# Is the boundary of an arbitrary bounded simply-connected open set in $\mathbb{R}^2$ the image of a Jordan curve?

I am trying to work out what the prerequisites are to the theorem that surfaces can always be triangulated. The key theorem seems to be Schoenflies' theorem, which depends on Caratheodory's theorem.

The Wikipedia entry for Caratheodory's theorem states that if a map $f$ maps the unit disc conformally onto a a bounded open set $U$, then $f$ has a continuous injective extension to the closed disc if and only if $\partial U$ is a Jordan curve.

At least, that's my interpretation. I'm fairly convinced the article has a (repeated) typo, since it says the unit disk $D$ is what must have $\partial D$ be Jordan.... but discs trivially have such boundaries.

My question is then, have I misunderstood the conditions? Or is there an example of a bounded, non-empty, simply-connected, open set whose boundary is not the image of a Jordan curve.

Or, if I have misunderstood something, and in fact every bounded, non-empty simply-connected open does have such a boundary, how would you go about proving it?

No, there are open, bounded, simply connected subsets $U$ of $\mathbb{R}^2$ such that $\partial U$ is not a Jordan curve. Take for example $$U = (-1,1)^2 \setminus \bigl( [0,1] \times \{0\} \bigr),$$ roughly speaking an open square with a segment removed. Then its boundary is $$\partial U = \partial \bigl( [0,1]^2 \bigr) \cup \bigl( [0,1] \times \{0\} \bigr),$$ i.e. the boundary of the square together with a small "hair". This is not the image of a Jordan curve, because it's not homeomorphic to $S^1$.
• I think do you mean "$\partial U=\partial\bigl(\boldsymbol{[-1,1]^2}\bigr)\cup\bigl([0,1]\times\{0\}\bigr)$". Apr 15, 2017 at 6:21
• Hi, thanks for your example. I have not studied homeomorphisms, I only know the definition. I see intuitively that you probably can't continuously deform the boundary in your example into $S^1,$ but is there a more solid way of thinking about it?