Suppose that we have a connected $T_1$ space $X$ and two disjoint, closed subsets $A, B$. Then if $X$ is also locally connected, it is a theorem that there is a connected subset $C \subset X \setminus (A \cup B)$ such that the closure of $C$ intersects both $A$ and $B$.

I am wondering if something stronger is true. Suppose that we are in a connected, locally connected set as before, and $D$ is an arbitrary, connected subset of $X$ intersecting both $A$ and $B$. Is it true that $D$ contains a connected subset $E$ also intersecting $A$ and $B$ such that its intersection with $X \setminus (A \cup B)$ is connected?

An instructive case is where $D$ is a circle and $A, B$ are two of its points. Then the complement of these two points in $D$ is not connected (it is two arcs), but we can pick one of them as our $E$.

Note that by the cited theorem, if we could always find a locally connected subcontinuum contained in $D$ intersecting both then we would be done. But this is not true, since a 'two-sided' topologist's sine curve can connect distinct points in the (connected, locally connected) unit disc.

I am also wondering the same thing, but for 'connected, locally connected' replaced by 'connected, compact, metric.'

The theorem and my question are not true in general spaces. For example, take two copies of the Knaster-Kuratowski Fan (a classic example of a connected metric space with a point whose removal leaves a totally disconnected space) attached along the ends of the 'intervals'. if we let $A$ and $B$ be the two explosion points we get a bad-as-possible counterexample.


My intuition is that it is true in the compact metric case, but I am very unsure in the lc case. It might also be true in the compact Hausdorff setting; it seems like a statement that will either be true for all of them or false for all of them to me.

  • $\begingroup$ Couldyou please give a reference for the theorem that you quote in the beginning? $\endgroup$ – erz Jun 27 '18 at 7:08
  • $\begingroup$ It's called the Cut Wire Theorem. The lc version is in Wilder, the continuum version in Nadler. $\endgroup$ – John Samples Jul 11 '18 at 23:58
  • $\begingroup$ Still wondering about this one. $\endgroup$ – John Samples Aug 10 '18 at 1:28
  • $\begingroup$ I do not understand the role of $X$ in the desired stronger result. Set $A' = A \cap D, B' = B \cap D$. Then $A', B'$ are nonempty disjoint closed subsets of $D$. To find a connected $E \subset D$ such that $E \cap (X \setminus (A \cup B))$ is connected is the same as to find a connected $E \subset D$ intersecting $A',B'$ such that $E \cap (D \setminus (A' \cup B'))$ is connected. This is a statement about a more or less arbitrary connected $D$, since local connectivity of $X$ does not imply that of $D$. $\endgroup$ – Paul Frost Oct 12 '18 at 8:23

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