Is the complement of a set with vanishing $(d-2)-$dimensional Hausdorff measure simply connected? In the same vein as this question, I want to ask whether $\mathbb{R}^d\setminus A$ is simply connected if $A\subseteq \mathbb{R}^d$ has vanishing $(d-2)-$dimensional Hausdorff measure, i.e. ${\cal H}^{d-2}(A)=0.$ If the answer is 'no' in general, what if we assume $A$ to be closed as well?
 A: I'll sketch an affirmative proof for the case where $A$ is closed.
Abbreviate $U=\mathbb{R}^d\setminus A$. Let $\gamma:[0,1] \to U$ be a closed curve. Since $U$ is open, it is easy to homotope (remaining inside $U$ during the homotopy) $\gamma$ into a piecewise linear closed curve $\tilde{\gamma}:[0,1]\to U$. The next part of the proof is to reduce the number of vertices of $\tilde{\gamma}$ repeatedly. Let $abcd$ be 4 successive vertices of $\tilde{\gamma}$ ($d$ could be equal to $a$ in one of the final steps. The case where $\tilde{\gamma}$ already has $<3$ vertices is trivial). We will construct a homotopy which moves $c$ to some point $c'$ very close to the original point $c$, deletes the vertex $b$ and leaves $a$, $d$ and $\tilde{\gamma}|_{[0,1]\setminus [\tilde{\gamma}^{-1}(a),\tilde{\gamma}^{-1}(d)]}$ unchanged. The first part of that homotopy is to look for the 'right' $c'$. The stereographic projection $\pi_{a\to H}$ from $a$ onto some $(d-1)-$dim. plane $H\supseteq cd$ that doesn't contain $a$ is Lipschitz continuous if we exclude from its domain a small ball $B(a,\varepsilon)$ which for sufficiently small $\varepsilon$ is contained inside $U$ anyway. So ${\cal H}^{d-2}(\pi_{a \to H}(A))=0$. Hence, there exist points $x$ arbitrarily close to $c$ so that the line $ax$ has no intersection with $A$. Then choose $c'$ to be such a point so that moreover the homotopy consisting of moving $c$ to $c'$ along an affine function $c(t)$ and moving the line segments $bc(t)$ and $c(t)d$ along the way stays nicely inside $U$ (this can be achieved since $U$ is open and $c'$ can be found arbitrarily close to $c$). The next part of our homotopy is to move $b$ along some continuous function $b(t)$ to the midpoint $y$ between $a$ and $c'$, all the while keeping the curve connected through the segments $ab(t)$ and $b(t)c'$. For that purpose, take some $(d-1)-$dim. plane $K$ containing $b$ and $y$ but not containing $a$ nor $c'$ (such a plane exists if $a$, $b$ and $c'$ are not collinear. If, on the other hand, $a$, $b$ and $c'$ are already collinear, $b$ is already not a "vertex" and nothing needs to be done). The stereographic projections $\pi_{a\to K}$, $\pi_{c'\to K}$ with domain restricted to $A$ are again Lipschitz continuous so that ${\cal H}^{d-2}(\underbrace{\pi_{a\to K}(A)\cup \pi_{c'\to K}(A)}_{=:S})=0$. A variation of the first answer to the question that I linked implies that (we use the isometry between $K$ and $\mathbb{R}^{d-1}$) $K\setminus S$ is a (relatively) open (path-)connected set that contains both $b$ and $y$. So the function $b(t)$ alluded to before can be constructed inside the plane $K$.
