# Decomposition with given closures [closed]

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?

• @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. Apr 29, 2019 at 17:40
• @Ramino de la Vega Why not make this an answer so that the question can be marked as answered? Apr 29, 2019 at 18:03