# Is this interior-closure identity true? $(C\cap \mathrm{Cl}(A))\cup(C\cap \mathrm{Cl}(X\setminus A))=C\cup (C\cap\partial A)$

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.

• 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. – Christoph Jan 21 at 11:49

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 $$C \cup (C \cap \delta A)= C$$.
• 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. – DanielWainfleet Jan 21 at 20:18
involving $$\partial$$(A $$\cup$$ B) and $$\partial$$(A $$\cap$$ B).
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.