TITLE: Must a closed totally path-disconnected subset of the sphere have connected complement?
QUESTION [8 upvotes]: This question (which is more a curiosity than a research problem) originates from these two:

https://math.stackexchange.com/questions/720254/is-there-a-nonempty-open-bounded-subset-of-plane-whose-boundary-contains-no-1-di
complement of a totally disconnected closed set in the plane

The first question basically asks: does there exist a non-dense, open subset of $S^2$ whose boundary contains no image of injective paths? I think that this reduces to asking for an open, non-dense set whose boundary is totally path-disconnected. 
If we remove the word "path", then the answer is given in question 2. Is the answer easy/known/unknown including the word "path"?

REPLY [10 votes]: A circular version of the pseudo-arc (where you construct it out of "circular chains" whose ends connect up to each other) is a counterexample.  It is connected and totally path-disconnected, and its complement has two components.  This example seems to be due to Bing (Example 2 of this paper).