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It seems that I do not understand thoroughly the axioms of ZFC.

I am thinking of "is really the axiom of power set independent from the other axioms, and if it is, how to prove that?".

In other words, how to prove that ZFC without the axiom of power set is not equal to ZFC?

Grešnik
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2 Answers2

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The independence of the axiom of powerset from the remaining axioms can be shown by constructing a model of $\mathsf{ZFC}^-$, that is $\mathsf{ZFC}$ without the axiom of powerset, which is not also a model of $\mathsf{ZFC}$.

The standard examples for such models are constructed as follows: fix an uncountable regular cardinal $\kappa$ and consider $H(\kappa)=\{x\mid |\mathrm{trcl}(x)|<\kappa\}$. It can be shown that for every $\kappa$ $H(\kappa)\subseteq V_\kappa$, so that the $H(\kappa)$ are sets, furthermore for a regular $\kappa$ it can be shown that $H(\kappa)$ is the set of sets "hereditarily of cardinality less than $\kappa$".

Now it's not hard to prove that $H(\kappa)$ is a model of $\mathsf{ZFC}^-$, and it can also be proved that the following are equivalent for a regular $\kappa$:

  • $H(\kappa)$ is a model of $\mathsf{ZFC}$.
  • $H(\kappa)=V_\kappa$.
  • $\kappa$ is strongly inaccessible.

So in particular for $\kappa=\aleph_1$ we have that $H(\kappa)$ is a model of $\mathsf{ZFC}^-$ but not of $\mathsf{ZFC}$.

Alessandro Codenotti
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Consider the set $H(\kappa)$ consisting of sets $x$ which satisfy $|\text{tc}(x)| < \kappa$. As mentioned in this answer, this set $H(\kappa)$ is a model of all axioms of ZFC except power set. In fact, if one takes $\kappa$ to be a successor cardinal (such as $\aleph_1$), then one can verify that the axiom of power set if false in $H(\aleph_1)$.

Now it follows that the axiom of power set cannot be proven from the other axioms of ZFC, for if this were the case, the power set axiom would be true in $H(\aleph_1)$.

Dirk van Bree
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  • You can't just pick any $\kappa$. For example, $H(\omega)$ does model the axiom of Power Set, and additionaly fails the Axiom of Infinity. – Vsotvep Jul 07 '19 at 13:46
  • And $H(\aleph_\omega)$ does not model replacement either for an explicit example showing why regularity is needed, since $\omega\in H(\aleph_\omega)$ but $F[\omega]\not\in H(\aleph_\omega)$, where $F(\alpha)=\aleph_\alpha$ for $\alpha\in\mathsf{Ord}$. – Alessandro Codenotti Jul 07 '19 at 13:49
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    You are both right. I was a bit stressed out because this is my first answer and forgot to check the details. Thanks! – Dirk van Bree Jul 07 '19 at 15:00