Does every non-singleton connected metric space $X$ contains a connected subset (with more than one point) which is not homeomorphic with $X$ ?

Also ; does every connected metric space $X$ contains a connected subset which is homeomorphic with $X$ ?

UPDATE : So as noticed by @orangeskid ; the answer to the 2nd question is "no" by considering $X=S^1$ . The first question still remains unanswered

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    $\begingroup$ See the Knuster-Kuratowski fan $\endgroup$ – Teri Dec 21 '15 at 15:44
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    $\begingroup$ @Teri : How does that help ? $\endgroup$ – user228168 Jan 3 '16 at 8:31
  • $\begingroup$ interesting question. I can show it is true for compact Hausdorff spaces. my intuition says it should be true. I am working on a proof by cases. $\endgroup$ – Forever Mozart Jan 10 '16 at 5:24
  • $\begingroup$ @ForeverMozart : You can show it for compact connected Hausdorff spaces ? Could you please take the trouble to write that out in comment or as an answer ; It would be very very helpful . Thanks in advance $\endgroup$ – user228168 Jan 10 '16 at 5:38
  • $\begingroup$ @SaunDev See my question (and answer) here: math.stackexchange.com/questions/1572687/… . Every cpt Haus. space is the union of two disjoint connected subsets. At least one must be non-compact, thus not homeomorphic to the entire space. $\endgroup$ – Forever Mozart Jan 10 '16 at 5:59

Take $X$ a $1$-dimensional circle. $X$ does not contain any proper subspaces homeomorphic to a circle, since any connected proper subspace is a segment.

${\bf Added:}$

The answer to the first question is yes for spaces that contain a segment. It seems a lot of connected metric spaces contain a segment.

  • $\begingroup$ Okay , so that answers the 2nd question ; what about the first one ? $\endgroup$ – user228168 Dec 21 '15 at 13:12
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    $\begingroup$ @Saun Dev: This post math.stackexchange.com/questions/796635/… seems to be related to your question $\endgroup$ – orangeskid Dec 24 '15 at 15:46

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