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A continuum is a compact connected metric space.

The continuum $X$ is called a Peano continuum if it is locally connected.

A chain in the topological space $X$ is a collection $U_1,U_2,\ldots ,U_n$ of subsets of $X$ such that $U_i\cap U_j\neq \emptyset \iff |i-j|\leq 1$. An $\varepsilon$-chain is a chain in $X$ such that $diam(U_i)<\varepsilon, \forall i\in \{ 1,2,\ldots ,n \}$. The space $X$ is a chain continuum if for any $\varepsilon >0$, there exists and $\varepsilon$-chain open cover of $X$.

There are Peano continua that are not chain continua (triode) and chain continua that are not Peano continua (closure of a topological sine curve). A segment is both Peano and chain continuum.

My question is: Must every continuum be either Peano or chain? If not, is there a classification of all continua?

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    $\begingroup$ Just stick a segment on the other end of the closed topologist's sine curve, giving both modes of failure in the same space. $\endgroup$ Commented Oct 24, 2021 at 12:05
  • $\begingroup$ @DavidHartley nice example! So, there exist continua that are not chain nor Peano. I wonder if there is a special classification for such continua? $\endgroup$ Commented Oct 24, 2021 at 12:24
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    $\begingroup$ there are classes of continua with nice properties (dendroids, dendrites, tree-like, planar etc.) but there are many continua, a classification seems hopeless $\endgroup$ Commented Oct 24, 2021 at 12:38
  • $\begingroup$ @DavidHartley can you provide us with a reference for this example? Thanks in advance! $\endgroup$ Commented Jan 11, 2022 at 16:22
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    $\begingroup$ I have no reference. You state yourself that the closed topologist's sine curve is not locally connected and the triode is not chain-connected. It easily follows that a continuum that contains both - nicely separated from each other - cannot have either property. (By the way, a Peano continuum is characterised by locally connected, not locally compact.] $\endgroup$ Commented Jan 11, 2022 at 19:50

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