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Let $X$ be a topological space. Let $A,C\subseteq X$. Then, is it true that $$(C\cap \mathrm{Cl}(A))\cup(C\cap \mathrm{Cl}(X\setminus A))=C\cup (C\cap\partial A)$$

I've shown (if I'm not wrong) that $$C\cap \mathrm{Cl}(X\setminus A)=C\setminus \mathrm{Int}(A)$$ but I don't know what else to do. Any help would be appreciated.

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    $\begingroup$ Note that any set $A\subset X$ partitions $X$ into three sets: the interior points of $A$, the boundary points of $A$, and the exterior points of $A$. The closure is interior + boundary. This idea helps for a lot of identities involving interiors, boundaries and closures. $\endgroup$ – Christoph Jan 21 at 11:49
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By the distributive law: $$(C\cap \overline{A}) \cup (C \cap \overline{(X\setminus A)})= C \cap (\overline{A}\cup \overline{X\setminus A})= C \cap X = C$$

Your statement is not wrong since $C \cap \delta A$ is a subset of C, thus $C \cup (C \cap \delta A)= C$.

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  • $\begingroup$ My pedantic style would be $ C\supseteq (C\cap \overline A)\cup (C\cap \overline {(X\setminus A)}\,)$ $=C\cap (\overline A\cup \overline {(X\setminus A)}\,)\supseteq$ $ \supseteq C\cap (A\cup (X\setminus A))=$ $C\cap X=C$........+1. $\endgroup$ – DanielWainfleet Jan 21 at 20:18
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Yes, correct on both. Derive a complicated formula
involving $\partial$(A $\cup$ B) and $\partial$(A $\cap$ B).

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Since $C\cap \partial A$ is a subset of $C$, the RHS of the identity if just $C$. The LHS simplifies to $$ C \cap \left( \operatorname{Cl}(A) \cup \operatorname{Cl}(X\setminus A) \right). $$ Since $\operatorname{Cl}(A) \cup \operatorname{Cl}(X\setminus A)$ contains all the points of $A$ and all the points of $X\setminus A$ it is equal to $X$. Hence the LHS is $C\cap X = C$ and agrees with the RHS.

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