Another example of a connected but non path connected set

Solution 1:

Yes, the closure isn't path connected. To connect a point with $r > 1$ in the set to a point with $r = 1$ in the closure, you'd need a path winding around the unit disk infinitely many times, but that isn't possible since a path is uniformly continuous, and hence a path in the closure can only wind finitely often around the unit disk.

Let us denote $\Gamma = f([\pi/2,+\infty))$. Then $\overline{\Gamma} = \Gamma \cup S^1$ where the union is disjoint. Let $\alpha \colon [0,1] \to \overline{\Gamma}$ be a path with $r_0 := \lVert \alpha(0) \rVert > 1$. Since $\Gamma$ intersects each circle $\{(x,y) : x^2+y^2 = r^2\}$ with $r > 1$ in exactly one point, we can write the polar angle as a function of the radius while $\alpha(t)\in\Gamma$,

$$\varphi(t) = \frac{1}{\log \lVert\alpha(t)\rVert}.$$

Thus to reach the radius $0 < r < r_0$, the polar angle must traverse the entire interval $\left[ \frac{1}{\log r_0}, \frac{1}{\log r}\right]$ and hence $\alpha$ must wind at least

$$n(r) = \left\lfloor \frac{1}{2\pi}\left(\frac{1}{\log r} - \frac{1}{\log r_0}\right)\right\rfloor$$

times around the unit disk. But each full winding around the unit disk contains points where the polar angle differs by $\pi$, and therefore points whose distance is $> 2$.

But, since $[0,1]$ is compact, $\alpha$ is uniformly continuous, so there is a $\delta > 0$ such that for $\lvert t -s\rvert \leqslant \delta$ we have $\lVert \alpha(t) - \alpha(s)\rVert \leqslant 2$. Thus $\alpha$ can wind at most

$$k(\alpha) = \left\lceil \frac{1}{2\delta}\right\rceil$$

times around the unit disk, and therefore

$$n(\lVert\alpha(t)\rVert) < k(\alpha)+1,$$

which implies

$$\lVert \alpha(t)\rVert > \exp \left(\frac{1}{(\log r_0)^{-1} + 2\pi(k(\alpha)+2)}\right) > 1,$$

and hence a path starting in $\Gamma$ never can reach the unit circle.