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Let $(X,\mathcal T)$ be a topological space. About the subsets $A,B,C$ of $X$ it is known that $$\mathrm {cl} (C)= A \cup B\,, \quad \mathrm {cl} (A) = A\,, \quad \mathrm {cl} (B) = B\,.$$

Does it always imply that there exist two subsets of $X$, $C_A$ and $C_B$ such that $$C = C_A \cup C_B\,, \quad \mathrm {cl} (C_A) = A\,, \quad \mathrm {cl} (C_B) = B \; ?$$

If this is not true in general, is it true at least for Euclidean spaces?

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    $\begingroup$ @JayKangel, that only works if $C$ is closed. If $C$ is not closed and $p \in \mathrm{cl}(C) \setminus C$ then $A=\mathrm{cl}(C)$ and $B=\{p\}$ is always a counterexample. $\endgroup$
    – Ramiro de la Vega
    Apr 29, 2019 at 17:40
  • $\begingroup$ @Ramino de la Vega Why not make this an answer so that the question can be marked as answered? $\endgroup$
    – Mark Fischler
    Apr 29, 2019 at 18:03

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