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

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

• See the Knuster-Kuratowski fan – Teri Dec 21 '15 at 15:44
• @Teri : How does that help ? – user228168 Jan 3 '16 at 8:31
• 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 5:24
• @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 – user228168 Jan 10 '16 at 5:38
• @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. – 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.
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