I've found a solution to the title here on page $71$
But why the arc give a separation in the solution "Then we can find some $p\in (1−ε,1)$ such that the arc of $A∩N$ containing $q$ is separated from $S^1$ by the circle $\{(r,θ) \vert r = p\}$. This gives a separation $U=\{(r,θ)\vert r<p\}\cap N$ and $V=\{(r,θ)\vert r>p\}\cap N$ of$N′$. Thanks.
2 Answers
$\begingroup$Let $u:[0,1]\to W$ be continuous with $u(0)\in S^1.$
For any $x\in [0,1]$ such that $u(x)\in S^1$ we may take an open disk $E_x$ in $\Bbb R^2,$ centered at $u(x), $ of sufficiently small radius that for any $p\in E_x\cap (W\setminus S^1)$ there exist $disjoint$ open subsets $U_{p,x}, V_{p,x}$ of $\Bbb R^2$ such that $$(*)\quad u(x)\in U_{p,x}\;\land \; p\in V_{p,x}$$ and such that $$(**)\quad U_{p,x}\cup V_{p,x}\supset E_x\cap W.$$ There exists $d_x>0$ such that $J(x)=\{u(y): y\in [0,1]\cap (-d_x+x,d_x+x)\} \subset E_x.$ Now $[0,1]\cap (-d_x+x,d_x+x)$ is connected and $u$ is continuous, so $J(x)$ is connected. But if $p\in J(x)\setminus S^1$ then by $(*)$ and $(**),\; $ $J(x)$ would not be connected.
Therefore no $p\in E_x\cap (W\setminus S^1)$ belongs to $J(x).$ So, with $I(x)=[0,1]\cap (-d_x+x,d_x+x),$ we have $I(x)\subset u^{-1}S^1.$
Therefore $u^{-1}S^1=\cup \{I(x): u(x)\in S^1\}$ is OPEN in the space $[0,1].$ But $S^1$ is closed in $W$ and $u$ is continuous so $u^{-1}S^1$ is CLOSED in $[0,1].$
So $u^{-1}S^1$ is open-and-closed in the connected space $[0,1]$ and is not empty (because $0\in u^{-1}S^1).$ So $u^{-1}S^1=[0,1].$ So $u$ cannot be a path from $u(0)$ to any member of $W\setminus S^1.$
Remark: If the radius of $E_x$ is small enough, and if $p=f(\theta)=\frac {\theta}{1+\theta}(\cos \theta,\sin \theta)\in E_x\cap (W\setminus S^1),$ (with $\theta \in [0,\infty)\,),$ then there exist $t,t'$ with $\theta<t<t', $ such that $f(t)$ and $f(t')$ lie on the boundary (in $\Bbb R^2)$ of $E_x,$ and such that $f(\psi)\not \in E_x$ for $t<\psi<t'.$ Then the sets $\{f(z): z\in [0,t]\}$ and $S^1\cup \{f(z):z\ge t'\}$ are closed and disjoint in $\Bbb R^2,$ so they are completely separated by some disjoint open $U_{p,x},V_{p,x}$ that will (automatically) satisfy $(*)$ and $(**)$.
$\endgroup$ 2 $\begingroup$$A$ is not properly defined - it is not the graph of $f(\theta) = \left(\dfrac{\theta}{1+\theta},\theta \right)$. Note that in the proof of Claim 2 one can see that the domain of $f$ is $[0,\infty)$.
First of all, the author means the image of $f$ which is a subset of the plane, whereas the graph is a subset of $[0,\infty) \times \mathbb R^2 \subset \mathbb R^3$. Moreover, the image of $f$ is certainly unbounded and doesn't produce the depicted set $A$.
The correct definition is $f(\theta) = \dfrac{\theta}{1+\theta}e^{i\theta} = \left(\dfrac{\theta}{1+\theta}\cos \theta, \dfrac{\theta}{1+\theta}\sin \theta \right)$. Then $A$ is a spiral asymptotic to $S^1$ and it is homeomorphic to $[0,\infty)$. Note that $A = W \setminus S^1$ is open in $W$.
The proof given in the book is certainly correct, but I think it is too complicated. I suggest to do it as follows.
Let $z \in S^1$. The radial line segment $R(z)$ connecting $z$ and $0$ is compact, thus $W(z) = W \setminus R(z)$ is an open subset of $W$. Writing $z = e^{it}$ with $z \in [0,2\pi)$, we see that $A \cap R(z) = \{f(\theta) \mid \theta = t + 2k \pi , k = 0,1,2 ,\ldots \}$, thus $A^*(z) = A \setminus R(z)$ is the countable union of pairwise disjoint open subsets $U_k \subset A$ which are the homeomorphic images under $f$ of the intervals $J_k = (t+2k\pi, t+2(k+1)\pi)$, $k \ge 0$, and for $t \ne 0$ the interval $J_{-1} = [0,t)$. All $U_k$ are closed in $A^*(z)$ because their complements are open. Each $U_k$ is contained in a compact disk $D$ of radius $< 1$ with center $0$, thus the closure of $U_k$ in $W(z)$ cannot contain points of $S^1$. This shows that the $U_k$ are closed in $W(z)$. Hence they are open, closed and connected subsets of $W(z)$ and therefore components of $W(z)$.
Now assume that $W$ is path connected. Then there exists a path $u : [0,1] \to W$ such that $u(0) \in S^1$ and $u(1) \in A$. Let $\tau = \sup\{t \in [0,1] \mid u(t) \in S^1 \}$. By contuinity $w = u(\tau) \in S^1$ because $S^1$ is closed in $W$ and by definition $u(t) \in A$ for $t > \tau$. Let $z = -w \in S^1$. Then $W(z)$ is an open neighborhood of $w$ and we find $\tau < \tau' \le 1$ such that $u([\tau,\tau']) \subset W(z)$. But $w' = u(\tau') \in A^*(z)$, thus $w' \in U_k$ for some $k$. Since $U_k$ is a component of $W(z)$, the connected set $u([\tau,\tau'])$ must be contained in $U_k$, a contradiction.
