The question was motivated by this question 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, 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 $p<\infty$?

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    $\begingroup$ An easy observation is that if we have an example of a non-almost $\ell_p$-path connected continuum for some $p$, then we can turn it into one which is not almost $\ell_q$-path connected for every $q<\infty$. Specifically if we take a convex function that is 'sharp enough' at 0 and compose it with the metric we'll get the required counterexample. So for instance letting the new metric be $\rho = \sum_{n>0} 2^{-n} d^{1/n}$ (after scaling the diameter of $X$ to be $1$) will work. $\endgroup$ – James Hanson Aug 14 '18 at 16:31
  • $\begingroup$ @JamesHanson Very good! Thanks. So, it remains to consider some specific metrics like the metrics inherited from the Euclidean spaces. $\endgroup$ – Taras Banakh Aug 14 '18 at 17:34

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.

  • $\begingroup$ Great! This example opens a door to considering $\ell_p$-connected components, $\ell_p$-metrics on the space of such components etc. $\endgroup$ – Taras Banakh Aug 16 '18 at 20:13
  • $\begingroup$ Can such a pseudoarc be realized in some Euclidean space $\mathbb R^n$? I means a pseudoarc which is not $\ell_1$-connected (since any continuum in $\mathbb R^n$ is $\ell_p$-connected for $p>n$). $\endgroup$ – Taras Banakh Aug 16 '18 at 20:17
  • $\begingroup$ @TarasBanakh I also would like to know if it can be realized. By the way, you might be interested in the following geometric reformulation: Given a compact set $P\subset\mathbb{R}^k$, denote by $\mathbb{R}^k/P$ the quotient space with quotient metric. If there is a pseudoarc $P$ that is not $\ell_1$-almost path connected, then the length-metric induced by the map $ \mathbb{R}^k\to \mathbb{R}^k/P$ coinsides with the original metric on $\mathbb{R}^k$. $\endgroup$ – Anton Petrunin Aug 16 '18 at 20:46
  • $\begingroup$ I am afraid that the situation with reformulation is not so optimistic. The $\ell_1$-almost path-connectedness of $P$ is equivalent to the triviality on $P$ of the length-metric induced by the quotient map.p $\endgroup$ – Taras Banakh Aug 16 '18 at 21:34
  • $\begingroup$ @TarasBanakh Right, but what do you mean by "optimistic" (?) I simply want to understand the induced length-metric. According to your argument, in dimesion 2 the length metric for the quotient map is isometric to the quotient (assuming that the inverse images are connected +...). In dimension 3 it is unknown, and I want to know. $\endgroup$ – Anton Petrunin Aug 16 '18 at 23:20

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