0

At the 2th chapter of the "Elementos de Topología General" by Fidel Casarrubias Segura and Ángel Tamariz Mascarúa it is written that the interior operator $\eta$ induces a topology and to prove this it say that defining the operator $$\kappa:\mathcal{P}(X)\owns{E}\rightarrow{X\setminus{\eta(E)}}\in\mathcal{P}(X)$$ it is possible to demonstrate that $\kappa$ is a closure operator and so $\eta$ induce a topology on X that is the same of closure operator $\kappa$.

Well since $\eta$ is an interior operetor I know that:

  1. $\eta(X)=X$;
  2. $\eta(E)\subseteq{E}$ for any $E\subseteq{X}$;
  3. $\eta(\eta(E))=\eta(E)$ for any $E\subseteq{X}$;
  4. $\eta(A\cap{B})=\eta(A)\cap{\eta(B)}$ for any $A,B\subseteq{X}$.

So to demonstrate that $\kappa$ is a closure operetor I must demonstrate that:

  1. $\kappa(\varnothing)=\varnothing$;
  2. $E\subseteq\kappa(E)$ for any ${E}\subseteq{X}$;
  3. $\kappa(\kappa(E))=\kappa(E)$ for any ${E}\subseteq{X}$;
  4. $\kappa(A\cup{B})=\kappa(A)\cup\kappa(B)$ for any $A,B\subseteq{X}$.

Howewer first it seems to me that $\kappa(\varnothing)=X\setminus\eta(\varnothing)=X\setminus\varnothing=X\neq\varnothing$ and then I can't prove the other points so I only see that

  1. $\eta(E)\subseteq{E}\Rightarrow{X\setminus{E}}\subseteq{X\setminus\eta(E)}=\kappa(E)$;
  2. $\kappa(\kappa(E))=\kappa(X\setminus\eta(E))=X\setminus\eta(X\setminus\eta(E))$;
  3. $\kappa(A\cup{B})=X\setminus\eta(A\cup{B})$.

Could someone help me?

Furthermore how to demonstrate that the identity $$ \operatorname{int}E=\eta(E) $$ holds?

Antonio Maria Di Mauro
  • 4,812

2 Answers2

2

If $\eta$ is an interior operator, you have to define $\kappa(E)= X\setminus \eta(X\setminus E)$ and then e.g.

$\kappa(E) = X\setminus \eta(X\setminus E) \supseteq X \setminus( X \setminus E) = E$ etc. Use that $\setminus$ reverses inclusions and changes $\cap$ into $\cup$ and vice versa by de Morgan's laws.

Henno Brandsma
  • 242,131
0

Another possible approaches is to show that the collection $$ \mathcal T:=\big\{Y\in\mathcal P(X):Y=\eta(Y)\big\} $$ is a topology such that $$ \operatorname{int}Y=\eta(Y) $$ for any $Y\in\mathcal P(X)$ so that we let to do this.

So first of all by the inclusion $2$ we observe that $$ \eta(\emptyset)\subseteq\emptyset $$ so that by the identity $1$ we conclude that $\emptyset,X\in\mathcal T$.

Moreover, by the identity $4$ we conclude that $\mathcal T$ is closed by finite intersection.

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 conclude that $$ \eta(Y_1)=\eta(Y_1\cap Y_2)=\eta(Y_1)\cap\eta(Y_2)\subseteq\eta(Y_2) $$ so staht if $\mathcal A$ is a subcollection of $\mathcal T$ then for any $A_0\in\mathcal A$ the inclusion $$ A_0=\eta(A_0)\subseteq\eta\Biggl(\bigcup_{A\in\mathcal A}A\Biggl) $$ holds and thus also the inclusion $$ \bigcup_{A\in\mathcal A}A\subseteq\eta\Biggl(\bigcup_{A\in\mathcal A}A\Biggl) $$ holds; howerver, by the inclusion $2$ it must be $$ \eta\Biggl(\bigcup_{A\in\mathcal A}A\Biggl)\subseteq\bigcup_{A\in\mathcal A}A $$ So we finally conclude that $$ \bigcup_{A\in\mathcal A}A=\eta\Biggl(\bigcup_{A\in\mathcal A}A\Biggl) $$ which means that $\mathcal T$ is closed for arbitrary unions.

So we conclude that $\mathcal T$ is a topology on $X$. Now the interior is open and moreover any set contains its interior, that is for any $Y\in\mathcal P(X)$ the inclusion $$ \operatorname{int}Y\subseteq Y $$ holds so that we conclude that the inclusion $$ \operatorname{int}Y=\eta(\operatorname{int}Y)\subseteq\eta(Y) $$ holds; however, the identity $3$ implies that $\eta(Y)$ is open so that by the inclusion $2$ we conclude that $$ \eta(Y)\subseteq\operatorname{int}Y $$ so we finally conclude that $$ \operatorname{int}Y=\eta(Y) $$

Antonio Maria Di Mauro
  • 4,812