The title might be sort of confusing. The set is an infinite set like $$ A = \{ \emptyset, \{ \emptyset \}, \{ \{ \emptyset \} \}, ... \{ \emptyset, \{ \emptyset \} \}, \{ \{ \emptyset \}, \{ \{ \emptyset \} \} \}, ... ... \} $$ And it could be defined by the following rules

  • #1: ∅ ∈ A
  • #2: if a ∈ A, {a} ∈ A
  • #3: if a ∈ A and b ∈ A, a ⋃ b ∈ A

Consider a subset of 2 elements of A, like {x, y} ⊂ A, say, x ∈ A and y ∈ A, so {x} ∈ A and {y} ∈ A (#2), then {x} ⋃ {y} ∈ A (#3), as a result {x, y} is also an element of A.

Similar deductions could be applied to subsets of 3 or even more elements. But that should have no chance according to Cantor's theorem. What mistakes I've made and what is the power set of A?

  • $\begingroup$ Cantor's th is about a set $A$ and its power set $\mathcal P(A)$, i.e. the set containing all and only the subsets of $A$. It seems to me that you are trying to "build up" at the same time bot $A$ and $\mathcal P(A)$ ... I'm not sure you will be able to do it. $\endgroup$ Commented Mar 4, 2014 at 14:30
  • $\begingroup$ There is a sort of "circularity" in your definition: you will take a subset $\{ x, y \}$ of $A$ and you will define a new element of $A$ (i.e.$\{ x \} \cup \{ y \} = \{ x, y \}$). $\endgroup$ Commented Mar 4, 2014 at 17:15
  • $\begingroup$ @MauroALLEGRANZA Do you mean the rule #3 goes against axiomatic systems like ZFC? $\endgroup$
    – neuront
    Commented Mar 5, 2014 at 2:31
  • $\begingroup$ Exactly (see answer below); I'm not sure that you will be able to prove that the "construction" (rule #3) will give you a set ... $\endgroup$ Commented Mar 5, 2014 at 8:42
  • $\begingroup$ @MauroALLEGRANZA: It doesn't contradict ZFC, and in fact the inductive set given by the axiom of infinity actually satisfies all the desired properties. $\endgroup$
    – user21820
    Commented Jul 21, 2014 at 16:09

1 Answer 1


I understand that the problem is that the "set" A appears to contain all subsets of itself and therefore P(A) ⊂ A so that |P(A)| <= |A| which would certainly upset Cantor.

Taking ZFC as the axiomatic framework to work in, then I don't think that your "set" is a valid contruction. The ZFC axioms were laid down to exclude pathological cases created by arbitrary definition, like Russell's "set" R of all sets which are not members of themseleves (is R ∈ R ?????): by the axioms of ZFC Russell's R is not a set, and nor do I think is yours.

You would have to demonstrate the construction of A quoting from the 10 ZFC axioms as you go along why each step is vaild.

1) ∅ the empty set exists by the axiom of the empty set.

2) {∅} esists by axiom of pairing

3) {∅, {∅}} exists by axiom of paring

4) there exists a set containing ∅ and the "successor" of each of it's elements by the axiom of infinity where the successor of a is {a}

Bit of a problem though to find axioms that will support your third rule. You could construct a set which contains the union of two specific elements or even a finite number of such, but not I think an infinite collection that satisfied this condition.

On the other hand, if you start from the powerset (which exists by axiom of the powerset) P(A) then your rule three is valid, but not rule 2: the powerset does not contain the sucessor of all of its elements.

  • $\begingroup$ I see. I think the point is #3 and the deductions are valid only when any finite number of elements are selected to construct a union set, but axiom of union does not allow a union of infinite sets at once. $\endgroup$
    – neuront
    Commented Mar 5, 2014 at 9:18
  • $\begingroup$ Axiom of union will allow the union of an infinite collection of sets into one union. What you need though is an infinity of unions of sets, which is another matter. $\endgroup$ Commented Mar 5, 2014 at 9:36
  • $\begingroup$ Actually the axiom of infinity guarantees a set that does have all three properties required, because each (un)natural number is both an element and a subset of its successor. $\endgroup$
    – user21820
    Commented Jul 21, 2014 at 16:05
  • $\begingroup$ Of course, as both of you noticed, property (3) only guarantees that $A$ contains all finite subsets of $A$, which is nowhere as 'big' as the power set of $A$. $\endgroup$
    – user21820
    Commented Jul 21, 2014 at 16:07
  • $\begingroup$ @neuront: Better still, we can even start with any infinite set, and create a sequence of sets by repeatedly adding all finite subsets of the previous set, and finally by taking the union of all of them. Then the result will have the same cardinality as the original and yet satisfy all the three desired properties. $\endgroup$
    – user21820
    Commented Jul 21, 2014 at 16:13

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