(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.

  • $\begingroup$ (In general, we define connectedness for topological spaces, not for subspaces. But I understand, for the purposes of this question, we are defining it otherwise.) $\endgroup$ Dec 22, 2022 at 16:13
  • $\begingroup$ What have you tried? You would need to know that if $O_1,O_2$ cover $A$ with $A\cap O_1\cap O_2=\emptyset,$ then we can find $O_1',O_2'$ with $O_1'\cap O_2'=\emptyset.$ $\endgroup$ Dec 22, 2022 at 16:21
  • $\begingroup$ @Thomas Andrews: it says "$A$ is connected, if ...". Well perhaps it should be "iff", but apart from this, it is just an easy observation, where $A$ connected is to be understood in the usual sense. So, it's not a definition! $\endgroup$
    – Ulli
    Dec 23, 2022 at 8:02
  • $\begingroup$ Why did you call this property "brown"? Since it's quite an unusual name, did you find this notion somewhere? I know of "Brown spaces" (see, e.g., here, but I guess this has nothing to do with the above property? $\endgroup$
    – Ulli
    Dec 23, 2022 at 8:15
  • $\begingroup$ @Ulli: I chose "brown" at random, because I like the color. Since it is an ad hoc notion, I did not want to term it in a way that resembles connectedness, so as people do not think this definition is actually used. $\endgroup$
    – Eilon
    Dec 23, 2022 at 20:39

1 Answer 1


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.

  • $\begingroup$ Many thanks, @mihaild. $\endgroup$
    – Eilon
    Dec 22, 2022 at 17:13

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