# Axiom of Regularity

I am having difficulty understanding Axiom of Regularity :

Every non-empty set $\rm A$ contains an element $\rm B$ which is disjoint from $\rm A.$

So from my understanding if I have a set like:

$$\{1, 2, 3, 4, 5\}$$

Then one of $[1,2,3,4,5]$ is not an element in $\rm A?$ huh?

• It says that one of these contains none of 1,2,3,4,5 as an element. Commented Sep 30, 2012 at 13:45

In axiomatic set theory everything is a set, you don't work with other objects.

So even if you denote some things by $1$, $2$, $3$, $4$, $5$, they are in fact sets.

In fact, if we look at your example and use the standard construction of positive integers in ZFC, then we have
$0=\emptyset$,
$1=\{0\}$,
$2=\{0,1\}$,
$3=\{0,1,2\}$,
$4=\{0,1,2,3\}$ and
$5=\{0,1,2,3,4\}$.

Axiom of regularity says that one of the elements of the set $A=\{1,2,3,4,5\}$ is a set, which is disjoint with $A$. Indeed, $1$ is such set -- the only element of $1$ is $0=\emptyset$, which is not an element of $A$; hence $1\cap A=\emptyset$.

This might be unusual viewpoint for someone used to work in naive set theory, but once again, the basic idea is that: Everything is a set. We have some axioms, which allow us to create new sets from the sets we have already constructed. To work in this setting we try to create models of various things using these axioms. So each integer, rational number, real number will be modeled in ZFC as some set.

Let me also say that you probably don't need to worry too much about Axiom of Regularity if you are just beginning to study axiomatic set theory. You will only need this axiom much later (perhaps when you encounter cumulative hierarchy, where this axiom ensures that every set can be obtained by this repeated process of taking unions and power sets or when you encounter inductive sets in Axiom of Infinity and construction of natural numbers.) A lot of stuff can be done without this axiom.

• I don't think you need regularity for working with inductive sets, the smallest inductive set is automatically well-founded. Commented Sep 30, 2012 at 14:02
• @Michael Maybe I don't remember this correctly, but isn't regularity used as an argument, why an inductive set must be infinite? (The word infinite is used in an intuitive sense here, not as some precise mathematical notion.) So the name Axiom of Infinity is used, although it guarantees existence of an inductive set. (It is not called Axiom of Inductive Set.) Commented Sep 30, 2012 at 14:10
• The definition of inductive set I know says that $I$ is inductive if (a) $\emptyset\in I$ and (v) $x\in I$ implies $x\cup\{x\}\in I$. So we have $0=\emptyset\in I$, $1=\{\emptyset\}=\{0\}\in I$, $\{0,1\}=2\in I$ etc. So there are infinitely many elements in every inductive set. If we would change the definition so that (a) gets replaced by (a'): $I$ is nonempty, then $x=\{x\}$ would imply that $x$ is an inductive set, but with the standard definition, an inductive set must contain infinitely many elements. Commented Sep 30, 2012 at 14:17
• Would regularity ever be used in a formal development of, say, number theory or real analysis? I can't imagine it. Commented Oct 1, 2012 at 13:30
• @MartinSleziak Could we use "Zermelo Ordinals" instead of "standard construction"? Are they interchangeable for constructing numbers? Commented Oct 24, 2020 at 16:41

The axiom of regularity says that one of $1,2,3,4,5$ is disjoint from $\{1,2,3,4,5\}$, there is some $x\in\{1,2,3,4,5\}$ such that $x\cap\{1,2,3,4,5\}=\varnothing$.

It may sounds a bit weird, but in modern set theory everything is a set.

For example take the set $\{\varnothing\}$, it has one element and indeed $\varnothing\cap\{\varnothing\}=\varnothing$. It does not imply that $\varnothing\notin\{\varnothing\}$.

So the axiom tells us that every set $A$ either has $\varnothing\in A$, or it has some element $x$ which is not a subset of $A$ (in fact $x\cap A=\varnothing$, which is a stronger requirement).

• Now, after the correction, I still disagree. If you take $A=\{1,2,3,4,5\}$ (or any ordinal), every element of this set is a subset of it. (Sorry for nitpicking - I was not sure what you want to say, which is why I did not try to correct the typo myself.) Commented Sep 30, 2012 at 14:12
• @MartinSleziak: With the standard construction of the natural numbers, no element of this set is a subset, because all of them contain the empty set, which is not an element of $A$. You probably meant $A=\{0,1,2,3,4,5\}$ Commented Sep 30, 2012 at 14:24
• @Martin: No, $\{1,2,3,4,5\}\cap\{1\}=\varnothing$. Commented Sep 30, 2012 at 14:29
• @user1700890: That was a typo, it should have been $1$ rather than $\{1\}$. Commented Sep 17, 2018 at 17:48
• @lockedscope: Yes. Intersecting any set with the empty the set is empty. Commented Oct 24, 2020 at 17:06

Terry Tao gives the following example in his Analysis I book (section 3.2). Consider the following set of sets $$A=\big\{\; \{3,4\},\;\{3,4,\{3,4\}\}\;\big\}.$$ Then take $$x=\{3,4\}$$. We know that $$x\in A$$ and that $$x$$ is a set. Since $$x$$ is a set, we can ask: what is $$x\cap A$$? Computing $$x\cap A = \{y: y\in x\text{ and } y\in A\} = \{y: (y=3\text{ xor } y=4) \text{ and } y\in A\} = \emptyset$$ since neither $$3$$ nor $$4$$ are elements of $$A$$. Thus we've found an element $$x\in A$$ that is disjoint from $$A$$.

• Honestly, this was the most useful example. Commented Aug 25, 2022 at 9:56