Subspace homeomorphic to open disk

Question

I want to show that if a subscriber of $\Bbb R^3$ is homeomorphic to open disk $D2$ then the subspace isn't open in $\Bbb R^3$. How can I prove ? I thought that the subspace is open then exits a neighborhood in the subspace homeomorphic to open $D3$

Answer 1

Assume $U\subseteq \Bbb R^3$ is open and $f\colon U\to D^2$ a homeomorphism. Pick an open ball $B_r(x_0)\subseteq U$ and let $X=B_r(x_0)\setminus\{x_0\}$, $Y=f(X)$. Then every continuous map $S^1\to X$ can be retracted into a point (i.e., can be extended to continuous map $\overline{D^2}\to X$), hence the same property holds for $S^1\to Y$. But there exists $\rho>0$ such that $B_\rho(f(x_0))\subseteq Y\cup\{y_0\}$, and the map $S^1\to Y$, $z\mapsto f(x_0)+\frac\rho2z$ is not retractable (why?)