Removing a set of zero $(d-2)-$Hausdorff measure from a $(d-1)-$dimensional connected surface does not destroy connectedness?

Question

Suppose $A\subseteq \mathbb{R}^d$ is connected and open while its complement $A^c$ is connected too and has non-empty interior. The answers to this question imply that $\partial A$ is then connected too and intuitively that boundary looks like a (possibly rough) $(d-1)-$dimensional surface. Suppose then that $B\subseteq \mathbb{R}^d$ is a set whose $(d-2)-$dimensional Hausdorff measure vanishes. Is $\partial A \setminus B$ still connected then?

Note that the more straightforward question "does removing a set $B$ of zero $(d-2)-$dimensional Hausdorff measure from a connected set $C$ of non-zero $(d-1)-$dimensional Hausdorff measure" has a negative answer since the non-zero $(d-1)-$dimensional measure doesn't prohibit $A$ to have 'narrow bottlenecks'.

My interest in this question arises from a possible application of Dubovitski\v{i}'s Theorem to show something of the sort that for every $f\in \mathbb{C}^2(\mathbb{R}^d,\mathbb{R})$ and $g\in C^1(\mathbb{R}^d,\mathbb{R})$ for which $\forall x \in \mathbb{R}^d$: rank$((\nabla f)(x),(\nabla g)(x))\leq 1$, we have that $g$ is constant on every connected component of a.e. level set of $f$. See this related question for some more context.

Answer 1

The original question is quickly answered in the negative (for $d>2$) by taking $A^c$ the union of two disjoint closed balls $B_1$ and $B_2$ (both with non-zero radius) and a straight line $\ell$ connecting their centers. Then take $B$ as a singleton somewhere on the line $\ell \setminus (B_1 \cup B_2).$

However, suppose $A\subseteq \mathbb{R}^d$ is open and connected and $(\overline{A})^c$ is connected too, then $S:=\partial \left(\overline{A}\right)\setminus B$ is connected for any closed $B\subseteq \mathbb{R}^d$ with ${\cal H}^{d-2}(B)=0$.

To see why, suppose there were such a $S$ so that $S=S_1 \cup S_2$ with $S_1\cap \overline{S_2}=\emptyset = \overline{S_1}\cap S_2$. Define $$O_1:= \bigcup_{x\in S_1}B\left(x,\frac{1}{4}d(x,S_2)\right),\\ O_2:=\bigcup_{x\in S_2}B\left(x,\frac{1}{4}d(x,S_1)\right),\\ f:\mathbb{R}^d\setminus B \to S^1\subseteq \mathbb{C}: x \mapsto \begin{cases} \exp\left(i\pi \frac{d(x,O_1)}{d(x,O_1)+d(x,O_2)}\right)& \text{when }x\in \overline{A}\\ \exp\left(-i\pi \frac{d(x,O_1)}{d(x,O_1)+d(x,O_2)}\right)& \text{when }x\in (\overline{A})^c \end{cases}$$ $O_1$ and $O_2$ are open and disjoint and $f$ is continuous. $\underbrace{(\overline{A}\cup O_1\cup O_2)\setminus B}_{=:X_1}$ and $\underbrace{((\overline{A})^c\cup O_1 \cup O_2)\setminus B}_{=:X_2}$ are both open. $\overline{A}\cup O_1\cup O_2$ and $(\overline{A})^c\cup O_1 \cup O_2$ are both connected and removing $B$ from them does not destroy connectedness since $B$ is closed and $0\leq {\cal H}^{d-1}(B)\leq {\cal H}^{d-2}(B)=0$. So $X_1$ and $X_2$ are both ($C^1$) path-connected. Choose arbitrary $x_1\in O_1$ and $x_2\in O_2$ and let $\gamma:[0,1]\to X_1$ be a path from $x_1$ to $x_2$ while $\delta:[0,1] \to X_2$ a path from $x_2$ back to $x_1$. Let $H$ be a homotopy inside $\mathbb{R}^d\setminus B$ (this homotopy exists) from the closed curve $\delta * \gamma$ to some constant closed curve. Then $f\circ H$ is a homotopy from a non-trivial homology to the trivial homology, which is a contradiction.

Answer 2

Definitely no. For example think of a disk on the plane, so in our case $d=2$. Then its boundary is the circle which is a $d-1=1$-dimensional surface. Now from that remove two dinstict points, i.e. a $d-2=0$-dim. set. Then the set is not connected.