# 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.

• (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.) Dec 22, 2022 at 16:13
• 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 $O_1'\cap O_2'=\emptyset.$ Dec 22, 2022 at 16:21
• @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!
– Ulli
Dec 23, 2022 at 8:02
• 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?
– Ulli
Dec 23, 2022 at 8:15
• @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. Dec 23, 2022 at 20:39

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.

• Many thanks, @mihaild. Dec 22, 2022 at 17:13