# Disconnecting sets

If E is a metric space, I call a subset C of E a cut if E-C is not connected and if C is minimal for this property (which is obviously equivalent to "for every p in C, E-C union p is connected". The empty set is a cut of any disconnected space, so the obvious conjecture is : every space of more than one point possesses a cut (this is almost obviously true for manifolds, for instance)... but this conjecture is in fact probably wrong, as it is not obvious at all to find cuts in convoluted spaces like, for instance, the common boundary of three open sets in the plane ; on the other hand, i was not able to prove it in this case either. Any hint on such a proof ? (by the way, Zorn's lemma dont apply here ; I wonder is there is any interesting example of a situation where the hypothesis of the lemma fail (ie, like here, the intersection of a decreasing family of sets having the property P does not necessarily have it), but the conclusion (the existence of a minimal such set) is still valid ?

• With this definition the space is always a cut of itself. Aug 23, 2011 at 3:29
• I'm not clear on your example, "the common boundary of three open sets in the plane." What is the metric space? Aug 23, 2011 at 3:30
• Maybe is the OP referring to en.wikipedia.org/wiki/Lakes_of_Wada? Aug 23, 2011 at 5:28
• @Gjergji: Your remark applies only if the whole-space subset is "minimal for this property". However, the whole space does provide an element for the set in which the existence of a minimal element is sought. Aug 23, 2011 at 6:39
• On the last question: sure, whenever the union of an increasing family of sets having the property always has the property.
– user5810
Aug 23, 2011 at 6:46 Assume $E$ is what remains after the infinite number of the iterations above. This is an example of indecomposable continuum, so it can not be presented as a union of any two proper subcontinua.
Assume there is a cut $C\subset E$.
Then each connected component of $E\backslash C$ has $C$ as the boundary. In particular, $C$ is closed. Let us present $E\backslash C$ as a union of two disjoint open sets $A$ and $B$. Clearly both $\bar A =C\cup A$ and $\bar B=C\cup B$ are proper subcontinua of $E$, a contradiction.