Let $X$ be a set and $h:\mathcal{P}(X)\to\mathcal{P}(X)$ a function with the following properties:
(1) $h(\emptyset)=\emptyset$
(2) $A\subseteq hA$
(3) $hhA=hA$
(4) $h(A\cup B)=hA\cup hB$
for every $A,B\subseteq X$.
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: $\tau:=\{A\subseteq X| h(A)^c\subseteq X\}$
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:
$\emptyset\in\tau$. Because it is $X\subseteq hX\subseteq X$ by property (2) and $h$ mapping onto $\mathcal{P}(X)$. So $hX=X$ and $h(X)^c=\emptyset$.
$X\in\tau$, because it is $h(\emptyset)=\emptyset$ by property (1). And then $h(\emptyset)^c=X$.
Now let $A,B\subseteq X$ be elements of $\tau$.
I have to show, that $A\cap B\in\tau$.
So it has to hold $h(A\cap B)^c\subseteq X$, and this is kinda suspicious, because $h:\mathcal{P}(X)\to\mathcal{P}(X)$ 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.
Best Answer
Your definition of $\tau$ isn't right. By your definition $\tau$ is simply the powerset of $X$. Here is likely what you want:
$$\tau=\{X\setminus h(A)\mid A\subseteq X\}$$
Then $X\setminus h(\emptyset)=X\in\tau$ and $X\setminus h(X)=\emptyset\in\tau$.
If $A,B\in\tau$ then $A=X\setminus h(U)$ and $B=X\setminus h(V)$ for some $U,V\subseteq X$. Then
$$A\cap B=(X\setminus h(U))\cap(X\setminus h(V))=X\setminus(h(U)\cup h(V))=X\setminus h(U\cup V)$$
Therefore $A\cap B\in\tau$.
Finally if $\{A_{\alpha}\}_{\alpha\in I}$ is some collection of elements in $\tau$ then say that $A_{\alpha}=X\setminus h(U_{\alpha})$ for some $U_{\alpha}\subseteq X$. Then
$$\bigcup_{\alpha\in I}A_{\alpha}=\bigcup_{\alpha\in I}(X\setminus h(U_{\alpha}))=X\setminus\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
To finish this line we will need your third and second axioms.
$$\bigcap h(U_{\alpha})\subseteq h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
For each $\alpha\in I$ we have that $\bigcap_{\beta\in I}h(U_{\beta})\subseteq h(U_{\alpha})$
Thus by monotonicity and idempotence (axioms two and three) we have
$$h\left(\bigcap_{\beta\in I}h(U_{\beta})\right)\subseteq hh(U_{\alpha})=h(U_{\alpha})$$
for each $\alpha$. We then have that
$$h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)\subseteq\bigcap_{\alpha\in I}h(U_{\alpha})\subseteq h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
Establishing equality between the two sets. We then have that
$$X\setminus\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)=X\setminus h\left(\bigcap_{\alpha\in I}h(U_{\alpha})\right)$$
Therefore, the union over the family $\{A_{\alpha}\}_{\alpha\in I}$ is an element of $\tau$, establishing that $\tau$ is indeed a topology on $X$.