Confusion about topologist's whirlpool I've found a solution to the title here https://static1.squarespace.com/static/5aff705c5ffd207cc87a512d/t/5c950a447817f75bd2262096/1553271370324/Topology.pdf 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.



$A$ 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. 
