Does there exist a connected metric space $X$(with more than one point) such that for every connected subset $A$ (with more than one point) of $X$ , $X \setminus A$ is nowhere dense in $X$, i.e. $ \operatorname{int}\left( \overline {X \setminus A}\right)=\emptyset$ ?

  • $\begingroup$ Well $A$ could be a point. Or even empty... $\endgroup$ Dec 21 '15 at 12:54
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    $\begingroup$ @DustanLevenstein : Those are not considered ; any connected set is considered with more than one point $\endgroup$
    – user228169
    Dec 21 '15 at 12:56
  • $\begingroup$ Then my guess is still no, but I don't know how to prove it. $\endgroup$ Dec 21 '15 at 13:00
  • $\begingroup$ @DustanLevenstein : thank you for your comment ; please feel free to give any kind of input regarding the question . I would really like to see a proof or counterexample of the statement . Thanks $\endgroup$
    – user228169
    Dec 21 '15 at 13:05

There seem to be such a set, but it has to have some properties.

The requirements on such a set is that connected sets has to be either "large" or trivial. Let $D(T)$ be the diameter of a non-empty set in $X$ (which can be infinite) if theres a connected set $A$ such that $1<D(A)<D(X)$ then there will be a point $b\in X$ such that $d(b,X)>0$ and therefore there's an open set surrounding $b$ not intersecting $A$ which means that $X\setminus A$ is not nowhere dense.

If for example $X$ is somewhere weakly locally connected we could construct such a set $A$. If the criterion for weakly locally connected is fulfilled for $a$ then we can just form a open ball with radius $D(X)/3$ which will contain a connected subset $A$ and $D(A)\le 2D(X)/3$.

  • $\begingroup$ " If $B_r(a)$ is connected , then it would only contain $a$" ; why is that ? $\endgroup$
    – user228169
    Dec 21 '15 at 13:49
  • $\begingroup$ @user228169 I did the assumption that $A$ only contained one point and $A$ is the connected component of $B_r(a)$ that contains $a$. If $B_r(a)$ is connected this makes $A=B_r(a)$. The reasoning leads in both cases to $X$ being disconnected with contradicts the assumption. $\endgroup$
    – skyking
    Dec 21 '15 at 13:54
  • $\begingroup$ how do you get an open set $B$ disjoint from $A$ such that $B_r(a)=A \cup B$ ? Why is $X \setminus B_{r/2}(a)$ open ? $\endgroup$
    – user228169
    Dec 21 '15 at 14:08
  • $\begingroup$ @user228169 $B$ is disjoint from $A$ because that's under the assumption that $B_r(a)$ is not connected (then you can write it as disjoint union of two open sets). It should be $X\setminus C_{r/2}(a)$ that is after removing a closed ball from $X$ (I've corrected that), it's open then by the triangle inequality. $\endgroup$
    – skyking
    Dec 21 '15 at 14:21
  • $\begingroup$ But is $A$ open ? I think $A$ is closed that's for sure ; so how do you get $B$ to be open ? $\endgroup$
    – user228169
    Dec 21 '15 at 14:28

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