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Let $\mathcal{P}(X)$ denote the powerset of $X$, $\mathcal{P}^2(X)=\mathcal{P}(\mathcal{P}(X))$, and $\mathcal{P}^n(X)=\mathcal{P}(\mathcal{P}^{n-1}(X))$; $\mathcal{P}^0(X)=X$. It is trivial to show that $\mathcal{P}^{n-1}(X)\in\mathcal{P}^n(X)$ for all $n\in\mathbb{N}$.

(The following provides a loose justification for the subsequent uses of infinity and can be skipped if desired. For more rigorous justification, see the works of Abraham Robinson.)

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[ Let $S(m)=m+1$ denote the successor of $m$, $S^2(m)=S(S(m))$, and $S^n(m)=S(S^{n-1}(m))$; $S^0(m)=m$. It is trivial to show that, for an infinite hyperinteger $h$, $S^h(m)\geq x$ for all $x\in\mathbb{R}$. It is trivial to show that $S^k(m)-S^l(m)=k-l$ for all $k,l\in\mathbb{N}$.

By definition $\bigcup_{n=1}^\infty \left\{S^n(1)\right\}=\mathbb{N}$*

Let $\widehat{\mathbb{N}}$ denote the set of hypernatural numbers (that is, $\forall n\in\mathbb{N}.\forall h\in\widehat{\mathbb{N}}\setminus\mathbb{N}.h>n$, and $h\in\widehat{\mathbb{N}}\iff h+1\in\widehat{\mathbb{N}}$). It follows that for $h,k\in\widehat{\mathbb{N}}$ $S^h(k)\in\widehat{\mathbb{N}}$, regardless of whether $S^h(k)$ is finite or infinite.

For $h_1,h_2\in\widehat{\mathbb{N}}\setminus\mathbb{N}$ the difference between $h_1$ and $h_2$ is finite iff $|h_1-h_2|\in\mathbb{N}$. This follows from the fact that if $h_1$ and $h_2$ are of the same order, then $h_1=S^n(h_2)$ xor $h_2=S^n(h_1)$ for some $n\in\mathbb{N}$. ]


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Now, for a particular kind of infinity in the extended real numbers*, we have that $\infty-1=\infty$. This suggests that $\mathcal{P}^{\infty-1}(X)=\mathcal{P}^\infty(X)$, and thus $\mathcal{P}^\infty(X)\in\mathcal{P}^{\infty}(X)$. If $X$ is tentatively "the set of all sets," then the powerset of $X$, itself being a set, is an element of $X$ - leading to Cantor's Paradox. As $\mathcal{P}^{\infty-1}(X)=\mathcal{P}^\infty(X)$, $\mathcal{P}^\infty(X)$ circumvents this to a degree. (additionally, the cardinality of $\mathcal{P}^{\infty+1}(X)$, assuming that $\infty$ is the upper bound of the cardinal numbers, is $2^\infty=\infty$, thus $\mathcal{P}^\infty(X)$ has the same cardinality as that of its powerset).

However, the Axiom of Foundation states that no set can contain itself, so either $\mathcal{P}^\infty(X)$ is not a set, or $\mathcal{P}^\infty(X)$ does not contain itself. Given that the latter leads to a contradiction, let us suppose the former.

For all $n\in\mathbb{N}$, $\mathcal{P}^n(X)$ is a set if $X$ is a set. The transfer principle would suggest that for an infinite hypernatural $h$, $\mathcal{P}^h(X)$ is also a set. If $\infty$ is a hypernatural, then this would entail that $\mathcal{P}^\infty(X)$ is, in fact, a set, thus contradicting the Axiom of Foundation.

On the other hand, if one assumes that $\mathcal{P}^\infty(X)$ is not a set (presumably, it is a class), then this would mean that $\mathcal{P}^{\infty-k}(X)$ is not a set. However, if there is a $k\in\widehat{\mathbb{N}}$ such that $|\infty-k|\in\mathbb{N}$, then this would also mean that $\mathcal{P}^n(X)$ is not a set for any $n\in\mathbb{N}$.

How should I regard $\mathcal{P}^\infty(X)$ and similar questionably set-like objects so as to avoid these problems? And if $\mathcal{P}^\infty(X)$ is not a set, then how can I best accommodate it in the definitions of 'element' and 'powerset'?


I have come up with a few guesses (these are just guesses though, and I will gladly defer to a better-informed answer):

1) $\mathcal{P}^n(X)$ is a set for all $n\in\widehat{\mathbb{N}}$ and $\infty\notin\widehat{\mathbb{N}}$. This would also suggest that $\infty$ ('perpetual' infinity) is not a member of the hyperreals, surreals, or any constructible set/class. $\mathcal{P}^\infty(X)$ is a proper class for all sets $X$. For $h\in\widehat{\mathbb{N}}$, $h-1\neq h$, so the properties of the powerset are otherwise preserved - this is also the case if ordinals are used in place of hypernaturals, in which case $\infty$ is not an ordinal.

2) $\mathcal{P}^\infty(X)$ is a set which contains itself, and the assertion that no set contains itself is false (it remains open whether $\mathcal{P}^\infty(X)$ is the only set which contains itself or not).

3) $\forall a.\{a\}=a$, and set membership ($\in$) is replaced with more general inclusion ($\subseteq$). It follows that all sets contain themselves, and the set of all sets that do not contain themselves is empty.


*This, along with $a\in\mathbb{N}\implies a+1\in\mathbb{N}$ suggests that $\infty\in\mathbb{N}$. This is another issue entirely, but for the sake of this problem, it may be resolved by instead allowing $\bigcup_{n=0}^\infty=\widehat{\mathbb{N}}$

** Specifically, the upper bound of the reals which can be defined either as the 'largest' hyperreal infinity, the reciprocal of zero (see also: one point compactification and quotient space), or as the limit $\lim_{n\to\infty}{a_n}$ of a sequence $\{a_n\}_{n\in\widehat{\mathbb{N}}}$ which diverges in the hyperreals (i.e. which does not converge to any other hyperreal infinity).

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closed as unclear what you're asking by Lord Shark the Unknown, Morgan Rodgers, Lee Mosher, ancientmathematician, hardmath Feb 7 at 16:30

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  • $\begingroup$ your problem comes from the $\infty$: $\infty-1$ is not well-defined. Furthermore, there is no integer $k$ such that $\infty-k\in \mathbb N$. $\endgroup$ – Dog_69 Feb 4 at 21:22
  • $\begingroup$ Instead of $\infty$, you should work with ordinals. For more insight, see <en.wikipedia.org/wiki/Von_Neumann_universe>. $\endgroup$ – Alberto Takase Feb 4 at 21:35
  • $\begingroup$ You might have a point about $\infty-1$. As for $\infty-k$, $k$ is an infinity, not an integer. You could consider $k$ to be a hyperinteger. $\endgroup$ – R. Burton Feb 4 at 22:33
  • $\begingroup$ Subtraction of the form $\infty - k$ is just as undefined when $k$ is infinite as it is when $k=1$. Once you start making deductions on the basis of such nonexistent subtractions, nonsense sets in. This, I suspect, is the source of downvoting and close votes (as it is of mine). What might help you in your further perusals of this topic is to realize that the exponents on power set notation are ordinals (for which, as said, subtraction is not defined). $\endgroup$ – Lee Mosher Feb 4 at 22:57
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    $\begingroup$ "by all accounts ... $\infty-1=\infty$." Well, no, not really. It seems you're interested in making sense of $\mathcal{P}^N$ for $N$ a hypernatural. In that case, $N-1$ is very different from $N$. Basically, you're conflating a few different ideas about infinity here, and that's leading to all the confusion. $\endgroup$ – Noah Schweber Feb 5 at 1:21
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First, it doesn't follow from Russell's Paradox that no set is a member of itself. It follows from the Axiom of Foundation, which is independent of the axioms necessary to prove Russell's Paradox.

More to the point, you really need to use ordinals to discuss your desired iterations of the Power Set operation. The appropriate definition is $$\mathscr P^\omega (X) = \bigcup_{n \in \Bbb N} \mathscr P^n(x) .$$ With this as your definition, $\mathscr P^\omega (X)$ is a perfectly good set for any set $X$ (which will be countable if $X$ is finite), and in fact you can use transfinite induction to form $\mathscr P^\alpha (X)$ for any ordinal $\alpha$.

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