Prove that $\mathcal T$ is a topology 
Let a function $h:\mathcal P(X)\to\mathcal P(X)$ be defined by

*

*$h(\emptyset)=\emptyset$


*$h(A\cup B)=h(A)\cup h(B),\;\forall A,B\in\mathcal P(X)$


*$h(A)\supseteq A,\;\forall A\in\mathcal P(X)$


*$h\circ h=h$
Now setup $\mathcal T:=\{A^\complement\in\mathcal P(X): h(A)=A\}$. Prove that $\mathcal T$ is a topology on $X$.

Im stuck with this problem. I can show that $\emptyset,X\in\mathcal T$ and that the finite intersection of elements of $\mathcal T$ belong to $\mathcal T$.
But Im unable to prove that arbitrary union of elements of $\mathcal T$ belong to $\mathcal T$. I was playing around with the properties of $h$ but I cant conclude something about this last axiom to define a topology.
Some hint or solution will be appreciated. Thank you.
 A: We are first going to show that $h$ preserves inclusions. Suppose that $A\subset B$. This is equivalent to saying that there exists an $C\in \mathcal{P}(X)$ such that $A\cup C=B$. Applying $h$ to this equality, we find that $h(A\cup C)=h(A)\cup h(C)=h(B)$, here we used that $h$ goes through finite unions. Since $h(A)\cup h(C)=h(B)$, we have that $h(A)\subset h(B)$. Thus we showed that $A\subset B \Rightarrow h(A)\subset h(B)$. (The converse is not necessarily true).
Now choose $A_i^C\in \mathcal{T}$ arbitrarily. Then $\cap_i A_i\subset h(\cap_i A_i)$ by the third property of $h$. Notice that $\cap_i A_i\subset A_i$, applying $h$ to this inclusion, we get that $$h(\cap_iA_i)\subset h(A_i)=A_i$$ holds for all $i$. Here we used that $h(A_i)=A_i$ since $A_i^C\in \mathcal{T}$. Since the above property holds for all $i$, we conclude that $h(\cap_i A_i)\subset \cap_i A_i$. Thus $h(\cap_i A_i)=\cap_i A_i$, which we needed to show by my comment above.
Notice that we didn't use that $h\circ h=h$.
A: To show that any union of subsets of $T$ belongs to $T,$ it suffices to show that if $Y\subset P(X)$ and  $\forall B\in Y\;(h(B)=B)$ then $h(\cap Y)=\cap Y.$
(1).If $C\subset D\subset X$ then $h(C)\subset h(D)$ because $h(D)=h( C\cup (D$ \ $C))=h(C)\cup h(D$ \ $C).$
(2). If $Y\subset P(X)$ and $\forall B\in Y\;(B=h(B))$ then by (1), $$\forall B\in Y\; (h(\cap Y)\subset h(B)).$$ $$ \text  { Therefore }\quad  \cap Y\subset h(\cap Y)\subset \cap_{B\in Y}h(B)=\cap_{B\in Y}B=\cap Y.$$ 
A: Here is Mathematician 42's proof, but written down in a different form, with two goals.  First, I wanted to convince myself that OP's property 4, $\;h \circ h = h\;$, is indeed not used, and I found it difficult to understand some details of that earlier answer.  And second, to present the proof in a different order, making it clearer that '$\;h\;$ preserves $\;\subseteq\;$' is indeed a reasonable thing to look at.$%
\require{begingroup}
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\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\T}{\mathcal T}
\newcommand{\c}{\complement}
%$

First we note that, by the definition of $\;\T\;$,
$$
\tag{0}
B \in \T \;\equiv\; h(B^\c) = B^\c
$$ In other words, this topology are the sets whose complements remain unchanged under $\;h\;$.
So, given an arbitrary $\;V \subseteq \T\;$, we calculate when its union is in $\;\T\;$:
$$\calc
    \cup V \in \T
\op\equiv\hint{expand definition of $\;\T\;$ using $\Ref{0}$}
    h((\cup V)^\c) \;=\; (\cup V)^\c
\op\equiv\hint{OP's property 3 -- to reduce our proof burden}
    h((\cup V)^\c) \;\subseteq\; (\cup V)^\c
\op\equiv\hints{by $\Ref{1}$ below ror the RHS}\hints{-- we choose the RHS because we don't seem to have}\hint{enough information about the LHS}
    \tag{*} h((\cup V)^\c) \;\subseteq\; \cap \{ B^\c \mid B \in V\}
\op\equiv\hints{set theory: basic property of $\;\ldots \subseteq \cap \ldots\;$}\hint{-- the only way to make progress}
    \langle \forall B \in V :: \; h((\cup V)^\c) \subseteq B^\c \; \rangle
\op\equiv\hint{$V \subseteq \T$ so $B \in \T$ or by $\Ref{0}$ equivalently $h(B^\c)=B^\c$}
    \langle \forall B \in V :: \; h((\cup V)^\c) \subseteq h(B^\c) \; \rangle
\op\when\hint{by $\Ref{2}$ below -- this looks reasonable to try}
    \langle \forall B \in V :: \; (\cup V)^\c \subseteq B^\c \; \rangle
\op\equiv\hint{set theory: basic property of $\;\ldots \subseteq \cap \ldots\;$}
    (\cup V)^\c \;\subseteq\; \cap \{ B^\c \mid B \in V\}
\op\equiv\hint{by $\Ref{1}$ below}
    \true
\endcalc$$
Here we used twice that $$
\tag{1}
(\cup V)^\c \;=\; \cap \{ B^\c \mid B \in V\}
$$ from set theory.
Now we are left to prove
$$
\tag{2}
A \subseteq B \;\then\; h(A) \subseteq h(B)
$$
for any $\;A,B\;$.  For this we calculate as follows, starting at the most complex side:
$$\calc
    h(A) \subseteq h(B)
\op\equiv\hints{set theory -- prepare for OP's property 2, which seems}\hint{the only information about $\;h\;$ we can use}
    h(A) \cup h(B) = h(B)
\op\equiv\hint{OP's property 2}
    h(A \cup B) = h(B)
\op\when\hint{logic -- just about the only thing we can do}
    A \cup B = B
\op\equiv\hint{set theory}
    A \subseteq B
\endcalc$$
And that completes the proof.

We see that this proof indeed it uses only OP's properties 2 and 3, and not property 4.
Comparing the structure of this write-up with the earlier one, we see that Mathematician 42's comment corresponds to the first calculation up until $\Ref{*}$; and the rest is the second part of their answer followed by its first part.  Here, $\Ref{*}$ --expanded using $\Ref{1}$-- says that $\;h\;$ (i.e., taking the closure of a set) distributes over arbitrary intersections of complements of open sets (i.e., intersections of closed sets).
$%
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%$
