$IC(U\cap V)$ VS $IC(U)\cap IC(V)$

Question

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?

Answer 1

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 $$x\in IC(U\cap A).$$

Answer 2

Let $(X,\tau)$ be a topological space with interior operator $I$ and closure operator $C$.

Let $A$ be a subset of $X$.

For convenience lets make the following notations: $C(A)=^{def}A^{-}$ and $X\backslash A=^{def}A'$

There is conjugation $I(A)=X\backslash (C(X\backslash A))=(( A') ^{-})'$. We will need two following lemas.

Lemma 1: Let $A$ be a subset of $X$, then $IC(IC(A))=IC(A)$.

Proof: Direction right to left:$$IC(A)\subseteq C(IC(A))$$ By monotonicity of interior and openness of $IC(A)$ $$IC(A)=I(IC(A))\subseteq I(C(IC(A)))=IC(IC(A)).$$ Direction left to right: $$I(C(A))\subseteq C(A)$$ By monotonicity of closure and closeness of $C(A)$ $$C(IC(A))\subseteq C(C(A))=C(A)$$ By monotonicity of interior $$IC(IC(A))\subseteq IC(A).$$

Lemma 2: Let $A,B$ are subsets of $X$, if $A$ is open then $A\cap C(B)\subseteq C(A\cap B)$.

Proof: $ x\in A$ and $x\in C(B)$ iff $x\in A$ and for arbitrary $U_x$ an open neighborhood of $x$ $U_x \cap B\neq \emptyset$. Then for arbitrary $U_x$ an open neighborhood of $x$ $U_x \cap A\neq \emptyset$ is open neighborhood of $x$ and then $U_x \cap A\cap B\neq \emptyset$, iff $x\in C(A\cap B)$.

Now let us prove the main result:

Theorem: If $U,V$ are open subsets of $X$, then $IC(U)\cap IC(V) \subseteq IC(U\cap V)$.

Proof: $U$ is open, then by Lemma 2 $$U\cap V^{-} \subseteq (U\cap V)^{-}$$ By complementing this relation $${(U\cap V)^{-}}' \subseteq U' \cap {V^{-}}'$$ $${{(U\cap V)^{-}}'}^{-} \subseteq {U'}^{-} \cup {{V^{-}}'}^{-}$$ By complementing once more we get $$[{U'}^{-}]' \cap [{{V^{-}}'}^{-}]' \subseteq [{{(U\cap V)^{-}}'}^{-}]' $$ Notice that $U=(( U') ^{-})'$, then we get $$U \cap [{{V^{-}}'}^{-}]' \subseteq [{{(U\cap V)^{-}}'}^{-}]'$$ For convenience lets denote $(A^-)'$ by $A^{\bot}$ we get $$U \cap V^{\bot\bot} \subseteq (U\cap V)^{\bot\bot}$$ Now put $U^{\bot\bot}$ in place of $U$, $U^{\bot\bot}$ is open $$U^{\bot\bot} \cap V^{\bot\bot} \subseteq (U^{\bot\bot}\cap V)^{\bot\bot}$$ $V$ is open, then $U^{\bot\bot}\cap V\subseteq (U\cap V)^{\bot\bot}$ and then $(U^{\bot\bot}\cap V)\subseteq (U\cap V)^{\bot\bot\bot\bot}$ $$U^{\bot\bot} \cap V^{\bot\bot} \subseteq(U^{\bot\bot}\cap V)\subseteq (U\cap V)^{\bot\bot\bot\bot}$$ Now notice that $A^{\bot\bot}=IC(A)$, after rewriting and using Lemma 1 we get $$IC(U)\cap IC(V)\subseteq IC(IC(U\cap V))=IC(U\cap V).$$