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In the set theory NFU (described by M. Randall Holmes in "Elementary Set Theory with a Universal Set"), it is possible to define the set of all sets, and the set of all one-element sets. An object is a set if and only if $\emptyset$ is a subset of it, so we can define $V^*$, the set of all sets, as $\{x\ |\ \emptyset \subseteq x\}$. We can also define the set of all one-element sets, $V^1$, as $\{x\ |\ x \text{ has exactly one element}\}$.

Is there a bijection between $V^*$ and $V^1$? Clearly, since $V^1$ is a subset of $V^*$, we can define an injection $f : V^1 \to V^*, f(x) = x$. It's not obvious how one could define an injection $V^* \to V^1$. The "most obvious" candidate, $g(x) = \{x\}$, does not exist; its definition is not stratified.

For finite sets, we can define an injection easily enough:

$$\begin{align} g(\emptyset) &= \{(0, \textit{anything})\}\\ g(\{x\}) &= \{(1, x)\}\\ g(\{x,y\}) &= \{(2, x, y)\}\\ g(\{x,y,z\}) &= \{(3, x, y, z)\} \end{align}$$

And so on. But this definition cannot be extended to infinite sets.

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Could you explain why you think g's definition isn't (and can't be) stratified? – Ben Crowell Oct 7 '12 at 18:21
1  
The fact that the natural definition of $g$ isn't stratified does not, by itself, imply that $g$ doesn't exist. NFU sometimes manages to cleverly prove the existence of sets whose straightforward definition isn't stratified. In the present case, though, I'm pretty sure it's known, not only that $g$ doesn't exist, but that the set of one-element sets has strictly smaller cardinality than the set of all sets. – Andreas Blass Dec 24 '12 at 19:32
    
If the function $g$ exists, then we can define the set from Russell's paradox, $R$, as $\{x | g(x) \not\subseteq x\}$. Then $R \in R$ if and only if $g(R) \not\subseteq R$ if and only if $R \not\in R$, which is a contradiction. – Tanner Swett Apr 30 '13 at 18:08

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