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In https://en.wikipedia.org/wiki/Bornological_space it is said that

A bornology on a set X is a collection ℬ of subsets of X such that

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  • ℬ is stable under inclusions, i.e. if A ∈ ℬ and A′ ⊆ A, then A′ ∈ ℬ;

where "is stable under" sounds very much like the same as "is closed under" from https://en.wikipedia.org/wiki/Closure_(mathematics):

A set is closed under an operation if performance of that operation on members of the set always produces a member of that set.

When do I choose which expression and what is the difference between those expressions?

Is https://en.wikipedia.org/wiki/Invariant_(mathematics)#Invariant_set related to "is stable under" and if so, how?

Asaf Karagila
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Make42
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    I would say that a subset $A' \subseteq A$ cannot be said to have been produced by "operation on members of the set" – Ben Grossmann Sep 05 '20 at 10:35
  • My personal observation, based on mathematicians I have known. "Stable" is what you say if French is your native language. They may say "stable under countable unions" rather than the English way "closed under countable unions". – GEdgar Sep 05 '20 at 11:51

1 Answers1

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“Closure” is about operations. For instance, the set $\Bbb N$, as a subset of $\Bbb R$, is closed under multiplication (the product of two natural numbers is always a natural number), but not under division (the quotient of two natural numbers is not alwyas a natural number).

However, this doesn't apply to inclusions, since inclusion is not an operation. Being “stable under inclusions” means that if $A$ belongs to your set and if $A'$ is a subset of $A$, then $A'$ also belongs to your set.

José Carlos Santos
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  • So, "stable under X" means that X is a function with one input symbol and "closed under X" means that X is a function with two input symbols? (In both cases all symbols must be elements of the same set as the output of the respective functions.) – Make42 Sep 05 '20 at 10:42
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    No. Let $P$ be a property. Being stable under a relation $\sim$ means that if $A$ has the property $P$ and if $A\sim B$, then $B$ also has the property $P$. Being stable under an operation $o$ means that if both $A$ and $B$ have the propery $P$, then $o(A,B)$ also has the property $P$. – José Carlos Santos Sep 05 '20 at 10:52
  • I am not sure how this is different to what I wrote regarding functions. After all operations and stuff like unions are functions. I would say that a relation like "<" is also a function (where the boolean set is the codomain), but I am not sure. Can you give an example, maybe that helps me to understand what you mean. (Disclaimer: I am more an engineer / (so more applied math) than a pure mathematician.) – Make42 Sep 05 '20 at 10:58
  • If you don't mind to work with functions which map elements into boolean values, then, yes, what you wrote is equal to what I wrote. It's just a matter of taste. – José Carlos Santos Sep 05 '20 at 11:01
  • Ah, I start to see what you mean. In that case my first comment was wrong actually, because I said that "stable under" has one input symbol. Correct would be: "set S is closed under f" would bean that $f$ is a function $f: S \times S \rightarrow S$, while "set S is stable under f" would mean that f is a function $f: S \times S \rightarrow {0,1}$. Right? – Make42 Sep 05 '20 at 11:12
  • Let us be more precise. The set $X$ is closed under $f\colon S\times S\longrightarrow S$ if $A,B\in X\implies f(A,B)\in X$. And it is stable under $f\colon S\times S\longrightarrow{0,1}$ if$$A\in X\text{ and }f(A,B)=1\implies B\in X.$$ – José Carlos Santos Sep 05 '20 at 11:15
  • Isn't the relation between $X$ and $S$ missing in those? Is $X\subseteq S$ in each case? Why not substitute $X$ with $S$ in the formulas? – Make42 Sep 05 '20 at 11:17
  • I was assuming that $X\subset S$. – José Carlos Santos Sep 05 '20 at 11:19
  • But then, couldn't I just substitute $S$ with $X$ in the function definition? So just write $f: X \times X \rightarrow X$ and $f X \times X \rightarrow {0,1}$ instead? What would be the point of introducing the symbol $S$ at all? – Make42 Sep 05 '20 at 11:24
  • Because what we are talking about is when a subset $X$ of a set $S$ is “stable” or “closed” with respect to something. – José Carlos Santos Sep 05 '20 at 11:36
  • Let us use the case of bornological space as an example, so "$S$ is stable under inclusion", with $f$ is being the inclusion. So it would be $f: S \times S \rightarrow {0,1}$ and $A \in S \wedge f(A,B) = 1 \Rightarrow B \in S$. I do not see the need for an additional set like $X$. Of course we could also have written $f: X \times X \rightarrow {0,1}$ and $A \in X \wedge f(A,B) = 1 \Rightarrow B \in X$ instead and then there would not be need a for a symbol $S$. – Make42 Sep 05 '20 at 11:48
  • Let us continue this discussion in chat. – José Carlos Santos Sep 05 '20 at 13:22
  • Not sure if you get a notification, when I write in the chat... (which I did). – Make42 Sep 05 '20 at 14:51