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Let $X$ be a metric continuum (compact + connected) which is the one-to-one continuous image of the interval $[0,\infty)$. Such an $X$ is called a linear continuum.

It seems like $X$ should be chainable (as defined in the 2nd paragraph here). Equivalently, for every $\epsilon>0$ there should be a mapping $f:X\to [0,1]$ such that the pre-image of each point in $[0,1]$ has diameter of $<\epsilon$. This property is also discussed here: Beautiful examples of arc-like continua.

Is this true, and it so why?

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  • $\begingroup$ @NateEldredge I think you must first define an $\epsilon$-chain for the interval of condensation. This can be done so that the remainder (complement of this partial chain) is just an interval which enters one member of the chain, and leaves another. Thus the chain can be extended to cover all of $X$. $\endgroup$
    – Forever Mozart
    Commented Jun 16, 2018 at 1:03

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The circle is a linear continuum according to the definition provided in the question. But the circle is not chainable, by the Borsuk-Ulam Theorem.

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  • $\begingroup$ Ok. I think I mis-understood the term chainable. According to my (incorrect) interpretation the circle was chainable. $\endgroup$
    – Forever Mozart
    Commented Jun 16, 2018 at 5:21
  • $\begingroup$ @ForeverMozart You can ask if each linear continuum is $\Gamma$-like for some fixed graph $\Gamma$ and if yes, which graphs $\Gamma$ can appear as images of linear continuous under $\epsilon$-mappings. But it seems that for any connected graph $\Gamma$ there is a linear $\Gamma$-like continuum. $\endgroup$
    – Taras Banakh
    Commented Jun 16, 2018 at 5:37
  • $\begingroup$ The Warsaw Circle is another counterexample and perhaps more in the spirit of the definition of linear continua. As well, any linear, triodic treelike continuum will do. You need to get out of the plane, I bet. $\endgroup$
    – John Samples
    Commented Jun 17, 2018 at 7:07

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