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Suppose $X$ is a topological space containing subsets $A$ and $B$ such that

$$\tag1ikA=ikiA=iA\subsetneq A=kA=kikA=kiA$$ $$\tag2kiB=ikiB=iB\subsetneq B\subsetneq kB=kikB=ikB$$

where $k$ is closure and $i$ is interior. These relations, as well as those implied for the complements $cA$ and $cB$, are represented by the following Hasse diagrams where sets in a given diagram are equal iff they have the same color:

Hasse diagrams Kuratowski closure complement 14 sets

Suppose $A\cup B$ satisfies $(1)$ and $A\cap B$ satisfies $(2){:}$

Hasse diagrams Kuratowski unions intersections

Finally, suppose $A\cup cB$ is clopen (this might already be implied).

Does it follow that $kiA\setminus ikiA=ki(A\cap cB)\setminus iki(A\cap cB)?$

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  • $\begingroup$ Why did you also assume (1) and (2) for the complements? $\endgroup$
    – Berci
    Mar 2, 2019 at 23:54
  • $\begingroup$ @Berci the relations satisfied by $cA$ (which are similar to $(1)$, but slightly different) and $cB$ ($(2)$) follow from the conditions put on $A$ and $B$. All four sets are relevant to the problem, so all four diagrams seemed worth displaying. $\endgroup$ Mar 3, 2019 at 4:08
  • $\begingroup$ I don't see any difference in the diagrams. Why would the same hold for the complement? $\endgroup$
    – Berci
    Mar 3, 2019 at 8:41
  • $\begingroup$ The conditions on $B$ are inconsistent. $\endgroup$ Mar 3, 2019 at 14:19
  • $\begingroup$ @Berci the diagrams (for $A$ and $cA$) are different because their colorings are different. The Hasse diagram for $cS$ $(\text{any }S\subset X)$ is obtained by flipping the diagram for $S$ upside down and interchanging $ki$ and $ik,$ because $i(cS)=(ckc)cS=c(kS),$ $iki(cS)=(ckc)k(ckc)cS=c(kikS),$ $ik(cS)=(ckc)kcS=c(kiS),$ $ki(cS)=cck(ckc)cS=c(ikS),$ $kik(cS)=cck(ckc)kcS=c(ikiS),$ and $k(cS)=cckcS=c(iS).$ $\endgroup$ Mar 3, 2019 at 20:35

1 Answer 1

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Yes, the diagrams alone imply the equation. They also imply $A\cup cB$ is clopen.

Proof. Let $\textsf{id}$ denote the identity operator. For each $\{\sigma,\tau\}\subset\{\textsf{id},i,ki,iki,k,ik,kik\},$ sets $E$ satisfying $\sigma E=\tau E$ are characterized in A Further Note on Closure and Interior Operators (written by Thomas A. Chapman when he was an undergrad). Two theorems from this paper will be useful here. We cite and prove both.

Theorem 2. A subset $E$ of a topological space $X$ satisfies $kiE=kE$ iff $E=iE\cup F$ where $F$ has a void interior in the subspace $E$ of $X.$

Proof: $(\Rightarrow)$ Define $F=E\setminus iE.$ Let $o_E$ denote the operator $o$ in the subspace $E$ of $X.$ Note that $$\eqalign{k_E(iE)&=&\bigcap\{D\cap E:kD=D\supset iE\}\cr&=&E\cap\bigcap\{D:kD=D\supset iE\}\cr&=&E\cap kiE=E\cap kE=E.\cr}$$ Thus $$\eqalign{i_EF&=&c_Ek_Ec_E(E\setminus iE)\cr&=&c_Ek_E(iE)=c_EE=\varnothing.}$$ Clearly, $E=iE\cup F.$

$(\Leftarrow)$ Let $q\in kE$ and $U$ be any open neighborhood of $q.$ Every neighborhood of $q$ intersects $E,$ hence $U\cap E\neq\varnothing.$ Since $U\cap E$ is open in $E$ and $i_EF=\varnothing,$ we have $(U\cap E)\cap F=\varnothing.$ Since $E=iE\cup F,$ it follows that $U\cap iE\neq\varnothing.$ Since $U$ is arbitrary, this implies $q\in kiE.$ Thus $kE\subset kiE.$ Since the opposite inclusion always holds, conclude $kiE=kE.$

Theorem 5. A subset $E$ of a topological space $X$ satisfies $kiE=iE$ iff $E=iE\cup F$ where $iE$ is clopen and $iE\cap F=\varnothing=iF.$

Proof: $(\Rightarrow)$ The hypothesis $kiE=iE$ implies that $iE$ is clopen. Define $F=E\setminus iE.$ Since $iE\cup iF$ is an open set contained in $A,$ we have $iE\cup iF=iE.$ Since $iE\cap iF=\varnothing,$ this implies $iF=\varnothing.$ Clearly, $iE\cap F=\varnothing.$

$(\Leftarrow)$ The equation $kiE=iE$ holds since $iE$ is clopen.

By Theorem 2 and the diagrams, it follows that $$\tag1A=iA\cup V$$ where $A$ is closed, $iA\neq\varnothing,$ $V\neq\varnothing,$ and $iA\cap V=\varnothing=i_AV.$

Similarly, $$\tag2A\cup B=i(A\cup B)\cup W$$ where $A\cup B$ is closed, $i(A\cup B)\neq\varnothing,$ $W\neq\varnothing,$ and $i(A\cup B)\cap W=\varnothing=i_{A\cup B}W.$

By Theorem 5 and the diagrams, it follows that $$\tag3B=iB\cup Y$$ where $iB$ is clopen (possibly empty), $Y\neq\varnothing,$ and $iB\cap Y=\varnothing=iY.$

Similarly, $$\tag4cB=icB\cup Z$$ where $icB$ is clopen (possibly empty), $Z\neq\varnothing,$ and $icB\cap Z=\varnothing=iZ.$

Claim 1. $i(A\cup B)=iA\cup iB$ and $i(A\cup cB)=iA\cup icB.$

Proof. Since $A$ is closed, we have $$\eqalign{i(A\cup B)\setminus A&=&i(A\cup B)\cap cA\cr&=&i(A\cup B)\cap icA\cr&=&i[(A\cup B)\cap cA]\cr&=&i(B\cap cA)\cr&=&iB\cap icA\cr&=&iB\cap cA\cr&=&iB\setminus A.}$$ It follows that $i(A\cup B)\subset A\cup iB.$ Since $iB$ is clopen, $i(A\cup B)\setminus iB$ is an open set contained in $A.$ Hence $i(A\cup B)\setminus iB\subset iA.$ Thus $i(A\cup B)\subset iA\cup iB.$ The opposite inclusion always holds, so $i(A\cup B)=iA\cup iB.$ The second equation holds by the same argument with $B$ replaced by $cB.$

Claim 2. $Y\cup Z$ is a clopen subset of $iA.$

Proof. Suppose $Y\cap cA\neq\varnothing.$ Since $Y\subset ciB$ and $ciB$ is clopen, this implies that $cA\cap ciB$ is a nonempty open set. Thus, since $i_{A\cup B}W=\varnothing,$ the set $(cA\cap ciB)\cap(A\cup B)$ must intersect $i(A\cup B)=iA\cup iB.$ But this is impossible, since $(cA\cap ciB)\cap (iA\cup iB)=\varnothing.$ This contradiction implies $Y\subset A.$

Replacing $B$ with $cB$ and $Y$ with $Z$ in the above argument yields $Z\subset A.$ Hence, $Y\cup Z\subset A.$

Note that $X=B\cup cB=(iB\cup Y)\cup(icB\cup Z).$ Thus, $kB=cicB=iB\cup(Y\cup Z)$ and $kcB=ciB=icB\cup(Y\cup Z).$ Hence, $kB\cap kcB=Y\cup Z.$ Since $kB$ and $kcB$ are each clopen, this implies that $Y\cup Z$ is a clopen subset of $A,$ which further implies that $Y\cup Z\subset iA.$

Aside. Since $\varnothing\subsetneq Y\cup Z\subset iA\subsetneq X,$ Claim 2 implies $X$ is not connected.

Corollary 1. $iA\cap kB$ is clopen.

Proof. Claim 2 and the diagrams imply that $iA\cap kB$ is a union of two clopen sets: $$\eqalign{iA\cap kB&=&iA\cap[iB\cup(Y\cup Z)]\cr&=&(iA\cap iB)\cup[iA\cap(Y\cup Z)]\cr&=&i(A\cap B)\cup(Y\cup Z).}$$

Claim 3. $k(iA\setminus kB)=(iA\setminus kB)\cup V$ and $ik(iA\setminus kB)\cap V=\varnothing.$

Proof. By $(1),$ Corollary 1 and the diagrams, we have $$\eqalign{(iA\cap kB)\cup (iA\setminus kB)\cup V&=&A\cr&=&kiA\cr&=&k[(iA\cap kB)\cup(iA\setminus kB)]\cr&=&k(iA\cap kB)\cup k(iA\setminus kB)\cr&=&(iA\cap kB)\cup k(iA\setminus kB).}$$ Since $c(iA\cap kB)$ is a closed set containing $iA\setminus kB,$ we have $k(iA\setminus kB)\subset c(iA\cap kB),$ hence $(iA\cap kB)\cap k(iA\setminus kB)=\varnothing.$ Thus, since we also have $iA\cap V=\varnothing,$ the equation above implies the first equation in the claim. Applying $i$ to both sides of the inclusion $k(iA\setminus kB)\subset A$ yields $ik(iA\setminus kB)\subset iA.$ The second equation follows.

Corollary 2. $ki(A\cap cB)\setminus iki(A\cap cB)=V.$

Proof. Note that $i(A\cap cB)=iA\cap icB=iA\setminus kB.$ By Claim 3, we conclude that $$\eqalign{ki(A\cap cB)\setminus iki(A\cap cB)&=&k(iA\setminus kB)\setminus ik(iA\setminus kB)\cr&=&[(iA\setminus kB)\cup V]\setminus ik(iA\setminus kB)\cr&=&[(iA\setminus kB)\setminus ik(iA\setminus kB)]\cup [V\setminus ik(iA\setminus kB)]\cr&=&[i(iA\setminus kB)\setminus ik(iA\setminus kB)]\cup V\cr&=&\varnothing\cup V\cr&=&V.}$$

Since $kiA\setminus ikiA=A\setminus iA=V,$ this proves that $$kiA\setminus ikiA=ki(A\cap cB)\setminus iki(A\cap cB).$$

This answers the question, but for completeness, we also show that the diagrams imply $A\cup cB$ is clopen.

Claim 4. $iB\setminus iA$ is clopen.

Proof. Since $iB$ and $i(A\cap B)$ are both clopen, it follows that $iB\setminus iA=iB\setminus (iA\cap iB)=iB\setminus i(A\cap B)$ is also clopen.

Claim 5. $iB\cap V=\varnothing.$

Proof. Since $iB\cap V\subset iB\setminus iA,$ if $iB\cap V\neq\varnothing,$ then Claim 4 implies $iB\setminus iA$ is a nonempty open set such that $(iB\setminus iA)\cap A\subset V,$ contradicting $i_AV=\varnothing.$ Thus $iB\cap V=\varnothing.$

Claim 6. $A\cup cB$ is clopen.

Proof. Claim 5 implies $iB\setminus iA=iB\setminus A.$ But $iB\setminus A=(B\setminus Y)\setminus A=B\setminus A$ since $Y\subset A.$ Thus $B\setminus A$ is clopen by Claim 4. Hence $c(B\setminus A)=A\cup cB$ is also clopen. $\blacksquare$

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