$IC(U\cap V)$ VS $IC(U)\cap IC(V)$ Let $(X, \tau)$ be a topological space, $U,V\subseteq X$ any subsets of $X$ . Let $I(A)$ denotes a interior of subset $A$ and $C(A)$ denotes closure of subset $A$. It is clear that $IC(U\cap V)\subseteq IC(U)\cap IC(V)$ Because $U\cap V\subseteq U$ and $U\cap V\subseteq U$ then $C(U\cap V)\subseteq C(U)$ and $C(U\cap V)\subseteq C(V)$; then $IC(U\cap V)\subseteq IC(U)$ and $IC(U\cap V)\subseteq IC(V)$ ; then $IC(U\cap V)\subseteq IC(U)\cap C(V)$. In terms of modal logic $S4: \Box \Diamond(\Box p\wedge \Box q)\rightarrow\Box\Diamond \Box p\wedge \Box \Diamond \Box q$ Now let $U,V\in \tau$ aere two open sets of $X$. Is it true another direction: $IC(U)\cap IC(V)\subseteq IC(U\cap V)$? (In terms of modal logic $S4: \Box\Diamond \Box p\wedge \Box \Diamond \Box q\rightarrow \Box \Diamond(\Box p\wedge \Box q)$) If it is, what will be the proof of this? If it isn't what is counterexamlpe of this?

**It is true more general fact.**

**Proposition.** If $U$ is open and $A$ an arbitrary subset of $X$, then $IC(U)\cap IC(A)\subseteq IC(U\cap A)$.

> **Proof**. Suppose $$x\in IC(U)\cap IC(A)$$. $x\in IC(U)\cap IC(A)$ iff there exists $U'$ an open neighborhood of $x$ such that $U'\subseteq C(U)\cap C(A)$.
>
> Let us denote $$U'\cap U=V_1$$ and $$U'\cap A=V_2.$$ $$U'\subseteq C(U'\cap C(U))\subseteq C(U'\cap U)=C(V_1)$$ $$U'\subseteq C(U'\cap C(A))\subseteq C(U'\cap A)=C(V_2)$$ $$U'\subseteq C(V_1)=C(U'\cap U)=C((U'\cap U)\cap U') \subseteq C((U'\cap U)\cap C(V_2))= \ = C((U'\cap U)\cap C(U'\cap A))\subseteq C((U'\cap U)\cap (U'\cap A))=C(V_1 \cap V_2)$$ We get $$U'\subseteq C(V_1 \cap V_2)=C(U'\cap (U\cap A))\subseteq C(U\cap A)$$ then $$x\in IC(U\cap A).$$