How to rigorously prove the following result?
Let $X$ be a non-empty set and let $k \colon \mathscr{P}(X) \longrightarrow \mathscr{P}(X)$ satisfy the following Kuratowski Closure Axioms:
[K$_1$] $k(\emptyset) = \emptyset$,
[K$_2$] $A \subset k(A)$,
[K$_3$] $k( A \cup B) = k(A) \cup k(B)$,
[K$_4$] $k\big( k(A) \big) = k(A)$.
Then there exists one and only one topology $\mathscr{T}$ on $X$ such that $k(A)$ will be the $\mathscr{T}$-closure of $A \subset X$.
From the above four axioms, we can also prove the following:
Let $A$ and $B$ be any subsets of $X$ such that $A \subset B$. Then $k(A) \subset k(B)$ also.
Proof:
As $A \subset B$, so we have $B = A \cup (B \setminus A)$, and hence \begin{align} k(B) &= k \big( A \cup (B \setminus A) \big) \\ &= k(A) \cup k(B \setminus A) \\ &\supset k(A), \end{align} thus showing that $k(A) \subset k(B)$.
And, from the last result, we can prove the following:
Let $\mathscr{A}$ be any (non-empty) collection of subsets of $X$. Then we have $$ k \left( \cap_{A \in \mathscr{A} } A \right) \subset \cap_{ A \in \mathscr{A} } k(A). $$
Proof:
Let us put $$ S := \cap_{A \in \mathscr{A} } A. $$ Then we have $S \subset A$ for every $A \in \mathscr{A}$, which implies that $k(S) \subset k(A)$ for every $A \in \mathscr{A}$, which in turn implies that $$ k(S) \subset \cap_{A \in \mathscr{A} } k(A), $$ as required.
Now for the proof of our main result.
My Attempt:
Let us define the collection $\mathscr{T}$ of subsets of $X$ as follows: $$ \mathscr{T} := \big\{ U \colon U \subset X \mbox{ and } k( X \setminus U) = X \setminus U \big\}. \tag{1} $$ We show that this collection $\mathscr{T}$ is a topology on $X$.
Since by definition of the mapping $k$, we have $k(A) \subset X$ for every subset $A$ of $X$, therefore we must have $k(X) \subset X$. And, by [K$_2$] we have $X \subset k(X)$. Thus we have $k(X) = X$. Now as $\emptyset \subset X$ and as $$ k(X \setminus \emptyset) = k(X) = X = X \setminus \emptyset, $$ we can conclude that the empty set $\emptyset$ is in $\mathscr{T}$.
Am I right?
As $X \subset X$ and as by [K$_1$] we also have $$ k(X \setminus X) = k(\emptyset) = \emptyset = X \setminus X, $$ so we can conclude that the set $X$ itself is also in $\mathscr{T}$.
Now let $\mathscr{A}$ be any subcollection of $\mathscr{T}$, and let us put $$ S := \cup_{A \in \mathscr{A} } A. $$ We show that this set $S$ is in our collection $\mathscr{T}$. We note that each set $A \in \mathscr{A}$ satisfies $A \subset X$ and $k(X \setminus A) = X \setminus A$. So we have $S \subset X$ and \begin{align} k( X \setminus S) &= k \left( X \setminus \cup_{A \in \mathscr{A} } A \right) \\ &= k \left( \cap_{A \in \mathscr{A} } (X \setminus A) \right) \\ &\subset \cap_{A \in \mathscr{A} } k ( X \setminus A ) \\ &= \cap_{A \in \mathscr{A} } ( X \setminus A ) \\ &= X \setminus \cup_{A \in \mathscr{A} } A \\ &= X \setminus S. \end{align} Thus we have $$ k ( X \setminus S) \subset X \setminus S. \tag{2} $$ For each set $A \in \mathscr{A}$, since $A \subset S$, we have $( X \setminus A) \subset (X \setminus S)$, which implies that $k ( X \setminus A) \subset k(X \setminus S)$, which in turn implies that $$ \cup_{A \in \mathscr{A} } k ( X \setminus A) \subset k(X \setminus S). \tag{3} $$ Therefore we have \begin{align} X \setminus S &= X \setminus \cup_{A \in \mathscr{A} } A \\ &= \cap_{A \in \mathscr{A} } (X \setminus A ) \\ &\subset \cup_{A \in \mathscr{A} } (X \setminus A ) \\ &= \cup_{A \in \mathscr{A} } k (X \setminus A ) \\ &\subset k( X \setminus S ). \qquad [ \mbox{ by (3) above } ] \end{align} Thus we have $$ X \setminus S \subset k ( X \setminus S). \tag{4} $$ From (2) and (4) we obtain $$ k(X \setminus S) = X \setminus S, $$ thus showing that $S$ is in $\mathscr{T}$.
Am I right?
Let $S_1 \ldots, S_n$ be any sets from $\mathscr{T}$, and let us put $S \colon= \cap_{i=1}^n S_i$. For each $i = 1, \ldots, n$, as $S_i \subset X$ and $k\left( X \setminus S_i \right) = X \setminus S_i$, so we have $S \subset X$ and \begin{align} k ( X \setminus S) &= k \left( X \setminus \cap_{i=1}^n S_i \right) \\ &= k \left( \cup_{i=1}^n \left( X \setminus S_i \right) \right) \\ &= \cup_{i=1}^n k \left( X \setminus S_i \right) \qquad [ \mbox{ from [K$_3$] } ] \\ &= \cup_{i=1}^n \left( X \setminus S_i \right) \\ &= X \setminus \cap_{i=1}^n S_i \\ &= X \setminus S, \end{align} thus showing that our set $S$ is in $\mathscr{T}$.
Thus our collection $\mathscr{T}$ in (1) above is indeed a topology on $X$.
Next, we show that, for every subset $A$ of $X$, we have $k(A) = \overline{A}$, where $\overline{A}$ denotes the $\mathscr{T}$-closure of set $A$ in $X$.
Let $\mathscr{C}$ be the collection of all the $\mathscr{T}$-closed supersets of $A$. Then $X$ is in $\mathscr{C}$ trivially so that $\mathscr{C}$ is a non-empty collection of sets. We show that $$ k(A) = \cap_{C \in \mathscr{C} } C. \tag{5} $$ We of course have $$ A \subset \cap_{C \in \mathscr{C} } C. \tag{6} $$ For each $C \in \mathscr{C}$, as $X \setminus C$ is in $\mathscr{T}$, so we have $$ k(C) = k\big( X \setminus (X \setminus C) \big) = X \setminus (X \setminus C) = C, $$ that is, $$ k(C) = C. \tag{7} $$ Therefore we have \begin{align} k(A) &\subset k \left( \cap_{C \in \mathscr{C} } C \right) \qquad [ \mbox{ by (6) above } ] \\ &\subset \cap_{C \in \mathscr{C} } k(C) \\ &= \cap_{C \in \mathscr{C} } C \qquad [ \mbox{ by (7) above } ]. \end{align} Thus we have $$ k(A) \subset \cap_{C \in \mathscr{C} } C. \tag{8} $$ Morevoer, by [K$_2$] above we have $A \subset k(A)$, and since $k(A) \subset X$, we also have $$ \big( X \setminus k(A) \big) \subset X $$ and
\begin{align} k \big( X \setminus \big(X \setminus k(A) \big) \big) &= k \big( k(A) \big) \\ &= k(A) \\ &= X \setminus \big(X \setminus k(A) \big), \end{align} which shows that $X \setminus k(A)$ is in $\mathscr{T}$, thus showing that $k(A)$ is a $\mathscr{T}$-closed superset of $A$ and hence $k(A)$ is also in $\mathscr{C}$, and therefore we have $$ \cap_{C \in \mathscr{C} } C \subset k(A), $$ which together with (8) above yields (5).
Am I right?
Finally, let $\mathscr{T}^\prime$ be any topology on $X$ such that, for each subset $A$ of $X$, the $\mathscr{T}^\prime$-closure of $A$ equals $k(A)$. We show that $$ \mathscr{T}^\prime = \mathscr{T}. \tag{9} $$
Let $U$ be any set of $\mathscr{T}$. Then $U \subset X$ and $k(X \setminus U) = X \setminus U$. [Refer to (1) above.] But then sinc $(X \setminus U ) \subset X$, the $\mathscr{T}^\prime$-closure of $X \setminus U$ equals $k(X \setminus U) = X \setminus U$, thus showing that $X \setminus U$ is $\mathscr{T}^\prime$-closed, which implies that $U$ is in $\mathscr{T}^\prime$, from which it follows that $$ \mathscr{T} \subset \mathscr{T}^\prime. \tag{10} $$
Am I right?
Conversely, suppose that $V$ is in $\mathscr{T}^\prime$. Then the set $X \setminus V$ is a $\mathscr{T}^\prime$-closed subset of $X$ and therefore the $\mathscr{T}^\prime$-closure of $X \setminus V$ equals $X \setminus V$, which implies that $k( X \setminus V) = X\setminus V$ by our hypothesis about $\mathscr{T}^\prime$, which in turn implies that $V \in \mathscr{T}$ and therefore we have $$ \mathscr{T}^\prime \subset \mathscr{T}, $$ which together with (10) above yields (9).
Am I right?
Is my proof solid enough in each and every detail? Or, are there issus therein of correctness or clarity?
