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This question (which is more a curiosity than a research problem) originates from these two:

  1. https://math.stackexchange.com/questions/720254/is-there-a-nonempty-open-bounded-subset-of-plane-whose-boundary-contains-no-1-di

  2. complement of a totally disconnected closed set in the plane

The first question basically asks: does there exist a non-dense, open subset of $S^2$ whose boundary contains no image of injective paths? I think that this reduces to asking for an open, non-dense set whose boundary is totally path-disconnected.

If we remove the word "path", then the answer is given in question 2. Is the answer easy/known/unknown including the word "path"?

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  • $\begingroup$ What do totally path connected and totally path disconnected mean? Totally disconnected according to Engelking's General topology means for each $x$ the quasi-component (= intersection of all clopen sets containing $x$) is $\{x\}$. Please include definitions or a reference. $\endgroup$
    – Mirko
    Commented Mar 24, 2014 at 13:00
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    $\begingroup$ I don't know it there is standard terminology. I would say that A is totally path disconnected if every continuous function $[0,1]\to A$ is constant. $\endgroup$
    – user126154
    Commented Mar 24, 2014 at 13:05
  • $\begingroup$ Sorry I realized that I stated badly question 1) $\endgroup$
    – user126154
    Commented Mar 24, 2014 at 13:07

1 Answer 1

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A circular version of the pseudo-arc (where you construct it out of "circular chains" whose ends connect up to each other) is a counterexample. It is connected and totally path-disconnected, and its complement has two components. This example seems to be due to Bing (Example 2 of this paper).

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  • $\begingroup$ I am so upset ... I knew the answer but my internet connection broke :) So anyway, just give a link to a recent paper about the pseudo-circle (as defined in continuum theory), and with more current references see arXiv or Proc.AMS $\endgroup$
    – Mirko
    Commented Mar 24, 2014 at 15:06
  • $\begingroup$ @Eric Wofsey you say that the pseudo arc is totally path-disconnected. Is it true that constant functions form $[0,1]$ to the pseudo-arc are constant? It follows form the fact that it is ereditary indecomposable, and that the continuous image is a decomposable continuum? $\endgroup$
    – user126154
    Commented Jul 11, 2022 at 16:20
  • $\begingroup$ @user126154: That's correct. $\endgroup$
    – Eric Wofsey
    Commented Jul 11, 2022 at 17:04

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