Let $X$ be a topological space and $A$ be a disconnected subspace of $X$. Is there an example such that there does not exist a pair of nonempty open subsets $U,V$ in $X$ satisfying $U\cap V=\emptyset$ and $(U\cup V)\cap A=A$?

I was marking general topology homework and I found one student assume the existence of such a pair $(U,V)$, but I think this is false in general. (It should be assumed that $U\cap V\cap A=\emptyset$ instead of $U\cap V=\emptyset$ to gurantee the existence of such $(U,V)$). However, I am having trouble finding a counterexample for this.


Let $U,V$ be two topological spaces and consider $X=U\cup V\cup\{p\}$ with a basis of open sets given by $O\cup\{p\}$ with $O$ open either in $U$ or $V$ and the singleton $\{p\}$. Then $A=U\cup V\subset X$ is a counterexample.

  • $\begingroup$ What's the corresponding pair of subsets of $X$? Is it $U$ and $V \cup \{ p \}$? $\endgroup$ – Ian Oct 16 '17 at 16:56
  • $\begingroup$ @Ian W Not really. The idea is that in any way you might try to cover $U\cup V$ by two non-empty open sets, you will have to include the added $p$ to both open sets (by the definition of open sets in $X$, and therefore the intersection of the two sets cannot be empy. $\endgroup$ – Daniel Robert-Nicoud Oct 16 '17 at 20:15

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