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Define a family $\{C_i\}_{i\in I}$ of continua, that is compact connected metrizable spaces, to be hereditarily unembeddable (a name I just made up) iff for all $i\neq j$ no nontrivial subcontinuum of $C_i$ embeds in $C_j$.

Is there an uncountable family of continua which is hereditarily unembeddable? Is there such a family made of planar continua?

Embarrassingly I can only construct such families with three (maybe four) elements!

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  • $\begingroup$ I can see how to get a family with two (the arc and the pseudo-arc). How do you get three? $\endgroup$
    – James Hanson
    Mar 17, 2022 at 5:56
  • $\begingroup$ @JamesHanson There are hereditarily decomposable continua that contain no arcs (and of course no pseudoarcs either), such a continuum is constructed in Exercise 2.27 of Nadler's Continuum Theory (I wrote maybe four in the question because it is not clear to me whether the continuum constructed in 2.32 of the same book can be added to this family) $\endgroup$
    – Alessandro Codenotti
    Mar 17, 2022 at 6:01
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    $\begingroup$ Consider so called Cook continuum: A nondegenerate continuum $C$ is called a Cook continuum if every continuous mapping $f$ of a subcontinuum $K$ of $C$ into $C$ is either a constant mapping or identity embedding on $K$. Using this single object (which can be constructed even in the plane), by considering disjoint subcontinua, you get a family with no nontrivial embeddings. Such a family can be easily made countable infinite using Boundary Bumping Theorem. At the moment, I dont see how to prove that there is such an uncountable family. $\endgroup$
    – Benjamin Vejnar
    Mar 17, 2022 at 6:29
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    $\begingroup$ There is a nice construction of a Cook continuum in the book by Pultr and Trnkova: Combinatorial, algebraic and topological representations of groups, semigroups and categories, page 320. Perhaps using the additional properties an uncountable collection of disjoint nondegenerate continua can be constructed. $\endgroup$
    – Benjamin Vejnar
    Mar 17, 2022 at 6:42

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