Let $(X,d)$ be a compact connected metric space with the property that for any distinct points $a,b$, $X\backslash \lbrace a,b\rbrace$ is disconnected. Clearly the unit circle has this property. Is there any other example (up to isomorphism) ??

$\begingroup$ Is a circle with a twist (so $\infty$) a "compact connected metric space"? $\endgroup$– Mark HurdAug 3 '13 at 17:22

1$\begingroup$ Yes, but it is not disconnected by any two points. And, for a more difficult attempted counterexample, the one point compactification of the closure of the graph of $\sin(1/x)$ (a cousin of the so called Warsaw circle), isn't either. $\endgroup$– BS.Aug 3 '13 at 20:40
This property indeed characterizes the circle, but this is not obvious.
This was shown by R. L. Moore, according to Sam Nadler's Continuum Theory p. 156.
Added: the precise reference is [522] in this historical survey of continuum theory.


$\begingroup$ The fact that the circle is metrizable has to be used here. There are compactifications of "long lines" that also have the property of being disconnected by deletion of any two points. But they're not metrizable. $\endgroup$ Aug 6 '13 at 4:17
Yes and this is a nontrivial result, see e.g. J. G. Hocking, G. S. Young, "Topology", page 55, theorem 228. See also their theorem 227 for topological characterization of the interval.

$\begingroup$ @BS: Thanks for the interesting reference. The question is also related to the old PhragmenBrouwer property (PHB) of a space $X$: This is that $X$ is connected and and if $D,E$ are disjoint nonempty closed subsets of $X$ and $a,b$ are points of $X\backslash (D\cup E)$ which lie in the same component of $X \backslash D$ and in the same component of $X \backslash E$ then $a,b$ lie in the same component of $X \backslash (D \cup E)$. If $X$ does not have the PHB, and is path connected and locally path connected, then its fundamental group contains the integers as a retract! $\endgroup$ Aug 4 '13 at 16:08