This ties back to a previous question I asked, which was about the notion of "to define $X$, you must first define $Y$". The current question is about a specific application of that notion. Is it possible to define powersets without first defining the subset relation? To make my question more precise, suppose we are working in $ZFC$ set theory. Does defining the binary relation, "$X$ is the powerset of $Y$" require one to first define the relation "$X$ is a subset of $Y$"? In other words, we are really making a definitional extension of $ZFC$. Of course, I know that one can directly make the definitional extension by defining "$X$ is the powerset of $Y$" in terms of the primitive notion of $\in$, but my question is, in that definitional extension, must you always use, either explicitly or implicitly, the binary relation of subset?

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    $\begingroup$ I think this is currently too vague to admit an answer. In particular, since $\subseteq$ can be defined from $\mathcal{P}$ and $\in$ as $$A\subseteq B\iff A\in \mathcal{P}(B),$$ I think it will be very hard to identify a precise sense in which we can get to $\mathcal{P}$ without "in effect" going through $\subseteq$. $\endgroup$ Jan 27 at 21:03
  • $\begingroup$ Like @NoahSchweber, I think your question is too unclear to have a definite answer. Perhaps one could firm it up by asking if one can axiomatise ZFC using the relation "$X$ is the power set of $Y$" as the only primitive. My guess is not, but I don't see how to prove that just now. $\endgroup$
    – Rob Arthan
    Jan 27 at 22:04
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    $\begingroup$ @RobArthan Powerset alone has very little power (hehehe): it just describes a total injective function on the universe whose image misses infinitely many things. That gives a decidable theory! $\endgroup$ Jan 27 at 22:07

1 Answer 1



∀xy(x ⊆ y ⟺ x∩y = x)

So, instead of talking about subsets one can talk about intersections.

This motivates an alternative definition of the Powerset:

The Powerset of y is the set of all sets whose intersection with y is themself.

𝒫(y) = { x | x∩y = x }


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