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Motivated by this complex dynamics question:

Let $X$ be a compact, path-connected metric space. Suppose there exist an integer $N\geq 2$ and distinct points $p_1,\dots,p_N\in X$ such that no proper path-connected subset of $X$ contains all $p_i$'s. Then, is it true that $X$ should be homeomorphic to a tree with at most $N$ leaves?

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  • $\begingroup$ With exactly $N$ leaves, no: your condition holds for three points on a line. $\endgroup$
    – AGenevois
    Commented May 14 at 4:48
  • $\begingroup$ @AGenevois Thanks, I changed to at most $N$ leaves. $\endgroup$
    – KhashF
    Commented May 14 at 4:49

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The answer seems to be yes. Just take a path between $p_1$ and $p_2$, and then, on each next $i$th step ($i\geq 3$), add a path from $p_i$ to already constructed space (that is, you take a path from $p_i$ to $p_1$, and cut it after it hits the already constructed part). At every moment, you get a tree; and, when you finish, you get a path-connected space which is a tree and contains all the $p_i$. So you should get $X$, Moreover, al the leavers are among the $p_i$.

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  • $\begingroup$ I wonder if you have a rigorous proof. Even in the case of $N=2$, how to show that there is only one path between $p_1,p_2$ and that path is homeomorphic to an interval? $\endgroup$
    – KhashF
    Commented May 14 at 12:56
  • $\begingroup$ You don’t need uniqueness; it will follow at the end, from your condition. You can take any simple path (i.e., homeomorphic to an interval, yes), and it works. I assume the existence of such path is standard… $\endgroup$
    – Ilya Bogdanov
    Commented May 14 at 22:01
  • $\begingroup$ Yes, you are right. There is a classical result indicating that a Hausdorff path-connected space is arcwise connected. Would you mind adding the fact that the paths can taken to be topological embeddings of $[0,1]$ to your answer? Then, I will accept it. $\endgroup$
    – KhashF
    Commented May 18 at 18:40

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