Is there a set in ZFC that can not be obtained from ordinals (defined according the Von Neumann definition) via set operations (union, intersection, set difference) and power set?

  • $\begingroup$ Can you do the set of even natural numbers? $\endgroup$
    – GEdgar
    May 3 '20 at 10:42

That depends on what you mean.

If you just iterate the power set operation starting from $\varnothing$, and taking unions at limit ordinals, you will generate the whole universe. But you will not "obtain" every set, because every set will be an element of some point in the hierarchy, rather than a step in that hierarchy.

But even here, when you reach a limit step, you'd need something that will let you collect the previous steps.

With just what you describe, the answer is very much negative. You will not have $V_\omega$, the set of hereditarily finite sets. To see that, note that $V_\omega$ is not an ordinal, and it is not a power set of an ordinal, or a second-power set of an ordinal, etc., because it is in fact the $\omega$th power set of any finite ordinal (i.e. $\bigcup\{\mathcal P^n(\varnothing)\mid n<\omega\}$). And so it is in fact not a subset of any finite power set either.

You can conclude from the above that it is not a union or an intersection of any such sets, because power sets, in general, are closed under unions and intersections.

  • $\begingroup$ My question was simpler than you thought and at the same times a bit ambiguous. I just would like to know an example of a set that you can not express as a combination of the set operations and the power set operator applied to ordinals also infinitely many times (and not just as an element of some other set) ? $V_\omega = ⋃\{\mathcal{P}^n(∅)∣n<ω\}$ is a set that you can build by combinations of set operations and power set over ordinals. $\endgroup$
    – holmes
    May 3 '20 at 14:35
  • $\begingroup$ Why is the set $\{\mathcal P^n(\varnothing)\mid n<\omega\}$ such a set, is my point. $\endgroup$
    – Asaf Karagila
    May 3 '20 at 14:36
  • $\begingroup$ You are telling me that the counterexample is exactly the following set $\{\mathcal{P}^n(∅)∣n<ω\}$ which I can rewrite like this $\bigcup_{n<ω} \{\mathcal{P}^n(\emptyset) \}$. Is it because $\{ \mathcal{P}^n(\emptyset)\}$ can not be built? What if you additionally allow the operation $S \rightarrow \{S\}$ $\endgroup$
    – holmes
    May 3 '20 at 14:54
  • $\begingroup$ But when you write $\bigcup X$, you need have created $X$ before. So now tell me how you plan to get $\{\{\mathcal P^n(\varnothing)\}\mid n<\omega\}$. $\endgroup$
    – Asaf Karagila
    May 3 '20 at 15:01
  • 1
    $\begingroup$ Yes, it is the union of $\omega$ different sets. But the union operator is defined on a single set, so when we write $\bigcup_{i\in I} X_i$ we mean $\bigcup\{X_i\mid i\in I\}$, and for that, we need $\{X_i\mid i\in I\}$ to be a set. So in order for you to take the union of a set, the set needs to already exist. In order words, show me $\{\{\mathcal P^n(\varnothing)\}\mid n<\omega\}$. $\endgroup$
    – Asaf Karagila
    May 3 '20 at 15:13

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