary". But what about the set $$ S := \{x \subset \mathbb R^2 \colon 0 < |x| < 1\}? $$

I have two questions regarding the segment condition (see bottom for the definition as in Adams):

1. Does the above set fulfill the segment condition?
2. Where exacly can I observe that the fact that $\Omega$ does not fulfill the segment condition leads to a contradiction in the following example? From what I see is that the problem should be the approximation of the weak derivative of $u$ as it is no problem to approximate the "step" function $u(x,y)$ by a $C_0^\infty(\mathbb R^2)$ function in $L^p$-norm.*
3. (addendum to 2.) Strictly speaking, $\Omega$ does not fulfill the segment condition because it is no domain. Why do we restrict the definition of "segment condition" to domains anyway?

The following example in Adams -- Sobolev Spaces (3.20, p.68) tries to motivate the segment condition:

Where does this example go wrong? Apparently $\partial\Omega$ only has $2$ points, namely $(0,1)$ and $(0,0)$ failing the segment condition:

Answer 1

1) The segment condition is a way of saying that the boundary is locally the graph of a function. The punctured disk fails this condition at its center $x$ because no matter what $U_x$ and $y_x$ are, choosing $z = -\delta y_x$ with small $\delta>0$ results in $z\in \Omega\cap U_x$, yet $z + \delta y_x = x\notin \Omega$.

2) is a weird example because, as you observed, it's disconnected. A better example would be the function $u(z)=\arg z$ in a slit annulus such as $\Omega = A\setminus (-2,-1)$ where $A=\{z\in\mathbb C : 1<|z|<2 \} $. This is a domain failing the segment condition. If $u$ was a limit of functions in $C^1(\overline{\Omega})$ with respect to $W^{1,p}$ norm, than this limit would be a $W^{1,p}$ function in the entire annulus $A$ (because $A\subset \overline{\Omega})$ that would have to agree with $u$ in $\Omega$. This is impossible because the absolute continuity would have to fail across the negative real axis.

(Incidentally, this argument also works for $\Omega$ in the example you cited.)

To answer the specific question "Where exacly can I observe..." - it's in the fact that $\Phi(b)-\Phi(a)$ is related to integration along segments from $(a,y)$ to $(b,y)$, which is a bunch of segments crossing the boundary of $\Omega$ and coming right back to $\Omega$.

3) We don't really have to: much of what is proved about Sobolev functions on domains could be proved on more general open sets. But (a) this does not bring any new insights; we just do whatever we do on each component separately; (b) some key results like the Poincaré inequality $$ \|u-u_\Omega\|_p \le C\|\nabla u\|_p $$ absolutely require connectedness.