# Axioms by Kuratowski, closure

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.

• What does $A^c$ mean? – Dog_69 Mar 7 '19 at 20:29
• @Dog_69 It means the complement of $A$ in $X$. So $A^c=X\setminus A$. – Cornman Mar 7 '19 at 20:30
• 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 $\tau$. – Dog_69 Mar 7 '19 at 20:33
• Did you translate that quoted block from German? – celtschk Mar 7 '19 at 20:51
• @celtschk Yes, I did. The task is taken from the book "Grundkurs Topologie" by Gerd Laures and Markus Szymik. – Cornman Mar 7 '19 at 20:56

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 $$X$$.

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