Examples for $\bar{A}\cap \bar{B}\neq\emptyset$, but $\bar{A}\cap B=A\cap\bar{B}=\emptyset$.

Are there examples for sets $$A, B\subset X$$, where $$X$$ is a topological spacce and $$A, B$$ are its nonempty subsets, satisfying $$\bar{A}\cap \bar{B}\neq\emptyset$$, but $$\bar{A}\cap B=A\cap\bar{B}=\emptyset$$.

I came up with this question when I was reading Basic Topology(M.A.Armstrong). I am trying to gain more intuitions about the difference between conditions for connectedness and separated from one another in $$X$$.

Note that a space $$X$$ is connected if whenever it is decomposed as the union $$A\cup B$$ of two nonempty subsests then $$\bar{A}\cap B\neq\emptyset$$ or $$A\cap \bar{B}\neq\emptyset$$.

And if $$A$$ and $$B$$ are subsets of a space $$X$$, and if $$\bar{A}\cap\bar{B}$$ is empty, we say that $$A$$ and $$B$$ are separated from one another in $$X$$.

In the ordinary real line, let $$A=(0,1)$$ and $$B=(1,2)$$, so that $$1 \in \overline{A} \cap \overline{B}$$.
• but $\overline{A}\bigcap B \not= \emptyset$ – Joel Pereira Jan 2 at 5:21
• @JoelPereira how so? $\overline{A}=[0,1]$. What belongs to this as well as $(1,2)$, which consists of things strictly bigger than $1$? – Randall Jan 2 at 5:22