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 $int(E)=\eta(E)$?

  • $\begingroup$ The correct formula is $\kappa(E)=X\setminus \eta(X \setminus E)$ $\endgroup$ Jan 30 '20 at 12:18

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.


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