I can't understand axiom of regularity....

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In summary, the axiom of regularity states that no set can be an element of itself, which means that the whole set cannot be considered as a single element. This is to avoid paradoxes such as the barber paradox. However, there are models of ZFC that do not satisfy this axiom and are still consistent. Therefore, the axiom of regularity is not essential to most of mathematics and can be omitted when studying sets that would violate it.
  • #1
ankit.jain
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It states that no set can be an element of itself..what does it exactly mean?why can't. We consider the whole set as a single element?
 
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  • #2
You have to exclude the possibility that a set can contain itself as an element because otherwise you will run into paradoxa: "The barber in this village shaves everyone except those who shave themselves."
So does the barber shave himself or not?

Of course you can consider sets which contain other sets as elements.
E.g. consider a set ##S=\{1,2,3\}##. Then the set of all subsets of ##S## is ##P(S) = \{∅,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}##.
Please regard that although ##S## is an element of ##P(S)## neither of both contains itself as an element. ##P(S)## is called power set of ##S##.
 
  • #3
fresh_42 said:
You have to exclude the possibility that a set can contain itself as an element because otherwise you will run into paradoxa: "The barber in this village shaves everyone except those who shave themselves."
So does the barber shave himself or not?

This is a very popular misconception. The axiom of regularity does not avoid any paradoxes. The barber paradox cannot be avoided by the axiom of regularity. It is the strengthening of the comprehension axiom that avoids the barber paradox.

Indeed, it has been shown that there are many nice models of ZFC- that are perfectly consistent (given ZFC- is consistent) and that do not satisfy the axiom of regularity. For an example of such models, see the nice book by Hrbacek and Jech.

As for the OP: it's just an axiom, and it's not an essential axiom. Almost all of math goes through perfectly fine if we choose to accept the axiom, or if we do not choose to accept the axiom. To 99% of mathematics, the axiom is irrelevant (and that can be given an informal argument). I myself choose not to accept the axiom of regularity because it is irrelevant. I never work with sets which are an element of itself anyway, even though I'm perfectly fine with them existing.
 
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  • #4
micromass said:
It is the strengthening of the comprehension axiom that avoids the barber paradox.
Thank you for clarification.

The axiom of regularity does not avoid any paradoxes. The barber paradox cannot be avoided by the axiom of regularity.
At this point I got stuck. If the axiom of regularity reads ∀x (x ≠ ∅ ⇒ ∃y ∈ x (y∩x = ∅) ) so why isn't the construction S = {S} forbidden by it, since x = y = S are the only elements and therefore in contradiction to the axiom. At which point am I wrong?
 
  • #5
fresh_42 said:
At this point I got stuck. If the axiom of regularity reads ∀x (x ≠ ∅ ⇒ ∃y ∈ x (y∩x = ∅) ) so why isn't the construction S = {S} forbidden by it, since x = y = S are the only elements and therefore in contradiction to the axiom. At which point am I wrong?

You are right that constructions of the form ##S = \{S\}## are forbidden by the regularity axiom. But that was not my point. My point is that it has been proven that the axiom of regularity is relative consistent with ZFC, but so is its negation. This means that if ZFC has a contradiction somewhere, then so will ZFC + regularity. The contradiction will just be slightly different, but it is there.

So if some kind of barber's paradox shows up without regularity, then some paradox will show up with regularity too.
 
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  • #6
Still not clear.
 
  • #7
ankit.jain said:
Still not clear.
What exactly is still not clear?
ankit.jain said:
It states that no set can be an element of itself..what does it exactly mean?why can't. We consider the whole set as a single element?
Yes, the whole set is considered as a single element. The axiom of regularity implies that for any set S, S∈S is not allowed.

As micromass explained, this axiom isn't terribly important for most of Mathematics.
 
  • #8
Yes. It isn't that important because development of the integers and arithmetic do not depend on it at all. It simply forces the notion of a "set" to agree with what our intuition says a set should represent. Infinite or circular inclusion chains do not make a whole lot of practical sense.
 
  • #9
How can you ever say that you have precisely defined a set that contains itself? I am curious if there is a well-defined example of such a thing. If not, I think it is irrelevant to mathematics.
 
  • #10
FactChecker said:
How can you ever say that you have precisely defined a set that contains itself? I am curious if there is a well-defined example of such a thing. If not, I think it is irrelevant to mathematics.

There are axioms of anti-regularity that deal with such things. So there are quite precise way of dealing with this. It's still true it's irrelevant to most mathematics though.
 
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  • #11
FactChecker said:
How can you ever say that you have precisely defined a set that contains itself? I am curious if there is a well-defined example of such a thing. If not, I think it is irrelevant to mathematics.
From my understanding, you don't have to define them. You simply omit the axiom of regularity/foundation and study the hypothetical "sets" that would violate it. [itex]\in[/itex] is just a binary predicate relating two objects we happen to call "sets" in set theory. Every object in the domain of the first-order system of ZFC is called a "set". It is the axioms themselves that demand "sets" operate as we would envision collections of "things" in the real world. Of course, we can have other things in the universe for which the extensionality axiom doesn't apply. You then have to add an "is a set" predicate and preface the extensionality axiom with it
 
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  • #12
mbs said:
From my understanding, you don't have to define them. You simply omit the axiom of regularity/foundation and study the hypothetical "sets" that would violate it. [itex]\in[/itex] is just a binary predicate relating two objects we happen to call "sets" in set theory. Every object in the domain of the first-order system of ZFC is called a "set". It is the axioms themselves that demand "sets" operate as we would envision collections of "things" in the real world. Of course, we can have other things in the universe for which the extensionality axiom doesn't apply. You then have to add an "is a set" predicate and preface the extensionality axiom with it

Of course we can just leave out the regularity axiom, but that doesn't get us far. All that negating the foundation axiom does, is saying there is one set for which the foundation axiom fails. You don't even have any information regarding the set. And even if you do. For example, take a set ##x## for which ##x=\{x\}##. Is such a set ##x## unique? How do we check equality with such sets? The extensionality axiom is useless here.
So just leaving out the regularity axiom is not useful, to have something useful, we should replace it with a better axiom for which we can actually prove things with. For example https://en.wikipedia.org/wiki/Aczel's_anti-foundation_axiom
 
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  • #13
mbs said:
From my understanding, you don't have to define them. You simply omit the axiom of regularity/foundation and study the hypothetical "sets" that would violate it. [itex]\in[/itex] is just a binary predicate relating two objects we happen to call "sets" in set theory. Every object in the domain of the first-order system of ZFC is called a "set". It is the axioms themselves that demand "sets" operate as we would envision collections of "things" in the real world. Of course, we can have other things in the universe for which the extensionality axiom doesn't apply. You then have to add an "is a set" predicate and preface the extensionality axiom with it
I think I understand. So leaving it out does not necessarily mean that there is, in fact, a set that contains itself. It just means that you can not use that axiom (no set contains itself) in proofs.
 
  • #14
FactChecker said:
I think I understand. So leaving it out does not mean that there is, in fact, a set that contains itself, it just means that you can not use that axiom (no set contains itself) in proofs.

You're correct of course, but I interpreted his post as assuming the negation of regularity.
 
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  • #15
micromass said:
You're correct of course, but I interpreted his post as assuming the negation of regularity.
I guess you could hypothetically define an example, just by saying that the set X has itself as an element. Although I would like to say that X is not well-defined, I would need to use the axiom to say that.
 
  • #16
micromass said:
Of course we can just leave out the regularity axiom, but that doesn't get us far. All that negating the foundation axiom does, is saying there is one set for which the foundation axiom fails. You don't even have any information regarding the set. And even if you do. For example, take a set ##x## for which ##x=\{x\}##. Is such a set ##x## unique? How do we check equality with such sets? The extensionality axiom is useless here.
So just leaving out the regularity axiom is not useful, to have something useful, we should replace it with a better axiom for which we can actually prove things with. For example https://en.wikipedia.org/wiki/Aczel's_anti-foundation_axiom
Okay. I don't think I said anything that contradicts anything you're saying. A set for which the axiom of foundation fails is simply a (non-empty) set S in which every member of S contains another member of S. Simply removing the foundation axiom keeps such sets purely hypothetical. You would need addition axioms to formally construct the existence of all such sets (much like the axiom of infinity constructs the integers). Simply having one non-regular set isn't enough to prove anything.
 
  • #17
micromass said:
What a fascinating area of study - I couldn't understand why I had not seen this before, then realized that I finished full-time academic study two years before Aczel's paper. Can you point me to a good, relatively discursive book? I was never very fond of books containing long formal proofs of theorem after theorem with no apparent motivation and 30 years out of study has not made me any more so.
 
  • #18
MrAnchovy said:
What a fascinating area of study - I couldn't understand why I had not seen this before, then realized that I finished full-time academic study two years before Aczel's paper. Can you point me to a good, relatively discursive book? I was never very fond of books containing long formal proofs of theorem after theorem with no apparent motivation and 30 years out of study has not made me any more so.

The book by Hrbacek and Jech contains a nice chapter on this anti-foundation thing. It's a really great book to read. https://www.amazon.com/dp/0824779150/?tag=pfamazon01-20
 
  • #19
micromass said:
The book by Hrbacek and Jech contains a nice chapter on this anti-foundation thing. It's a really great book to read. https://www.amazon.com/dp/0824779150/?tag=pfamazon01-20
Many thanks, ordered from the train on the way home and it will be with me on Sunday apparently - how did life work before Amazon?
 
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What is the axiom of regularity?

The axiom of regularity, also known as the axiom of foundation, is a fundamental principle in set theory that states that every non-empty set contains an element that is disjoint from the set itself. In other words, a set cannot be a member of itself.

Why is the axiom of regularity important?

The axiom of regularity is important because it ensures that sets do not contain circular or self-referential elements, which can lead to paradoxes and contradictions in mathematics. It also allows for a well-defined hierarchy of sets, which is essential for building a strong foundation for mathematical reasoning.

How does the axiom of regularity relate to other axioms in set theory?

The axiom of regularity is one of the axioms in the Zermelo-Fraenkel set theory, which is the standard system of axioms used in modern mathematics. It is closely related to other axioms such as the axiom of extensionality, which states that sets with the same elements are equal, and the axiom of choice, which allows for the creation of new sets from existing ones.

What are some examples of sets that violate the axiom of regularity?

One example is the set {1, {1}}, where the element {1} is also a subset of the set itself. Another example is the set of all sets, which contains itself as a member. These examples violate the axiom of regularity because they contain elements that are not disjoint from the set itself.

How can the axiom of regularity be used in mathematical proofs?

The axiom of regularity can be used in mathematical proofs to ensure that sets are well-defined and do not contain paradoxical elements. It can also be used to prove the existence of certain objects, such as the smallest element in a set, by showing that it is disjoint from the set itself.

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