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 $1 \in \overline{A} \cap \overline{B}$.

  • $\begingroup$ but $\overline{A}\bigcap B \not= \emptyset$ $\endgroup$ – Joel Pereira Jan 2 at 5:21
  • $\begingroup$ @JoelPereira how so? $\overline{A}=[0,1]$. What belongs to this as well as $(1,2)$, which consists of things strictly bigger than $1$? $\endgroup$ – Randall Jan 2 at 5:22
  • $\begingroup$ oh i read it as "A complement" as opposed to cl(A). my fault. $\endgroup$ – Joel Pereira Jan 2 at 5:24

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