Let $U \subseteq \mathbb{R}^n$.
We say that $U$ is path-connected if for every $x, y \in U$ there exists a path that connects them; i.e, if there exists a continuous function $\gamma:[0,1]\rightarrow U$ so that $\gamma (0)=x, \gamma (0)=y$. (Intuitively, any two points in the set can be connected through the set).
We sat that $U$ is set-connected if there are no open sets $A, B\subseteq \mathbb{R}^n$ so that: $U\subseteq A\cup B, \; U\cap A \cap B = \varnothing, \; U\cap A \neq \varnothing, \; U\cap B \neq \varnothing $ (intuitively, these sets "seperate" $U$ to non-empty disjoint sets)
I need to show that these to defintions are equavalent. I've shown that path-connectivity implies set-connectivity, so I need the other direction.
My idea was to try to divide $U$ into path-connected-components and show that there can be only one, but how can I show that?
I'd like to hear your suggestions. Thank you!
Hint: Topologist's sine curve.
2nd Hint: If $U$ is open they are equivalent, consider the points which are path-connected to some fixed point and the ones that are not.