Dtjytyk

Let $X$ be a set and $h:mathcal{P}(X)tomathcal{P}(X)$ a function with the following properties:

(1) $h(emptyset)=emptyset$

(2) $Asubseteq hA$

(3) $hhA=hA$

(4) $h(Acup B)=hAcup hB$

for every $A,Bsubseteq X$.
There exists exactly one topology on $X$ such that for every subset $A$ in $X$ the set $hA$ is the closure of $A$ with regards to that topology.

I tried to define this topology $tau$ by: $tau:={Asubseteq X| h(A)^csubseteq X}$

Now I want to show, that this is well-defined and indeed a topology.
The definition makes sense, as far as I can tell, because $hA$ has to be the closure of $A$ and therefor $h(A)^c$ has to be open.
By definition of $tau$ the sets $h(A)^c$ are open.

Now for the axioms of the topology:

$emptysetintau$. Because it is $Xsubseteq hXsubseteq X$ by property (2) and $h$ mapping onto $mathcal{P}(X)$. So $hX=X$ and $h(X)^c=emptyset$.

$Xintau$, because it is $h(emptyset)=emptyset$ by property (1). And then $h(emptyset)^c=X$.

Now let $A,Bsubseteq X$ be elements of $tau$.
I have to show, that $Acap Bintau$.

So it has to hold $h(Acap B)^csubseteq X$, and this is kinda suspicious, because $h:mathcal{P}(X)tomathcal{P}(X)$ and my definition of $tau$ might be bad...

Is the definition of $tau$ correct?
Hints are appreciated, I would like to try again on my own.

Thanks in advance.

Your definition of $tau$ isn't right. By your definition $tau$ is simply the powerset of $X$. Here is likely what you want:

$$tau={Xsetminus h(A)mid Asubseteq X}$$

Then $Xsetminus h(emptyset)=Xintau$ and $Xsetminus h(X)=emptysetintau$.

If $A,Bintau$ then $A=Xsetminus h(U)$ and $B=Xsetminus h(V)$ for some $U,Vsubseteq X$. Then

$$Acap B=(Xsetminus h(U))cap(Xsetminus h(V))=Xsetminus(h(U)cup h(V))=Xsetminus h(Ucup V)$$

Therefore $Acap Bintau$.

Finally if ${A_{alpha}}_{alphain I}$ is some collection of elements in $tau$ then say that $A_{alpha}=Xsetminus h(U_{alpha})$ for some $U_{alpha}subseteq X$. Then

$$bigcup_{alphain I}A_{alpha}=bigcup_{alphain I}(Xsetminus h(U_{alpha}))=Xsetminusleft(bigcap_{alphain I}h(U_{alpha})right)$$

To finish this line we will need your third and second axioms.

$$bigcap h(U_{alpha})subseteq hleft(bigcap_{alphain I}h(U_{alpha})right)$$

For each $alphain I$ we have that $bigcap_{betain I}h(U_{beta})subseteq h(U_{alpha})$

Thus by monotonicity and idempotence (axioms two and three) we have

$$hleft(bigcap_{betain I}h(U_{beta})right)subseteq hh(U_{alpha})=h(U_{alpha})$$

for each $alpha$. We then have that

$$hleft(bigcap_{alphain I}h(U_{alpha})right)subseteqbigcap_{alphain I}h(U_{alpha})subseteq hleft(bigcap_{alphain I}h(U_{alpha})right)$$

Establishing equality between the two sets. We then have that

$$Xsetminusleft(bigcap_{alphain I}h(U_{alpha})right)=Xsetminus hleft(bigcap_{alphain I}h(U_{alpha})right)$$

Therefore, the union over the family ${A_{alpha}}_{alphain I}$ is an element of $tau$, establishing that $tau$ is indeed a topology on $X$.