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I'm trying to show that if $\overline A\cap B = A\cap \overline B = \varnothing$, $A\cup B$ is disconnected. First of all, I think I have to assume that $A$ and $B$ are nonempty, or else the statement would not be true if I just let $A = \varnothing$ and let $B$ be a connected set.

I'm working in a metric space where the definition of a set $S$ being open is that $\forall x\in S$, $\exists \varepsilon > 0: B(x,\varepsilon)\subseteq S$, and the definition of a set $T$ being closed is that $T$ is the complement of an open set.

So assuming $A$ and $B$ are nonempty and $\overline A\cap B = A\cap \overline B = \varnothing$, to demonstrate that $S = A\cup B$ is disconnected I must find open sets $U_1$ and $U_2$ such that

  • $U_1\cap U_2 = \varnothing$
  • $S = (S\cap U_1) \cup (S\cap U_2)$
  • $S\cap U_1\neq \varnothing$ and $S\cap U_2\neq \varnothing$.

At first I thought of taking $U_1 = (\overline A)^c$ and $U_2 = (\overline B)^c$, but I don't necessarily know that these are disjoint. My second thought was this: Let $x\in A$. Then since $x\in (\overline B)^c$ and this set is open, $\exists \varepsilon_x > 0: B(x,\varepsilon_x)\subseteq (\overline B)^c$. Then I wanted to define $U_1 = \cup_{x\in A} B(x,\varepsilon_x)$ so that $A\subseteq U_1 \subseteq (\overline B)^c$, and then similarly define $U_2$ so that $B\subseteq U_2\subseteq (\overline A)^c$. However, I still don't think this works, since $U_1$ and $U_2$ could just share a point that is neither in $A$ nor $B$, but is in $(\overline A)^c \cap (\overline B)^c$. Any suggestions?

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\begin{align*} U_1 &= \{x : \mathop{\text{dist}}(x,A) < \mathop{\text{dist}}(x,B) \} \\ U_2 &= \{x : \mathop{\text{dist}}(x,A) > \mathop{\text{dist}}(x,B) \} \end{align*}

(The use of this kind of distance trick is suggested by the fact that the result is not true in general topological spaces; a nice counterexample is the co-finite topology on $\mathbb N$, which produces exactly the problem you were worried about where $U_1$ and $U_2$ end up having to share points.)

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  • $\begingroup$ Okay... if $x\in A\cup B$, then $x\in A$ or $x\in B$ (but not both). If $x\in A$, $\text{dist}(x,A) = 0$ and $\text{dist}(x,B) > 0$ since $x\notin \overline B$, so $x\in U_1$. Likewise, if $x\in B$, $x\in U_2$. So doesn't this just imply that $U_1 = A$ and $U_2 = B$? $\endgroup$ – justin May 21 '15 at 2:42
  • $\begingroup$ Urk, yes, sorry. I'll fix. $\endgroup$ – user21467 May 21 '15 at 10:49
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First, note that a separation on a subspace of a topological space consists of two non-empty, disjoint sets that are open relative to the subspace. Here, you need two non-empty, disjoint sets that are open in $A \cup B $. It is an equivalent problem to prove the existence of two non-empty disjoint sets that are closed relative to $A \cup B$. The sets $\overline{A} \cap (A \cup B)$ (which equals A) and $\overline{B} \cap (A \cup B)$ (which equals B) satisfy this second condition.

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  • $\begingroup$ Thank you! This works with the definition of being disconnected in a general topological space, where we simply require that $S\cap U_1\cap U_2 = \varnothing$. However, and I should have been more clear about this, I was trying to satisfy the additional condition that $U_1\cap U_2 = \varnothing$. $\endgroup$ – justin May 22 '15 at 1:46

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