Path-connectedness of the boundary of a set?

Question

Is the boundary of a simply connected set of the plane bounded and with non-empty interior a path-connected set?

Can I consider as counterexample the area between the x-axis and the topologist's sine curve?

If not, besides the counterexample I would also appreciate a similar (but true) result concerning to the path-connectedness of set boundaries.

Answer 1

Another easy counterexample: $\Bbb R\times[0,1]$ has two separated lines as boundary.

And the boundary of $\Bbb R^2\setminus\{\,(tn,tm)\mid n,m\in \Bbb Z, t\ge 1, (n.m)\ne 0\,\}$ even has infnitely many connected components.

But of course your example (or simply the complement of the topologists sine curve) has the advantage of having a more "interesting" reason for failed path-connectedness.

Answer 2

I don't have a similar result for you, but the interval $[0,1]$ is simply connected, but its boundary $\{0,1\}$ is not path-connected since it is discrete with more than one point.