$\newcommand\R{\mathbb R}$Let $U$ be an open subset of $\R^2$ such that the point $(0,0)$ is on the boundary of $U$. Let $f\colon[0,1]\to\R^2$ be the path that starts at $(0,0)$ and moves with a (say) constant speed, first horizontally right to $(2,0)$, then horizontally left to $(1,0)$, and finally goes counterclockwise twice over the unit circle. Let $g\colon[0,1]\to\R^2$ be another continuous path that also starts at $(0,0)$ and such that $g(t)\in U$ for all $t\in(0,1]$ and $|g(t)-f(t)|<1/16$ for all $t\in[0,1]$, where $|\cdot|$ is the Euclidean norm.


Can then the set $U$ be simply connected?



Can such a set be simply connected?
Let $I\_0$ be the interval where $f$ is a horizontal motion, and let $I\_1$ the interval where it is a circular motion. So up to reparametrisation (and adopting complex notation) $I\_0=[-3,0]$ and $I\_1=[0,4\pi]$, and $f(t)=\min(t+3,1-t)$ in $I\_0$ and $f(t)=e^{it}$ in $I\_1$.


Since $|g(3\pi/2)+i|<1/16$ and $|g(5\pi/2)-i|<1/16$ by connection there are points $t\_0\in I\_0$ and $ 3\pi/2< t\_1 < 5\pi/2$ such that $g(t\_1)=g(t\_0)$ **(\*)**.


This means that $g$ restricts to a closed arc on $[t\_0,t\_1]$, which is still $1/16$ close, up to reparametrisation, to the loop $f|[-2,2\pi]$, which is non contractible in $\mathbb C\setminus\{0\}$. Therefore so is the loop $g\_{|[t\_0,t\_1]}$, and a fortiori in $U\subset \mathbb C\setminus\{0\}$. Therefore $U$ is not simply connected.


**(\*)** This can be seen geometrically as a consequence of the Jordan curve theorem (note that $g(I\_0)$, as any image of a continuous curve, contains a simple curve with the same endpoints, and this can be enlarged to a simple closed curve bounding a domain containing $g(3\pi/2)$ and not $g(5\pi/2)$. Otherwise, without bothering about injectivity: $g\_{|I\_0}$ can be extended to a closed loop $\gamma$, juxtaposing to it an arc far from $g(I\_1)$, and such that $\text{ind}(\gamma,g(3\pi/2))=-1$ and $\text{ind}(\gamma,g(5\pi/2))=0$, which implies that $g\_{|[3\pi/2,5\pi/2]}$ must cross the curve $\gamma$ (necessarily in $g(I\_0)$) otherwise $\text{ind}(\gamma,g(t))$ would remain constant for $3\pi/2\le t \le 5\pi/2$


