Prove that $\iota(Y):=\operatorname{ext}(X\setminus Y)$ is an interior operator?

Question

So given a set $X$ in this question I ask to show that if $\iota$ is a function from $\mathcal P(X)$ to $\mathcal P(X)$ such that

$\iota(X)=X$
$\iota(Y)\subseteq Y$ for any $Y\in\mathcal P(X)$
$\iota\big(\iota(Y)\big)=\iota(Y)$ for any $Y\in\mathcal P(X)$
$\iota(Y_1\cap Y_2)=\iota(Y_1)\cap\iota(Y_2)$ for any $Y_1,Y_2\in\mathcal P(X)$

then the collection $$ \mathcal T:=\big\{Y\in\mathcal P(X):Y=\iota(Y)\big\} $$ is a topology on $X$ such that $$ \iota(Y)=\operatorname{int}Y $$

Analogous issue is here and here taken on for the derived operator and boundary operator so that I ask to me if it is possible to do an analogous thing for exterior operator so that given a function $\eta$ from $\mathcal P(X)$ to $\mathcal P(X)$ such that

$\eta(\emptyset)=X$
$\eta(Y)\subseteq X\setminus Y$ for any $Y\in \mathcal P(X)$
$\eta(X\setminus Y)\subseteq\eta\big(\eta(Y)\big)$ for any $Y\in\mathcal P(X)$
$\eta(Y_1\cup Y_2)=\eta(Y_1)\cap\eta(Y_2)$ for any $Y_1,Y_2\in\mathcal P(X)$

I tried to show that the collection $$ \mathcal T:=\big\{Y\in\mathcal P(X):Y=\eta(X\setminus Y)\big\} $$ is a topology on $X$.

So first of all by the identity $1$ we observe that $$ X=\eta(\emptyset)=\eta(X\setminus X) $$ which means that $X$ lies in $\mathcal T$. Now by the inclusion $2$ we observe that $$ \eta(X)\subseteq X\setminus X=\emptyset $$ so that we conclude that $$\eta (X)=\emptyset$$ So by the last identity we observe that $$\emptyset=\eta(X)=\eta(X\setminus\emptyset)$$ which means that $\emptyset$ lies in $\mathcal T$.
Therefore if $A_1$ and $A_2$ are two element of $\mathcal T$ then by the identity $4$ we observe that $$ A_1\cap A_2=\eta(X\setminus A_1)\cap\eta(X\setminus A_2)=\eta\big((X\setminus A_1)\cup(X\setminus A_2)\big)=\eta\big(X\setminus(A_1\cap A_2)\big) $$ which means that $A_1\cap A_2$ lies in $\mathcal T$.
Now we observe that if $Y_1,Y_2\in\mathcal P(X)$ are such that $$ Y_1\subseteq Y_2 $$ then by the identity $4$ we observe that $$ \eta(Y_2)=\eta(Y_1\cup Y_2)=\eta(Y_1)\cap\eta(Y_2)\subseteq\eta(Y_1) $$So we conclude that if $\mathcal A$ is a subcollection of $\mathcal T$ then for any $A_0\in\mathcal A$ the inclusion $$ A_0=\eta(X\setminus A_0)\subseteq\eta\Biggl(\bigcap_{A\in\mathcal A}X\setminus A\Biggl)=\eta\Biggl(X\setminus\bigcup_{A\in\mathcal A}A\Biggl) $$ holds and so this implies that $$ \bigcup_{A\in\mathcal A}A\subseteq\eta\Biggl(X\setminus\bigcup_{A\in\mathcal A}A\Biggl) $$ However by the identity $2$ also the inclusion $$ \eta\Biggl(X\setminus\bigcup_{A\in\mathcal A}A\Biggl)=X\setminus\Biggl(X\setminus\bigcup_{A\in\mathcal A}A\Biggl)=\bigcup_{A\in\mathcal A}A $$ holds. So we finally conclude that $$ \bigcup_{A\in\mathcal A}A=\eta\Biggl(X\setminus\bigcup_{A\in\mathcal A}A\Biggl) $$ which means that $\bigcup_{A\in\mathcal A}A$ lies in $\mathcal T$.

So we conclude that $\mathcal T$ is a topology on $X$ but unfortunately I was not able to prove that the identity $$ \eta(Y)=\operatorname{ext}Y $$ holds for any $Y\in\mathcal P(X)$ so that I observe that if the position $$ \tilde\iota(Y):=\eta(X\setminus Y) $$ would define a interior operator as here defined then the identity $$ \operatorname{ext}Y=\operatorname{int}(X\setminus Y)=\tilde\iota(X\setminus Y)=\eta\big(X\setminus(X\setminus Y)\big)=\eta(Y) $$ would be trivially true so that I tried to show that $\tilde\iota$ is an interior operator.

So first of all we observe that $$ \tilde\iota(X)=\eta(X\setminus X)=\eta(\emptyset)=X $$ so that the identity $1$ holds; then we observe that $$ \tilde\iota(Y)=\eta(X\setminus Y)\subseteq X\setminus(X\setminus Y)=Y $$ so that the inclusion $2$ holds; moreover we observe that for any $Y_1,Y_2\in\mathcal P(X)$ the idenity $$ \tilde\iota(Y_1\cap Y_2)=\eta\big(X\setminus(Y_1\cap Y_2)\big)=\eta\big((X\setminus Y_1)\cap(X\setminus Y_2)\big)=\\ \eta(X\setminus Y_1)\cap\eta(X\setminus Y_2)=\tilde\iota(Y_1)\cap\tilde\iota(Y_2) $$ holds so that the identity $4$ holds for $\eta$. Finally we observe that for any $Y\in\mathcal P(X)$ the inlcusion $$ \tilde\iota\big(\tilde\iota(Y)\big)=\eta\big(X\setminus\tilde\iota(Y)\big)\subseteq X\setminus\big(X\setminus\tilde\iota(Y)\big)=\tilde\iota(Y) $$ holds so that the identity $$ \tilde\iota\big(\tilde\iota(Y)\big)=Y $$ would follow by showing that $$ \tilde\iota(Y)\subseteq\tilde\iota\big(\tilde\iota(Y)\big) $$ but unfortunately I was not able to do it so that I thought to put a specific question: as you can see I never use the inclusion $$ \eta(X\setminus Y)\subseteq\eta\big(\eta(Y)\big) $$ so that I argue that I have to use it or rather I have to substitute it with another inclusion. So could someone help me, please?

I haven't tried looking through this (simply don't have time now, or in the near future), but possibly the references in my answer to definitionS of a topological space (reference) and my 3-part answer to General topological space with closure operation as in Russian translation of Hausdorff's 1914 and 1927 Mengenlehre will be of use and/or interest to you. — Dave L. Renfro, Jul 10 '22 at 15:43
@DaveL.Renfro Oh, thanks for the reference: I will read it. Anyway it seem to me that if I substitute the inclusion $$\eta(X\setminus Y)\subseteq\eta\big(\eta(Y)\big)$$ with the identity $$\eta(Y)=\eta\big(X\setminus\eta(Y)\big)$$ then the statement follows but unfortunately this do not implies that the statement cannot follow by the above inclusion. — Antonio Maria Di Mauro, Jul 10 '22 at 15:54
So the question is: what properties characterizes effectively the exterior of a set? — Antonio Maria Di Mauro, Jul 10 '22 at 16:43
Did you look at the paper by Miron Zarycki that I cited? I don't know if your use of the term "exterior" is the same as his use of the word "exterior", but Zarycki's paper should certainly be looked over I think. — Dave L. Renfro, Jul 10 '22 at 17:04
@DaveL.Renfro Oh, thanks for the nice reference!!! Reading this wiki article I presume that Zarycki use the word "exterior" just as I used it: indeed it seem that he want make a topology using exterior or boundary or closure sets and not only directely open (and so interior!) sets; moreover it seem that $IV_e$ point it is just the identity $$\eta(Y)=\eta\big(X\setminus\eta(Y)\big)$$ so that what I observed above would be correct: what can you say about? — Antonio Maria Di Mauro, Jul 10 '22 at 19:02

score 0 · Accepted Answer · answered Jul 15 '22 at 15:49

So it seems that the inclusion $$ \eta(X\setminus Y)\subseteq\eta\big(\eta(Y)\big) $$ for any $Y\in\mathcal P(X)$ does not works: indeed, although I did not find any couterexample, in this paper (mentioned into the comments by Dave L. Renfro) Miron Zarycki says that an exterior operator $\eta$ must be suc that $$ \eta(Y)=\eta\big(X\setminus\eta(Y)\big) $$ for any $Y\in\mathcal P(X)$ and in fact if the last equality holds then $$ \tilde\iota\big(\tilde\iota(Y)\big)=\tilde\iota\big(\eta(X\setminus Y)\big)=\eta\big(X\setminus\eta(X\setminus Y)\big)=\eta(X\setminus Y)=\tilde\iota(Y) $$ for any $Y\in\mathcal P(X)$ so that in this case effectively $\tilde\iota$ is an interior operator.

Prove that $\iota(Y):=\operatorname{ext}(X\setminus Y)$ is an interior operator?

1 Answers1