The question is originally from math stack exchange here.

Basically, what I am asking is if we can define ordered tuples of proper classes in NBG. My idea, for finite tuples of proper classes, was to define it as the set $\{0\} \times X_0 \cup \{1\} \times X_1 \times ... \times \{n\} \times X_n$. Here I am not assuming $(X_i)_{i < n+1}$ as an encoded structure since that would assume the conclusion. However, we can define this using recursion. The empty tuple is the empty set(i.e. $()=\emptyset$) and $(X_0, X_1, ..., X_n) = (X_0, ..., x_{n-1}) \cup \{n\} \times X_n$.

Now I would intuitively define $(X_\alpha)_{\alpha < \beta}$ for $ \beta $ an ordinal, by transfinite recursion, by taking $(X_\alpha)_{\alpha < \beta + 1} = (X_\alpha)_{\alpha < \beta} \cup \{\beta\} \times X_{\beta}$ and if $\beta$ is a limit ordinal, then $(X_\alpha)_{\alpha < \beta} = \bigcup_{\alpha < \beta} \left( (X_{\gamma})_{\gamma < \alpha} \right)$. However, there is a problem with the limit ordinals, since I am taking the "infinite union", the sum class, which is basically the union of its members, so it would require that $X_a \in T$, which is impossible since $X_a$ is a proper class.

However, I have also found this answer on stack exchange that uses exactly this method and it I don't know whether what I said above is correct or not.

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    $\begingroup$ You strategy seems correct, except for the union case you need to define it in a way that doesn't require the $X_\gamma$'s to be elements, since they can't if they were proper classes. Define: $\bigcup_{\alpha< \beta}((X_\gamma)_{\gamma < \alpha})= \{\langle \alpha, x \rangle \in X_\alpha \mid \alpha < \beta \}$ $\endgroup$ Commented Feb 26 at 12:06
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    $\begingroup$ As a general matter, recursions defining classes by induction (even for length $\omega$) cannot be undertaken in NBG. For example, the definition of first-order truth via the Tarski recursion is a recursion of length $\omega$, yet in NBG it is consistent that there is no class truth predicate. One is using the principle known as elementary transfinite recursion (ETR) when making such definitions by recursion. This principle is provable in KM, but not NGB. Meanwhile, you don't need recursion at all here, if you have already the classes $X_\alpha$ in a way that is definable from $\alpha$. $\endgroup$ Commented Feb 26 at 13:14
  • $\begingroup$ You can read more about ETR in cambridge.org/core/journals/journal-of-symbolic-logic/article/… and related references. $\endgroup$ Commented Feb 26 at 13:14
  • $\begingroup$ I should have said "by recursion" not "by induction" in my first comment above. $\endgroup$ Commented Feb 26 at 13:21

1 Answer 1


You can define infinite tuple of order $\omega$ as a subclass of the Kuratowski-Cartesian product of $\omega$ and some class, having the class of all first projections of elements of it (aka preimage; domain) being $\omega$ itself (if the empty set is not an entry, otherwise we don't need the latter condition but we need the domain which is already a subclass of $\omega$ to be of order type $\omega$).

Now we define projections of a tuple $t$ as:

$x \text{ is the } \alpha \text{ entry of } t \iff x=\{y \mid \langle \alpha, y \rangle \in t \}$

Clearly $x,t$ can be proper classes!

Of course if you demand all entries to be proper classes, then you add the condition that every $\alpha \in \omega$ entry of $t$ is a proper class.

You can do that for any infinite ordinal $\lambda$ to get an infinite tuple of order $\lambda$.

  • $\begingroup$ Can you elaborate on the "Kuratowski-Cartesian product"? I am not an expert in the study of set theory. $\endgroup$ Commented Feb 26 at 12:09
  • $\begingroup$ @Shthephathord23, the Kuratowski-Cartesian product between classes $A$ and $B$ (denoted as $A \times B$) is the class of all Kuratowski-ordered pairs $\langle a,b \rangle$ where $a \in A$ and $b \in B$. The Kuratowski ordered pair of $a$ and $b$ denoted as $\langle a,b \rangle$ is the set $\{\{a\},\{a,b\}\}$. So, to write it formally: $A \times B = \{ \langle a, b\rangle \mid a \in A \land b \in B \}$ $\endgroup$ Commented Feb 26 at 12:11
  • $\begingroup$ ok, but I still don't understand the construction. From what I can follow, is the class of all first projection the domain(i.e. $\{ x \vert (\exists y)( (x, y) \in A \times B) \}$)? $\endgroup$ Commented Feb 26 at 12:19
  • $\begingroup$ @Shthephathord23, Yes! Since you want an infinite tuple of proper classes, then each entry is a proper class and thus nonempty, so you need that domain to be $\omega$ itself, where $\omega$ is the set of all naturals. $\endgroup$ Commented Feb 26 at 12:38
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    $\begingroup$ @Shthephathord23, Yes! That is correct. Because of global choice a proper class is a class of cardinality $On$. I'm speaking within MK and not NBG. But for although all proper classes are of the same cardinality, namely $On$, yet working in MK, there are more proper classes than sets. For example you cannot have a tuple of all proper classes, because the largest tuple you can have is of cardinality $On$ while you have more proper classes than elements of $On$. $\endgroup$ Commented Feb 26 at 14:08

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