Let $S^n$ be the $n$-dimensional sphere and let $K\subseteq S^n$ be a compact, locally contractible subspace of real codimension $\geq 2$. Applying Alexander duality we find that

$$ \tilde{H}_{i}(S^n-K)\simeq \tilde{H}^{n-i-1}(K) $$ where $\tilde{H}$ denotes the reduced homology (cohomology) with coefficients in $\mathbf{Z}$ and $0\leq i\leq n-1$. In particular, taking $i=0$ we find that $$ \tilde{H}_{0}(S^n-K)\simeq \tilde{H}^{n-1}(K)=0. $$ In particular, we deduce that $S^n-K$ is connected and thus path connected (since $S^n-K$ is a euclidian open set and therefore locally path connected).

Q: Is there a simple (low-tech and/or geometrical proof) that $S^n-K$ is path connected ?

Let $S^n$ be the $n$-dimensional sphere and let $K\subseteq S^n$ be a compact, locally contractible subspace of real codimension $\geq 2$. Applying Alexander duality we find that

$$ \tilde{H}_{i}(S^n-K)\simeq \tilde{H}^{n-i-1}(K) $$ where $\tilde{H}$ denotes the reduced homology (cohomology) with coefficients in $\mathbf{Z}$ and $0\leq i\leq n-1$. In particular, taking $i=0$ we find that $$ \tilde{H}_{0}(S^n-K)\simeq \tilde{H}^{n-1}(K)=0. $$ Thus $S^n-K$ is connected and therefore path connected (since $S^n-K$ is a euclidian open set and therefore locally path connected).

Q: Is there a simple (low-tech and/or geometrical proof) that $S^n-K$ is path connected ?