Suppose that $X$ is a metric continuum irreducible between two points $p$ and $q$.

Suppose further that whenever $U$ is a connected open set missing $p$ and $q$, we have $X\setminus U$ has two connected components, one containing $p$ and the other containing $q$.

Is $X$ necessarily equal to (homeomorphic to) $[0,1]$?

EDIT: I don't have much context to add. I came up with this question while trying to solve a very different problem. Honestly I have to clue if the answer is yes or no, though if the answer is yes it would be useful to me.

  • $\begingroup$ Related: math.stackexchange.com/questions/322411/… $\endgroup$ – Moishe Kohan Jul 17 '16 at 23:02
  • $\begingroup$ @studiosus thanks, let me think about why every point other than $p$ and $q$ is a cut point... $\endgroup$ – Forever Mozart Jul 18 '16 at 1:49
  • $\begingroup$ It’s shocking that there are four votes to close. $\endgroup$ – Brian M. Scott Jul 18 '16 at 6:07
  • 2
    $\begingroup$ Yes, its quite ridiculous to close this question. $\endgroup$ – Moishe Kohan Jul 19 '16 at 4:55
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    $\begingroup$ Please improve this post by adding additional context. Questions which only pose a problem are often put on hold. $\endgroup$ – Carl Mummert Jul 20 '16 at 0:49

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