Is it always possible to make both pieces connected?

Question

Inside a path-connected set, there is a path-connected subset (Green) whose complement (Pink) is disconnected:

We would like to move a small piece from G to P, such that both of them are path-connected:

Is it always possible?

Formally, given two sets $G,P$ such that $G$ and $G\cup P$ are path-connected, and given $\epsilon>0$, we would like to find a subset $H\subseteq G$, such that:

• The area of $H$ is at most $\epsilon$;
• $G\setminus H$ is path-connected;
• $P\cup H$ is path-connected.

Is this always possible? If not, what conditions on $G,P$ are required to make it possible?

Answer 1

For a counterexample, let $$ G = \{0\}\times[0,1] \cup (0,1) \times ([0,1]\setminus \mathbb Q) \\ P = \{1\}\times([0,1]\setminus \mathbb Q) $$

$G\cup P$ is path-connected, but the only way to connect two different points of $P$ is to go all the way down to the $x$-axis, then horizontally and then up again. So $H$ has to be all of $G$, which has area (that is, Lebesgue measure) $1$.

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Or, for a less pathological example, consider

$$ G = (\mathbb R\setminus \{0\})\times (0,\infty) \cup \{(0,0)\} \\ P = (\mathbb R\setminus \{0\})\times (-\infty,0) $$

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Or, nicer yet,

$$ G = \mathbb R \times (0,1) \\ P = \mathbb R^2 \setminus G $$

Answer 2

To add to Henning Makholm's counterexamples, here's an example where $G$ has finite area and $P$ is open. Let $P = \{(x,y) : |y| < |x|\}$. Choose $f(x)$ to be a non-negative function with $f(0) = 0$, $f(x) > 0$ for $x \not= 0$ and $0 < \int_{0}^\infty f\;dx < \infty$. Then, let $$G = \{(x,y) : |x| \le |y| \le |x| + f(|x|)\}.$$ The area of $G$, $m(G)$, is equal to $4\int_0^\infty f\;dx$. If $H$ is a subset of $G$ such that $H \cup P$ is path connected, then $(0,0) \in H$. On the other hand, $G \setminus \{ (0,0)\} $ has four connected components of area $\int_0^\infty f\;dx$, so $H$ must contain three of these components, and $m(H) \ge 3\int_0^\infty f\;dx = \frac{3}{4}m(G) > 0$.