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.
 A: 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.
A: 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 $X=\emptyset$).
