From Wikipedia:

The axiom of regularity is an axiom of Zermelo–Fraenkel set theory that states that every non-empty set A contains an element that is disjoint from A.

I'm sure I'm grossly misunderstanding this, but it doesn't seem to make any sense. Two disjoint sets do not have elements in common, correct? So this says a set contains an element that it is disjoint with. So wouldn't this mean that the set doesn't contain that element if it is disjoint with it? To me, this axiom seems to be saying "Every non-empty set contains an element that it doesn't contain." This obviously doesn't make sense, so I'm definitely misunderstanding at least one aspect of this axiom, but which part?

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    $\begingroup$ To me, this axiom seems to be saying "Every non-empty set contains an element that it doesn't contain." But it doesn't say "an element that it does not contain" it says "an element that it does not intersect (meaning that it shares no elements with it.) Contain and intersect do not mean the same thing. The thing that is probably most confusing is the act of taking the intersection of a set with one of its elements (which is perfectly admissible, but a little disorienting compared to our everyday use of sets.) $\endgroup$
    – rschwieb
    Commented May 18, 2017 at 13:53
  • $\begingroup$ It's worth pointing out that the "it doesn't seem to make any sense" objection you're raising is the exact objection that Haskell's type system will raise if you try to do this the naive way (type Set x = [x], when you try to do intersect s1 s2 and s2 is an element of s1, it will complain that it can't unify the types Set x and Set (Set x) together). $\endgroup$
    – CR Drost
    Commented May 18, 2017 at 15:11
  • $\begingroup$ $\in\ne\subset$. $\endgroup$ Commented May 19, 2017 at 9:03

3 Answers 3


As an example, let $X = \{1,2\}$ (where $0=\varnothing, 1= \{0\}, 2=\{0,1\}$). Then $$X\cap 1 = \varnothing$$ $$X\cap 2 = \{1\}$$ Due to the first equality, our set satisfies the axiom of regularity. $X$ being disjoint with $1$ does not imply that $X$ does not contain $1$; $X\cap 1 = \varnothing$ is a completely different statement from $1\notin X$. Indeed, the former holds, as the only element of $1$ is $0$, which is not an element of $X$.

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    $\begingroup$ I suppose my problem was in how I was thinking of numbers as sets. I suppose I assumed that if an element is a number, that's equivalent to a set containing that number. However, I can't actually assume that, because while there is no one right way to define 1 or 2 as a set, if we take the axiom of regularity, we can't define a set to contain itself. $\endgroup$
    – RothX
    Commented May 18, 2017 at 13:56
  • 3
    $\begingroup$ Right, having sets contain themselves definitely contradicts regularity. $\endgroup$
    – florence
    Commented May 18, 2017 at 13:58
  • $\begingroup$ @florence: More to the point, disallowing sets to (indirectly) contain themselves is the whole purpose of the axiom. $\endgroup$
    – Kevin
    Commented May 19, 2017 at 1:52
  • $\begingroup$ @Kevin It's a bit of an exaggeration to say that it's the whole purpose of that axiom. It also disallows infinite $\in$-descending sequences of sets even though they don't involve any set indirectly containing itself. $\endgroup$
    – Adayah
    Commented Oct 11, 2018 at 19:26

In a nutshell, the axiom forbid "loops".

Consider he case $A= \{ a_0, a_1, a_2 \}$.

If the axiom does not hold, we have the possibility that: for all $i$ : $a_i \cap A \ne \emptyset$.

1) Consider $a_0$; if the intersection of $a_0$ and $A$ is not empty, it must be an element of $A$.

Thus, three possibilities: i) $a_0 \in a_0$: loop; ii) $a_1 \in a_0$; iii) $a_2 \in a_0$.

Two possibilities left to avoid loops.

2) Consider $a_1$; again, if the intersection of $a_1$ and $A$ is not empty, we have: i) $a_0 \in a_1$; ii) $a_1 \in a_1$: loop; iii) $a_2 \in a_1$.

But $a_0 \in a_1$ and the previous $a_1 \in a_0$ form a loop: $a_0 \in a_1 \in a_0$.

Thus what remain is: $a_2 \in a_0$ and $a_2 \in a_1$.

Now: 3) Consider $a_2$; we have: i) $a_0 \in a_2$; ii) $a_1 \in a_2$; iii) $a_2 \in a_2$: loop.

Again, with: $a_0 \in a_2$ and $a_2 \in a_0$ we have a loop and the same with $a_1 \in a_2$ and $a_2 \in a_1$.

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    $\begingroup$ Rather than loops, it forbids infinite descent of any kind (looping around or not). (Or at least, infinite descent you can actually see.) $\endgroup$
    – tomasz
    Commented May 18, 2017 at 14:22

Your confusion seem to be due to levels of containment.

The members of a set is only those elements that are directly contained in the set. By this I mean that a set may contain other sets, but that does not make the elements of that set a member of the outer set.

This is a bit in contradiction in how we think of containment in ordinary life. Let's for example say you have a box of chocolates in your kitchen drawer. Then in ordinary life you would say that the drawer contains chocolates, but in mathematics terminology it doesn't it only contains a box (which in turn contains chocolates). Now you see that in mathematical terms the kitchen drawer and the box of chocolates are disjoint (but not in ordinary language).

To imagine a set that violates the axiom of regularity would probably need you to have some form of self containment. That is a container that in some way contains itself - which is quite contra intuitive, it doesn't fit well with the way we think of things.

  • 2
    $\begingroup$ I think the canonical example of a set violating the axiom of regularity is a box containing only itself. In this case, the set's only element is itself, which it clearly isn't disjoint from. Since the only thing that describes the identity of the set is the set itself, the definition of the set isn't "founded" on anything else. I think this is why the axiom also goes by the name "axiom of foundation". $\endgroup$ Commented May 18, 2017 at 19:17
  • 1
    $\begingroup$ After combing the web for a plain English explanation of the axiom of regularity for the better part of 30 minutes, you've finally cracked the nut for me @skyking. I was quite stuck on assuming that a set contains the elements of its subsets, i.e., the drawer contains the chocolates. Thank you for clearing this up! $\endgroup$
    – krry
    Commented Jan 28, 2019 at 22:12

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