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Let $S$ be a topological space. It means that collection of open subsets of $S$ satisfying the following axioms:

  • $\varnothing$ and $S$ are open
  • any union of open sets is open
  • any finite intersection of open sets is open

Besides we may consider collection of closed sets of $S$. Question: can you give me an axioms of closed sets?

Thanks a lot!

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You might want to take a look at this related question : math.stackexchange.com/questions/189200/… –  Vishesh Oct 4 '12 at 16:43
    
This can be found in Wikipedia article on Topological space in the section Equivalent definitions. I'll add a link to the current revision - in case the article will change in the future. –  Martin Sleziak Oct 5 '12 at 7:57

3 Answers 3

Note that a set is closed if and only if its complement is open.

By this we have the following:

  1. $S$ and $\varnothing$ are closed.
  2. Any intersection of closed sets is closed.
  3. Finite unions of closed sets are closed.

All those follow immediately by DeMorgan laws and the first line in my answer.

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Sure:

  • $S$ and $\emptyset$ are closed
  • any intersection of closed sets is closed
  • any finite union of closed sets is closed
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Win for few seconds :) –  enzotib Oct 4 '12 at 16:41
1  
@Henning: I'm perplexed by the edit! –  Asaf Karagila Oct 4 '12 at 21:44
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@Asaf: The first axiom in the question is phrased "$\varnothing$ and $S$ are open". If we mechanically exchange "open" and "closed" and replace each of the named sets with its complement, then the one before the "and" becomes $S$ and the one after "and" becomes $\varnothing$. –  Henning Makholm Oct 4 '12 at 21:49
    
@Henning: Oh. I suppose that I should edit my post as well, then. Just when I thought I got "and" figured out as a commutative operator... :-) –  Asaf Karagila Oct 4 '12 at 21:50
    
@Asaf: It's not as if it's wrong -- I just thought I would complete Berci's effort to mimic the phrasing from the question exactly (even down to punctuation and capitalization). –  Henning Makholm Oct 4 '12 at 21:54

I would like to add something to this fruitful discussion.

Let $ X $ be a set, and denote its power set by $ \mathcal{P}(X) $. A function $ \text{cl}: \mathcal{P}(X) \rightarrow \mathcal{P}(X) $ is called an abstract closure operator on $ X $ if it satisfies the following four properties (called Kuratowski's Axioms):

  1. (Extensivity) $ \forall A \in \mathcal{P}(X): \, A \subseteq \text{cl}(A) $.
  2. (Idempotence) $ \forall A \in \mathcal{P}(X): \, \text{cl}(\text{cl}(A)) = \text{cl}(A) $.
  3. (Preservation of binary unions) $ \forall A,B \in \mathcal{P}(X): \, \text{cl}(A \cup B) = \text{cl}(A) \cup \text{cl}(B) $.
  4. (Preservation of nullary unions) $ \text{cl}(\varnothing) = \varnothing $.

Why did Kuratowski bother studying these axioms? The answer lies in the following observation. For any topology $ \tau $ on $ X $, we have a function $ \text{cl}_{(X,\tau)}: \mathcal{P}(X) \rightarrow \mathcal{P}(X) $ that maps a subset $ A $ of $ X $ to its closure with respect to $ \tau $. We call this function the topological closure operator on $ X $ with respect to $ \tau $. Kuratowski noticed that topological closure operators satisfy Kuratowski's Axioms (of course, he did not name the axioms after himself) and are therefore, by definition, abstract closure operators.

However, he went further and asked the following: Is every abstract closure operator a topological closure operator with respect to some topology $ \tau $? The answer turns out to be 'yes', and we even have that the corresponding $ \tau $ is unique. Here are the details. Given an abstract closure operator $ \text{cl} $ on $ X $, consider the set of fixed points of $ \text{cl} $: \begin{equation} \text{FixedPoints}(\text{cl}) := \{ C \in \mathcal{P}(X) \,|\, \text{cl}(C) = C \}. \end{equation} It is easily shown that $ \text{FixedPoints}(\text{cl}) $ satisfies the axioms for closed sets given by the two gentlemen in the earlier posts. In other words, $ \text{FixedPoints}(\text{cl}) $ forms a system of closed subsets of $ X $, which then yields a topology on $ X $. This topology is the $ \tau $ that we are looking for.

There is therefore a 1-1 correspondence between the set of abstract closure operators on $ X $ and the set of topologies on $ X $.

The main motivation for studying the axioms was therefore to isolate the essential properties that are satisfied by the topological closure operators of a topological space, which Kuratowski did with great success. This practice of abstracting mathematical properties from their usual context is highly important in mathematics, and it was precisely this sort of approach for which Alexandre Grothendieck was well-known.

The foregoing discussion can be generalized further. Suppose that we have a Boolean algebra $ \mathbb{B} = (B,+,\cdot,',0,1) $. Let $ \leq $ be the natural partial order of $ \mathbb{B} $. A closure operator on $ \mathbb{B} $ is defined to be a function $ f: B \rightarrow B $ that satisfies the following properties (also called Kuratowski's Axioms):

  1. $ \forall b \in B: \, b \leq f(b) $.
  2. $ \forall b \in B: \, f(f(b)) = f(b) $.
  3. $ \forall b_{1},b_{2} \in B: \, f(b_{1} + b_{2}) = f(b_{1}) + f(b_{2}) $.
  4. $ f(0) = 0 $.

With this further abstraction, we arrive at the subject of pointless topology. What pointless topology basically says is: Any theorem that is true for a general topological space and that is formulated in terms of subsets, their closures, their interiors and their boundaries can be formulated as a theorem about Boolean algebras equipped with closure operators. Hence, one can find a Boolean-algebraic proof of the topological theorem that does not involve points of a topological space at all.

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You can give a slightly better presentation of what is a closed set using the $\operatorname{cl}$ function. Either say that closed sets are exactly those in $\operatorname{rng}(\operatorname{cl})$; or better yet say that $C$ is closed if and only if $C=\operatorname{cl}(C)$. –  Asaf Karagila Oct 4 '12 at 22:51
    
Wikipedia is under a Creative Commons license, and it's not as if the language originated with them anyway. –  Kevin Carlson Oct 5 '12 at 1:41

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