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Question: Is every one dimensional peano continuum a countable union of arcs which pairwise intersect only in their end points?

Definitions:

  • Continuum - A connected, compact metric space
  • Peano continuum - A locally connected continuum
  • Arc - A continuum homeomorphic to $[0,1]$.
  • Graph - A continuum which can be written as the union of finitely many arcs which pairwise intersect only in their end points (none, one or both)
  • Arc-like continuum - A continuum for which every open cover has a refinement $\mathcal{U} = \{U_1,\cdots,U_n\}$ such that $U_i\cap U_j$ is nonempty if and only if $|i-j|\leq 1$.
  • Circle-like, tree-like etc have similar definitions.

It is obvious that not every one dimensional peano continuum is a graph. (This is because one can construct a continuum which is a countable union of arcs which pairwise intersect only in their end points.) However, I have shown that every arc-like peano continuum is an arc, and every circle-like peano continuum is a circle.

Therefore I have conjectured that every one dimensional peano continuum is essentially "a graph that may have countably many edges rather than finitely many". Unfortunately I'm unsure about an approach to prove this. Any help will be appreciated!

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Some Googling finds us:

Have you considered the Menger universal curve (which is a one-dimensional Peano continuum that contains a homeomorphic copy of all one-dimensional separable metric spaces. I don't think you can easily see it as something simply "graph-like".

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The conjecture is false, such a space is clearly HLC.

See e.g. "Graph-Like Continua, Augmenting Arcs and Menger's Theorem."

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