about the definition of connected sets in $\mathbb R^n$ (Edited to make reference to the topological space $X$ precise.)
A subset $A$ of a topological space $X$ is connected if there are no two open subsets $O_1$ and $O_2$ of $X$ such that (a) $A \subseteq O_1 \cup O_2$, (b) $A \cap O_1 \neq \emptyset$, (c) $A \cap O_2 \neq \emptyset$,
and (d) $A \cap O_1$ and $A\cap O_2$ are disjoint.
Call the subset $A$ brown if in the definition above, we replace (d) by (d') $O_1$ and $O_2$ are disjoint.
The definitions imply that every connected set is brown.
The opposite is not true: take the topological space $X = \{1,2,3\}$, where a subset of $X$ is open if it contains 1 or if it is the empty set. The set $A = \{2,3\}$ is not connected yet it is brown.
My question is whether for sets in the Euclidean space $X = \mathbb R^n$ (with the standard Euclidean topology), connectedness and being brown are equivalent.
 A: In at least metric space, any brown set is connected.
Let $A$ be not connected. Let $O_1$ and $O_2$ be open in $X$ sets such that $O_1 \cap A \neq \varnothing$, $O_2 \cap A \neq \varnothing$, $A = (O_1 \cap A) \sqcup (O_2 \cap A)$.
For $x \in A \cap O_1$ let $f(x)$ be such that $B_{f(x)}(x) \subset O_1$, for $x \in A \cap O_2$ let $f(x)$ be such that $B_{f(x)}(x) \subset O_2$.
Then $O_1' = \cup_{x \in A \cap O_1} B_{f(x) / 2}(x)$ and $O_2' = \cup_{x \in A \cap O_2} B_{f(x) / 2}(x)$ are disjoint and form cover of $A$, thus $A$ is not brown.
To prove disjointness, assume $p \in O_1' \cap O_2'$. Then for some $x \in A \cap O_1$ and $y \in A \cap O_2$ we have $\rho(x, p) < f(x) / 2$ and $\rho(y, p) < f(y) / 2$. Wlog assume $f(x) < f(y)$, then $\rho(x, y) < f(x) / 2 + f(y) / 2 < f(y)$, but then $x \in B_{f(y)}(y) \subset O_2$ and thus $x \in A \cap O_2$ also, which contradicts disjointness of $A \cap O_1$ and $A \cap O_2$.
I think some separation axiom should be also enough for this, but not sure.
