8

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

Najib Idrissi
  • 54,185
  • 1
    See the Knuster-Kuratowski fan – Teri Dec 21 '15 at 15:44
  • 2
    @Teri : How does that help ? –  Jan 03 '16 at 08:31
  • 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. – Forever Mozart Jan 10 '16 at 05:24
  • @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 –  Jan 10 '16 at 05:38
  • @SaunDev See my question (and answer) here: http://math.stackexchange.com/questions/1572687/compact-connected-space-is-the-union-of-two-disjoint-connected-sets . 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. – Forever Mozart Jan 10 '16 at 05:59

1 Answers1

4

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.

orangeskid
  • 53,909
  • Okay , so that answers the 2nd question ; what about the first one ? –  Dec 21 '15 at 13:12
  • 1
    @Saun Dev: This post http://math.stackexchange.com/questions/796635/every-metrizable-toronto-space-is-discrete seems to be related to your question – orangeskid Dec 24 '15 at 15:46