Puncturing a connected open subset of ${\mathbb{R}}^n$ preserves path-connectedness for $n\ge 2$

Fix $x,z\in \mathrm{\Omega }\setminus \{y\}$. If $x=z$, the constant path at $x$ lies in the punctured set. Otherwise [F1] gives a polygonal path $\gamma :[0,1]\to \mathrm{\Omega }$ from $x$ to $z$. Choose $r>0$ small enough that ${\bar{B}}_2(y,r)\subset \mathrm{\Omega }$ and $r<\min \{\mathrm{\mid }x-y\mathrm{\mid },\mathrm{\mid }z-y\mathrm{\mid }\}$. If $\gamma$ avoids $y$, it already gives the required path.

Suppose $\gamma$ meets $y$. The preimage $A={\gamma }^{-1}[{\bar{B}}_2(y,r)]$ is a finite union of closed intervals: on each of the finitely many affine pieces in [F1], the preimage of the convex closed ball is a closed interval, possibly empty or a point. Its connected components are therefore finitely many closed intervals $[a_j,b_j]$. Since the endpoints $x,z$ lie outside the closed ball, each component is contained in $(0,1)$, and continuity and maximality give $\gamma (a_j),\gamma (b_j)\in S_2(y,r)$. If $a_j=b_j$, that component is a single sphere point and cannot contain $y$.

For each component with $a_j<b_j$, use [F2] to choose a continuous path on $S_2(y,r)$ from $\gamma (a_j)$ to $\gamma (b_j)$, reparameterized on $[a_j,b_j]$. Replace $\gamma$ on those finitely many intervals by these sphere paths and retain it on the intervening closed intervals. The pieces agree at every endpoint, so [F3] gives a continuous path $\tilde{\gamma }:[0,1]\to \mathrm{\Omega }$. Every replacement lies on a sphere of positive radius and hence misses $y$; the retained portions lie outside the closed ball, apart from singleton components already on the sphere. Some component has positive length because $\gamma$ meets the interior point $y$. Thus $\tilde{\gamma }$ avoids $y$ and joins $x$ to $z$. This also covers any zero-length polygonal pieces and tangencies to the sphere.

The construction proves path-connectedness, including equal endpoints by the constant path in step 1.1. The set is open: for each $x\mathrm{\ne }y$ in $\mathrm{\Omega }$, openness of $\mathrm{\Omega }$ gives a ball about $x$ contained in $\mathrm{\Omega }$, and shrinking its radius below $\mathrm{\mid }x-y\mathrm{\mid }$ makes it avoid $y$. It is nonempty because a ball about $y$ contains $y+t{e}_1\mathrm{\ne }y$ for some sufficiently small $t>0$, where $e_1=(1,0,\dots ,0)$. By [F4] it is connected. The proof covers zero-length polygonal pieces, tangent singleton components and the case $x=z$; no infinite family of detours or iff claim is used. [F3, F4, given, step 1.1, step 3.1, algebra, cases, discharge-construct] \square