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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.

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  • $\begingroup$ What does $A^c$ mean? $\endgroup$ – Dog_69 Mar 7 '19 at 20:29
  • $\begingroup$ @Dog_69 It means the complement of $A$ in $X$. So $A^c=X\setminus A$. $\endgroup$ – Cornman Mar 7 '19 at 20:30
  • $\begingroup$ But then your topology is discrete. $h$ gives you an element of $P(A)$ and hence the complementary will be too. You need to modify your definition of $\tau$. $\endgroup$ – Dog_69 Mar 7 '19 at 20:33
  • $\begingroup$ Did you translate that quoted block from German? $\endgroup$ – celtschk Mar 7 '19 at 20:51
  • $\begingroup$ @celtschk Yes, I did. The task is taken from the book "Grundkurs Topologie" by Gerd Laures and Markus Szymik. $\endgroup$ – Cornman Mar 7 '19 at 20:56
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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 $X$.

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    $\begingroup$ Equivalently, $\tau = \{ A \subseteq X \mid h(A^c) = A^c \}$. $\endgroup$ – Daniel Schepler Mar 7 '19 at 20:54
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    $\begingroup$ You mean "For each $\alpha\in I$ we have $\bigcap_{\beta\in I} h(U_\beta)\subseteq h(U_\alpha)$", right? $\endgroup$ – Cornman Mar 7 '19 at 21:15
  • $\begingroup$ Yes I do, thank you. $\endgroup$ – Robert Thingum Mar 7 '19 at 21:16
  • $\begingroup$ How can one deduce, that $\tau$ is unique? Is it because it depends on $h$ only and the $h(A)^c$ beeing unique? $\endgroup$ – Cornman Mar 7 '19 at 21:24
  • $\begingroup$ You can show that $h(A)$ is the closure of $A$ in $\tau$ and of course the complements of the closed sets are uniquely determined. $\endgroup$ – Robert Thingum Mar 7 '19 at 21:27

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