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A continuum $X$ is called minimal if it is not a single point and is homeomorphic to all its nontrivial subcontinua. Here a trivial continuum is a single point.

What is an example of a minimal continuum not homeomorphic to the interval?

This question is motivated by the following post and its related linked questions.

Two consecutive continua

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    $\begingroup$ I think the pseudo-arc is minimal, but I don't have time to check now. $\endgroup$
    – Andreas Blass
    Commented Jan 12, 2017 at 14:42
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    $\begingroup$ @AndreasBlass: You are correct -- in fact, the pseudo-arc is homeomorphic to each of its nontrivial subcontinua. $\endgroup$
    – Will Brian
    Commented Jan 12, 2017 at 14:56
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    $\begingroup$ @AndreasBlass Thank you for this answer. $\endgroup$
    – Ali Taghavi
    Commented Jan 12, 2017 at 16:24
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    $\begingroup$ @WillBrian I thank you too for the answer. $\endgroup$
    – Ali Taghavi
    Commented Jan 12, 2017 at 16:24
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    $\begingroup$ @AliTaghavi: Since Andreas was the first to think of the pseudo-arc, I'll let him type it up as an answer if he wants to. But I looked up a reference for you if you want it: according to Wikipedia's page on the pseudo-arc, Moser first proved in 1948 that it is homeomorphic to each of its nontrivial subcontinua (the full reference is given there). $\endgroup$
    – Will Brian
    Commented Jan 12, 2017 at 16:31

2 Answers 2

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As Andreas points out in the comments, the pseudo-arc provides an example of a continuum like this. This fact was first proved by Moise in 1948, in

Edwin Moise, "An indecomposable plane continuum which is homeomorphic to each of its nondegenerate subcontinua," Transactions of the AMS 63 (1948), pp. 581-594.

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The term that's usually used is hereditarily equivalent.

As Will Brian pointed out, the pseudo-arc has this property. It is indecomposable, i.e., it is not the union of two of its proper subcontinua. G. W. Henderson showed that a hereditarily equivalent decomposable continuum is an arc (homeomorphic to $[0,1]$).

It is still a big open question as to whether the arc and the pseudo-arc are the only hereditarily equivalent metric continua.

Edit July 2020: The problem alluded to above has been solved for planar continua. The pseudo-arc is the only "minimal" (i.e. hereditarily equivalent) plane continuum other than the arc: https://doi.org/10.1016/j.aim.2020.107131.

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