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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?

Jiew Meng
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3 Answers3

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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.

Martin Sleziak
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    I don't think you need regularity for working with inductive sets, the smallest inductive set is automatically well-founded. – Michael Greinecker Sep 30 '12 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.) – Martin Sleziak Sep 30 '12 at 14:10
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    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. – Michael Greinecker Sep 30 '12 at 14:17
  • Would regularity ever be used in a formal development of, say, number theory or real analysis? I can't imagine it. – Dan Christensen Oct 01 '12 at 13:30
  • @MartinSleziak Could we use "Zermelo Ordinals" instead of "standard construction"? Are they interchangeable for constructing numbers? – lockedscope Oct 24 '20 at 16:41
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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).

Asaf Karagila
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  • 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.) – Martin Sleziak Sep 30 '12 at 14:12
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    @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 $A={0,1,2,3,4,5}$ – celtschk Sep 30 '12 at 14:24
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    @Martin: No, ${1,2,3,4,5}\cap{1}=\varnothing$. – Asaf Karagila Sep 30 '12 at 14:29
  • Now I see. If every element of $A$ is a subset of $A$, then we have $x\cap A=x$ for each $x\in A$ and, by regularity, there must be $x\in A$ such that $x=x\cap A=\emptyset$. – Martin Sleziak Sep 30 '12 at 14:36
  • Both answers are hard to grasp, since it is unclear what element is and what set is. Is every element also set? Then when we have intersection, is it operation on elements or on the sets? – user1700890 Sep 17 '18 at 15:38
  • @user1700890: Yes. Everything is a set. – Asaf Karagila Sep 17 '18 at 15:54
  • @AsafKaragila thank you, then how {1,2,3,4,5}\cap{1}=\varnothing. You do not have to explain, just point me to relevant reading – user1700890 Sep 17 '18 at 17:46
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    @user1700890: That was a typo, it should have been $1$ rather than ${1}$. – Asaf Karagila Sep 17 '18 at 17:48
  • When we add then $ {0,1,2,3,4,5} ∩ 0 = {0,1,2,3,4,5 } ∩ {} = ∅ $, is this right? – lockedscope Oct 24 '20 at 16:44
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    @lockedscope: Yes. Intersecting any set with the empty the set is empty. – Asaf Karagila Oct 24 '20 at 17:06
  • Thanks @AsafKaragila btw may be seems obvious but i meant to "add 0" but forget to type. – lockedscope Oct 24 '20 at 17:09
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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 $A$.

jjjjjj
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