The question was motivated by [this question](https://mathoverflow.net/questions/307534/running-most-of-the-time-in-a-connected-set) of Anton Petrunin.


By a *metric continuum* we understand a connected compact metric space.


Let $p$ be a positive real number. A metric continuum $X$ is called *$\ell\_p$-almost path-connected* if for any points $x,y\in X$ and any $\varepsilon>0$ here exists a family $\big((a\_n,b\_n)\big)\_{n\in\omega}$ of pairwise disjoint open intervals in the unit segment $[0,1]$ and a continuous map $\gamma:[0,1]\setminus\bigcup\_{n\in\omega}(a\_n,b\_n)\to X$ such that $\gamma(0)=x$, $\gamma(1)=y$ and $\sum\_{n=0}^\infty d\_X(\gamma(a\_n),\gamma(b\_n))^p<\varepsilon$.


It is easy to see that each almost $\ell\_p$-connected metric continuum is $\ell\_q$-almost connected for any $q\ge p$.


By my answer to the [question of Anton Petrunin](https://mathoverflow.net/questions/307534/running-most-of-the-time-in-a-connected-set), each plane continuum is almost $\ell\_1$-connected. By analogy it can be shown that each continuum in $\mathbb R^3$ is $\ell\_2$-connected.


**Problem.** Is there a metric continuum which is not almost $\ell\_1$-path connected? not almost $\ell\_p$-connected for every $p<\infty$? 



Is every metric continuum almost path-connected?
Yes there are such examples.


Assume the sequence $\varepsilon\_n$ is very fast converging to $0$.
Consider a sequence of short $\varepsilon\_n$-crooked maps between intervals $\mathbb{J}\_n\to \mathbb{J}\_{n-1}$. 
Its inverse limit is a pseudoarc $\mathbb{J}\_\infty$; denote by $\phi\_n\colon\mathbb{J}\_\infty\to \mathbb{J}\_n$ the projections.


Equip $\mathbb{J}\_\infty$ with the maximal metric such that 
$$|x-y|\_{\mathbb{J}\_\infty}\le \tfrac1{2^{n/p}}+|\phi\_n(x)-\phi\_n(y)|\_{\mathbb{J}\_n}.$$


Assume that $\mathbb{J}\_\infty$ is $\ell\_p$-almost path-connected.
Let $\gamma$ be the path that connects the ends of $\mathbb{J}\_\infty$.
It can not have more then one jump of size $1$; the jump brakes $\alpha$ in two arcs one of which has diameter at least $1-\varepsilon\_1$;
this arc can not have more than 2 jumps of length $\tfrac12$, so one of the subarcs has diameter at least $1-\varepsilon\_1-2\cdot \varepsilon\_2$ and so on. At the end of the day you see that $\mathbb{J}\_\infty$ contains a subarc of $\alpha$ of positive diameter. But $\mathbb{J}\_\infty$ has no arcs, a contradiction.


The given construction is nearly identical to Example 4.2 in [my paper on intrinsic isometries](https://arxiv.org/abs/1003.5621).


