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Suppose $f: \mathcal{P}(X) \to \mathcal{P}(X)$ is a function that

satisfies, for every set $A,B \subseteq X$

$(C_1): f(\emptyset) = \emptyset$

$(C_2): A \subseteq f(A)$

$(C_3): f(A \cup B) =f(A) \cup f(B)$

$(C_4): f(f(A)) = f(A)$

Prove that there exists a unique topology $\mathcal{T}$ on $X$ such that for all $A \subseteq X$, $\text{Cl}(A) = f(A)$, where $\text{Cl}(A)$ denotes the closure of the set $A$, with respect to the topology $\mathcal{T}$

My attempt:

I defined $\mathcal{T}:= \{A \subseteq X \mid f(X \setminus A) = X \setminus A\}$

and I managed to prove with $C_1, C_2, C_3$ that this set is a topology, and that it is unique, if it exists. So I only need to show that it exists, for which I have to prove that $\text{Cl}(A) = f(A)$ whenever $A \subseteq X$

I can show that $f(A) \supseteq \text{Cl}(A)$. I just need the other inclusion and I'm done!

Any hints?

Asaf Karagila
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    The closure $\mathrm{cl}(A)$ is the intersection of all closed sets that contain $A$. So it would be enough to show that $f(A)$ is closed. – Jan Bohr Feb 25 '18 at 16:21
  • "and that it is unique, if it exists" $-$ $\mathcal{T}$ definitely exists, you just defined it. – Arnaud Mortier Feb 25 '18 at 16:21
  • This is "Kuratowski closure axiom" – Qurultay Feb 25 '18 at 16:24
  • I made a mistake. I actually had trouble proving the other inclusion (I edited the post accordingly) –  Feb 25 '18 at 17:29
  • What do you mean by "I only need to show it exists." when clearly it is defined, is nof empty and exists. – William Elliot Feb 25 '18 at 21:25
  • It may help to show f is increasing. – William Elliot Feb 25 '18 at 21:28
  • Showing that a topology exists WITH the condition I want –  Feb 26 '18 at 08:25
  • No, we must prove that $f(A) = Cl_{\mathcal{T}}(A) := {x \in X \mid \forall V \in \mathcal{V}_{\mathcal{T}}(x): V \cap A \neq \emptyset}$. I don't see at all how this is obvious? –  Feb 26 '18 at 14:36
  • If you can show that $f(A)⊇\operatorname{Cl}(A)$ for all $A\in\mathcal{P}(X)$ then in particular substituting $f(A)$ in place of $A$, you have, $$\operatorname{Cl}(f(A))\subseteq f(f(A))=f(A)$$From this you can conclude that $f(A)$ is closed. Now follow @jabo's hint. –  Feb 26 '18 at 15:23
  • One thing that may be useful is to show that, if $A\subseteq B$, then $f(A)\subseteq f(B)$. (Hint: since $A\subseteq B$, we have $B=A\cup(B\setminus A)$. Apply $f$ to both sides.) – Akiva Weinberger Feb 26 '18 at 16:01
  • Also, (a) $A\subseteq{\rm Cl}(A)$. (b) Due to the way you've defined $\cal T$, we have that $f$ of a closed set is itself. – Akiva Weinberger Feb 26 '18 at 16:04

1 Answers1

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I see that you define the closure of $A$ as $\operatorname{Cl}_{\mathcal{T}}(A) = \{x \in X: \forall O \in \mathcal{T}: x \in O \to O \cap A \neq \emptyset \}$, i.e. the set of adherent points of $A$.

For closure operators it's more convenient (and dual to the way we can define interiors) to use

$$\operatorname{Cl}_{\mathcal{T}}(A) = \bigcap \{F \subseteq X: A \subseteq F, F \text{ closed }\} (\ast)$$

which says that the closure of $A$ is the smallest closed set that contains $A$ as a subset.

It's a classic excercise in introductory topology to show that these definitions are equivalent.

Also the closure axioms (as in the case of interior operators) imply that $f$ is monotonic: $A \subseteq B$ implies $f(A) \subseteq f(B)$:

Proof: $A \subseteq B$ implies $A \cup B = B$ so $(C_3)$ tells us that $f(B) = f(A \cup B) = f(A) \cup f(B)$ so that $f(A) \subseteq f(B)$ follows.

Also note that in your definition we could just as easily have defined $A$ is closed iff $f(A) = A$.

Now if $A$ is a subset of $X$ we have that $A \subseteq f(A)$ by $(C_2)$ and $f(A)$ is closed by $(C_4)$, as $f(f(A)) = f(A)$. So $f(A)$ is one of the closed sets we take the intersection of in $(\ast)$ so that $\operatorname{Cl}_{\mathcal{T}}(A) \subseteq f(A)$.

On the other hand, if $F$ is one of the closed sets in the right hand side of $(\ast)$, we know that $A \subseteq F$ and $F$ is closed so that by monotonicity we know that $f(A) \subseteq f(F) = F$, and as this holds for all $F$, $f(A) \subseteq \operatorname{Cl}_{\mathcal{T}}(A)$ and we have equality of $f$ and the closure operator of $\mathcal{T}$ for any $A \subseteq X$.

Note the duality with your earlier interior operator here. Given an interior operator $i$ we can define a closure operator $c(A) = X\setminus i(X\setminus A)$. Standard set theory gives that axioms $(I_1)-(I_4)$ there then are equivalent to $(C_1)-(C_4)$ above. We can reduce one excercise to the other, essentially. $i(A) = X\setminus c(X\setminus A)$ allows us to go from a closure operator to an interior operator in the same way.

Henno Brandsma
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  • Thanks for your answer! I just solved the problem using the hints in the comments, by showing that $f(B)$ is the smallest closed set that contains $B$, hence $f(B) = cl(B)$. Your answer shows that my attempt is correct! –  Feb 26 '18 at 20:00
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    @Math_QED Glad I could help. I've always loved these "basic" exercises to show the equivalence of different ways to define a topology (e.g. also by neighbourhood systems or filter convergence systems or rather obscurely by the boundary operator etc.) – Henno Brandsma Feb 26 '18 at 20:05
  • Is the converse of the claim statement true? That is, is it true that if $X$ be a non-empty set and $\tau$ be a topology on $X$ then there exists a unique closure operator on $X$ such that the topology induced by it is unique? –  Aug 13 '18 at 16:42
  • @user170039 yes, the standard closure induced by $\tau$ is such. – Henno Brandsma Aug 13 '18 at 16:44
  • Where can I find a proof of this fact? –  Aug 13 '18 at 16:45
  • I mean the proof of uniqueness of the closure operator of which you talked earlier. –  Aug 13 '18 at 16:49
  • @user170039 it's obvious. It follows from the above fact. – Henno Brandsma Aug 13 '18 at 16:50
  • If you don't mind then can you come to this room. I am having trouble to understand it and don't want to lengthen the comment chain here. –  Aug 13 '18 at 16:54
  • The part that I am having trouble is the following. Let $(X,τ)$ be a topological space and $\text{Cl}_τ$ be the closure operator induced by $τ$ and $\text{Cl}$ be any other closure operator which induces the same topology $τ$. Then how can I show that for all $A$ we have $\text{Cl}_τ(A)=\text{Cl}(A)$? –  Aug 13 '18 at 17:02
  • @user170039 both closures must equal the intersection of all $\tau$-closed sets that contain $A$. As $\tau$ is given, these are the same closures. – Henno Brandsma Aug 13 '18 at 17:48
  • Thanks. I think I figured it out. –  Aug 14 '18 at 14:57