Let $P$ be a compact connected set in the plane and $x,y\in P$.



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> Is it always possible to connect $x$ to $y$ by a path $\gamma$ such that the length of $\gamma\backslash P$ is arbitrary small?
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**Comments:** 


* Be aware of [pseudoarc](https://en.wikipedia.org/wiki/Pseudo-arc) --- it is a compact connected set which contains no nontrivial paths.
* If $P$ contains an everywhere dense curve, then the answer is "yes"; the same holds if $P$ contains a dense countable collection of curves.



Running most of the time in a connected set
The answer to this question is positive. A required path $\gamma$ can be constructed inductively using the following


**Lemma.** For any continuum $P\subset\mathbb R^2$, distinct points $x,y\in P$, and $\varepsilon>0$ there are continua $P\_1,\dots,P\_{k}\subset P$ of diameter $<\varepsilon$, and horizontal or vertical arcs $A\_0,\dots,A\_{k+1}$ in the plane such that


1. $\{x\}=A\_0$, $\{y\}=A\_{k+1}$;
2. $\sum\_{i=1}^k\lambda(A\_i\setminus P)<\varepsilon$;
3. for every $i\in\{1,\dots,k\}$ the continuum $P\_{i}$ intersects the arcs $A\_{i-1}$ and $A\_i$.


*Proof.* Replacing $P$ by a smaller subcontinuum, we can assume that $P$ is irreducible between points $x$ and $y$, which means that any subcontinuum of $P$ that contains $x$ and $y$ coincides with $P$.


The irreducibility of $P$ implies that for any rational numbers $a<b$ the (closed) set $\{x\in \mathbb R:\{x\}\times[a,b]\subset P\}$ is nowhere dense in $\mathbb R$. Consequently the set $$G\_1:=\{x\in\mathbb R:P\cap(\{x\}\times\mathbb R)\mbox{ is zero-dimensional}\}$$is dense $G\_\delta$ in $\mathbb R$. Choose an increasing sequence $a\_0,\dots,a\_n\in G\_1$ such that $P\subset [a\_0,a\_n]\times\mathbb R$ and $|a\_i-a\_{i-1}|<\frac12\varepsilon$ for all positive $i\le n$. For every $i\in\{0,\dots,n\}$ the inclusion $a\_i\in G\_1$ implies that the set $Z\_i:=\{y\in \mathbb R:(a\_i,y)\in P\}$ is zero-dimensional and hence nowhere dense in $\mathbb R$.


By analogy, the set $$G\_2:=\{y\in\mathbb R:P\cap(\mathbb R\times \{y\})\mbox{ is zero-dimensional}\}$$is dense $G\_\delta$ in $\mathbb R$. Then we can choose a monotone sequence $b\_0,\dots,b\_m\in\mathbb G\_2\setminus\bigcup\_{i=0}^nZ\_i$ such that $P\subset \mathbb R\times [b\_0,b\_m]$ and $|b\_i-b\_{i-1}|<\frac12\varepsilon$ for all positive $i\le m$.


For every $(i,j)\in\{1,\dots,n\}\times\{1,\dots,m\}$ consider the cube $Q\_{i,j}:=[a\_{i-1},a\_i]\times[b\_{j-1},b\_j]$ and its boundary $\partial Q\_{i,j}$ in the plane. The choice of the points $a\_{i-1},a\_i,b\_{j-1},b\_j$ guarantees that the intersection $P\cap\partial Q\_{i,j}$ is zero-dimensional and does not contain the vertices of the cube $Q\_{i,j}$.


Let $V:=\bigcup\_{i=0}^n\{a\_i\}\times [b\_0,b\_m]$ and $H:=\bigcup\_{j=0}^m[a\_0,a\_n]\times\{b\_j\}$ be the unions of vertical and horizontal sides of the cubes $Q\_{i,j}$. Let $\mathcal C$ be the family of connected components of the symmteric difference $\Xi:=(V\cup H)\setminus (V\cap H)$. Each component $C\in\mathcal C$ is a vertical or horizontal interval of length $<\frac12\varepsilon$. By the regularity of the Lebesgue measure on the real line, there is a disjoint finite family $\mathcal I$ of compact connected subsets of $\Xi\cup\{x,y\}$ such that


$\bullet$ $\{x\},\{y\}\in\mathcal I$;


$\bullet$ $P\cap\Xi\subset \bigcup\mathcal I$;


$\bullet$ $\lambda(\bigcup\mathcal I\setminus P)<\varepsilon$;


$\bullet$ each set $I\in\mathcal I$ meets the set $P$.


Now consider the finite graph $\Gamma$ whose set of vertices coincides with $\mathcal I$ and two distinct intervals $I,J\in\mathcal J$ form an edge of the graph $\Gamma$ if there exists a square $Q\_{i,j}$ and a connected component of $P\cap Q\_{i,j}$ that intersects both arcs $I$ and $J$. It can be shown that the graph $\Gamma$ is connected, so we can find a sequence $\{x\}=A\_0,A\_1,\dots,A\_{k+1}=\{y\}$ of pairwise distinct vertices of the graph $\Gamma$ such that for every $i\le k$ the unordered pair $\{A\_i,A\_{i+1}\}$ is an edge of $\Gamma$. The latter means that $A\_{i-1}$ and $A\_i$ are intersected by some subcontinuum $P\_i$ of $P$ that is contained in some square $Q\_{j,l}$ and hence has diameter $<\varepsilon$.


It is clear that the sequences $P\_1,\dots,P\_k$ and $A\_0,\dots,A\_{k+1}$ have the required properties.


**Remark.** Unfortunately, the proof of the Lemma essentially uses the planarity of the continuum $P$. It would be interesting to know if the problem of Anton Petrunin still has affirmative answer for continua in arbitrary (finite-dimensional) Banach spaces.


