When is the inside of a Jordan curve open? [closed]

I'm working purely on intuition here. The Jordan curve theorem states that a Jordan curve separates the plane into a bounded component and an infinite component. For toy curves, it seems like this bounded component is always open. But in pathological cases, like an Osgood curve which has positive measure, clearly the inside cannot be open since it does not contain an open ball (I think).

Are there examples of Jordan curves with measure $$0$$ that don't have an open inside? Do Jordan curves with positive measure never have an open inside? More importantly, if the inside is open, does it guarantee that the curve is "non-pathological"?

EDIT: Perhaps my intuition was wrong. According to MO user Timothy Chow in this post, "The Jordan curve theorem was strengthened by Schoenflies to the statement that the two components are homeomorphic to the inside and outside of a circle." By Brouwer's invariance of domain theorem, this implies that the inside component of a Jordan curve is open, if I understand everything correctly.

• the complement to a jordan curve is open. every component of an open set in the plane is open
– erz
Jun 29 '20 at 6:11
• The circle $\mathbb{S}^1$ is compact, and so is any continuous image of it in the plane, hence closed. So the complement of the image is open. Connected components of an open subset of a locally connected space are open (voting to close) Jun 29 '20 at 6:47
• Of course, I wasn't thinking. Thanks for clarifying this anyway. Jun 29 '20 at 6:51

In the linked example, the Osgood curve shown is actually an "Osgood arc". You should take four copies (say, rotate the given arc through multiples of $$\pi/2$$) and glue them end to end. Then you will have a Jordan curve with an inside and an outside. Both will be open.
• Complement.${}$ Jun 29 '20 at 6:55