Let $X$ be a topological space. Let $A$ and $B$ be sets in $X$ such that the intersection of $A$ and $B$ is empty. Suppose the union of $A$ and $B$ is open. Does it follow that both $A$ and $B$ are open?
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No. For example, let $X$ be $\mathbb{R}$ with the usual topology, $A$ be $(0,1]$, and $B$ be $(1,2)$. |
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Not even in the reals. Let $A$ be the rationals and $B$ the irrationals. There are many other examples. In general, $A$ and $B$ can be more or less arbitrarily badly (or well) behaved. |
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Expanding on the nice examples above… If $X$ is a topological space and if $A$ and $B$ are non empty subsets of $X$ with empty intersection, such that $A\cup B\supset C$ for some connected open subset $C$ of $X$. Assume also that both $A$ and $B$ intersects C. Then clearly $A$ and $B$ cannot be open, since that would imply that $C$ is not connected. |
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