$\forall S \subseteq \mathbb R^m, (S:\text{Connected}\land S:\text{Open in } \mathbb R^m \to S: \text{Path Connected})$
I searched on this website, and I could only find two proofs.
The first proof uses the concept of locally path connectedness, but I do not know this.
About the other proof, I cannot understand the second paragraph.
Let $x\in A$ (hence not empty); suppose $\exists \varepsilon >0 : B(x,\varepsilon)\subseteq U$. Since open balls are convex, it is path connected. Thus for any point $y$ in $B(x)$, there is a path from $x$ to $y$. Since $A$ is a set of points in $U$ that can join $a$, then there exist a path between $x$ and $a$. Since there is a path from $y$ to $x$ and $x$ to $a$; then, there is a path from $y$ to $a$ implying $y$ is in $A$. Since $y\in B(x,\varepsilon)$, we conclude $B(x) \subseteq A$, hence $A$ is open
Could you provide the easy proof?
