# Proving an operation of interior is a set of open sets.

I am looking at General Topology notes on Alternative Ways of Defining Topology and come up with this questions:

If $\iota : \mathscr P (X) \to \mathscr P (X)$ is an operation of interior, then $\tau = \{A \in \mathscr P(X) \ : \ \iota (A) = A \}$ is a set of open sets in $X$.

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For your convenience, a map $\iota : \mathscr P (X) \to \mathscr P (X)$ has to meet the followings to be an operation of interior:

($\mathscr I_1) : \iota (\emptyset) = (\emptyset )$

($\mathscr I_2) : \forall A \in \mathscr P(X), \iota (A) \subseteq A$

($\mathscr I_3) : \forall A \in \mathscr P(X), \iota (\iota(A)) = \iota( A)$

($\mathscr I_4) : \forall A, B \in \mathscr P(X), \iota (A \cap B) = \iota (A) \cap \iota (B)$

For your convenience again, $\tau \subseteq \mathscr P (X)$ is a set of open sets in $X$ if $\tau$ satisfies the following conditions:

$(\mathscr O_1)$ : $\emptyset \in \mathscr \tau$

$(\mathscr O_2)$ : $X \in \tau$

$(\mathscr O_3)$ : If $\{ U_i \}_{i \in I} \in \tau$, then $\cup _{i \in I} \in \tau$

$(\mathscr O_4)$ : If $U, V \in \tau$, then $U \cap V \in \tau$

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And here are my two questions:

(1) It is easy to solve $(\mathscr O_1)$ and $(\mathscr O_4)$ since they are straightforward. But when dealing with $(\mathscr O_2)$, I have to come up with $\iota (X) = X$ and I completely am stuck here, please help.

(2) For $(\mathscr O_3)$, suppose that $T_i = \{A_i \in \mathscr P(X) : \iota (A_i) = A_i \}$. Can I use $\iota ( \cup T_i ) = \cup \iota (T_i)$ because it is just a natural outcome of union of sets, or do I have to prove it first? Or perhaps there is another way without using it?

Thank you for your time and help.

EDIT ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I have been responded that there is maybe a mistake in the definition. Perhaps the error is here: On the next page of the same note, there is definition for Operation of Closure as follow:

$(\mathscr C_1) : \kappa (\emptyset ) = \emptyset$

$(\mathscr C_2) :$ For $\forall A \in \mathscr P(X), A \subseteq \kappa (A)$

$(\mathscr C_3) :$ For $\forall A \in \mathscr P(X), \kappa (\kappa (A)) = \kappa (A)$

$(\mathscr C_4) :$ For $\forall A, B \in \mathscr P(X), \kappa (A \cup B) = \kappa (A) \cup \kappa (B)$.

Since the Operation of Interior is a dual of Operation of Closure, I have been speculating perhaps there is a typo on $(\mathscr I_1)$, that it should not be $\iota (\emptyset) = \emptyset$ but it should be $\iota (X) = X$ instead. But if that is indeed the typo, I am now facing another dilemma: Can you solve the $(\mathscr I_1)$? Please help and as always, thank you so very much for your time and help.

• Note that $\mathscr I_1$ is an immediate consequence of $\mathscr I_2$ – Hagen von Eitzen Dec 13 '14 at 22:12

## 2 Answers

It should indeed say $\iota(X) = X$ as $\mathscr{I}_1$. See also this forum post. The duality is clear from defining $\kappa(A) = X\setminus \iota(X\setminus A)$. Then $\kappa$ is a closure operator (using your correct closure axioms) iff $\iota(X)$ is an interior operatator (with the corrected definition). Of course the reverse also goes, defining $\iota(A) = X\setminus \kappa(X \setminus A)$.

The duality is clear if we think of $\iota(A)$ as the largest open subset of $A$, and $\kappa(A)$ as the largest closed subset containing $A$, and knowing closed and open sets are related by complements.

Also, $\iota(\emptyset) = \emptyset$ is immediate from $\iota(A) \subseteq A$, as the only subset of $\emptyset$ is $\emptyset$. And the axioms are chosen to be independent: there are "almost"-interior operators that satisfy all but one of the conditions, for all conditions can appear as the odd one out. Taking $\iota(A) = \emptyset$ satisifies all but the correct $\mathscr{I}_1: \iota(X) = X$, as one example. There are other examples for the other conditions, showing them all to be necessary.

• Brandsme : Thank you! Now I am done with this problem! I have up-voted your answer. Thank you again. – Amanda.M Dec 15 '14 at 3:04

With the definition as you cite them, the claim is wrong. Indeed, $\iota(A):=\emptyset$ fulfills your definition of interior, but it makes $\iota(X)\ne X$ (unless $X=\emptyset$).

• Thanks, but how do I go about solving $\mathscr O_2$? Thanks again. – Amanda.M Dec 13 '14 at 22:25
• @A.Magnus As I said, you cannot show $\mathscr O_2$ because it is not necesssarily true. Maybe double-check your definitions. – Hagen von Eitzen Dec 13 '14 at 22:38
• The definition of $T_i$ doesn't make sense. You need to start with a collection $T \subseteq \tau$ and show that $\iota ( \cup T ) = \cup T$. Think about what $(\mathscr I_4)$ tells you when $A \subseteq B$. – David Hartley Dec 13 '14 at 22:55
• @HagenvonEitzen : To Hagen von Eitzen: Honestly, I have been suspecting that there is a typo on the definition of $(\mathscr I_1)$, for that reason I have posted an Edit. Could you please take another look, please? Thank you very much for your time and help. – Amanda.M Dec 14 '14 at 3:30
• @DavidHartley : I made the $T_i$ definition myself, the mistake is completely mine. Thanks for your pointing out. – Amanda.M Dec 14 '14 at 3:37