# Where is a non-empty open set connected iff path-connected?

I would to know how can the following claim can be generalized:

Theorem. A nonempty open set in the plane is connected if and only if any two of its points can be joined by a polygon which lies in the set.

An example of generalisation would be plane$$=\mathbb{R}^2$$ into $$\mathbb{R}^n$$ into metric space (or some special type of metric space).

"joined by polygon", I assume, could be generalised to "path-connected" (meaning a continuous map $$[0,1]\to X$$).

• Path connected implies connected as all that can be show. Feb 15, 2020 at 17:21
• @WilliamElliot If I understand correctly you are saying that "Every path-connected space is connected" (in any topological space), thus any path-connected set is connected. Therefore the only interesting part of the claim in the Q is the other direction. Feb 15, 2020 at 17:25
• A classic fact is that in a locally path-connected space, an open subset of $X$ is path-connected iff it is connected. This generalises the $\Bbb R^n$ and other locally convex spaces situation, where we can replace path-connectedness even by polygonally connected (which is handy for integrals in complex analysis e.g.). Feb 15, 2020 at 22:43
According to Exersice 6.3.10 from “General Topology” by Ryszard Engelking (2nd ed., Heldermann, Berlin, 1989), a topological space $$X$$ is locally pathwise connected if for every $$x\in X$$ and any neighbourhood $$U$$ of the point $$x$$ there exists a neighbourhood $$V$$ of $$x$$ such that for any $$y\in V$$ there exists a continuous mapping $$f: [0,1]\to U$$ satisfying $$f(0)=x$$ and $$f(1)=y$$. In the exercise is proposed the following. Verify that every locally pathwise connected space is locally connected and give an example of a locally connected subspace of the plane which is not locally pathwise connected (cf. Problem 6.3.11). Show that every connected and locally pathwise connected space [and so every open connected subspace of a locally pathwise connected space AR.] is pathwise connected.