what is a closure (hull) operator?

Question

Just that. what is a closure operator?

reading the wiki wasn't enough and i would like to know more.

I'd be happy if someone shared examples of closure operators so that i may further understand their significance.

edited for clarity

Answer 1

WARNING: LONG POST.

To anyone trying to understand these operators, consider the following:

There's a need to always work with certain functions when doing certain manipulations on a set. The attributes of these functions make sure our answers adhere to a coherent set of results. This is why we define them axiomatically and make sure they fulfill these axioms.

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Let's consider $f: P(A) \rightarrow P(A)$ and its definition $f =: \{n | n,m \in \mathbb{N}, n \text{ divides } m\}$

if you don't know how to read this:

$f: P(A) \rightarrow P(A)$ means "a function named f, that maps some subset of A, to another subset of A (since we're mapping from the power set to the power set)"

$f =: \{n | n,m \in \mathbb{N}, n \text{ divides } m\}$ means "a function named f, which is defined to return the resulting set of all n's that satisfy the condition: n and m being elements of the natural numbers and n divides m"

Now for a concrete example:

$f$ of 2 is the result set {1,2} (because we return all the n's that divide 2)

$f$ of 4 is the result set {1,2,4}.

hey, that seems kind of related, right? {1,2} is a subset of {1,2,4}, interesting. what does that give us? (if you don't understand this part, you probably need to read up on some basic naive set theory)

Well, how about we check the attributes of a hull operator, and see if $f$ is just that:

According to https://en.wikipedia.org/wiki/Closure_operator :

1) A hull operator is extensive: $\forall A(A \subseteq f(A)) $ alt. form $\forall x\in A (x \in A \wedge x \in f(x))$

what does that mean? it means that for all of A (all of the elements of A) these elements of A are included in the result set.

does this hold for us? Yes. consider $f$ divides 4.

$f$ of 4 results in {1,2,4}, and we see that 4 is indeed an element of the result set.

2) A hull operator is increasing: $\forall A\forall B(B\subseteq A \rightarrow f(B)\subseteq f(A)$ alt. form $\forall x\in A\forall x\in B(x\in B \wedge x\in A \rightarrow x\in f(B) \wedge x\in f(A)$

I personally understand this as "a result set of an element must also be a member of a set of a larger element".

this isn't random, these elements need to be related to one another, and in our case, in that one of them divides the other.

understand that $f(A)$ is the function applied to a set. $f(\{1,2,4\})$ and the attribute "increasing" means that we should be able to take any subset of $f(4)$ and find that it is included in the result of $f(4)$.

so let's choose the greater subset $f(21) := \{1,3,7,21\}$

we see now that $f(1),f(3) \text{ and }f(7)$ are subsets of $f(21)$ in such a way that respectively we have: $\{1\},\{1,3\},\{1,7\} \text{ and } \{1,3,7,21\}$ with:

• $\{1\} \subset f(21)$
• $\{1,3\} \subset f(21)$
• $\{1,7\} \subset f(21)$
• $\{1,3,47,21\} \subseteq f(21)$

I hope this is clear.

3) A hull operator is idempotent: $\forall B(f(f(B)) = f(B))$

This is probably the easiest attribute that we must understand. An idempotent operator is an operator whose result does not change beyond its first application, no matter how many times we apply the operator on the result.

To fully grasp this notion, understand that if we apply an operator on a set, we are applying it on every element of the set and union`ing each result set to return the final result.

Thus:

$f(f(4)) =$ $f(\{1,2,4\}) =$ $f(1) \cup f(2) \cup f(4) =$ $\{1\} \cup \{1,2\} \cup \{1,2,4\} = \{1,2,4\} = f(4) = f(f(4))$

fin.