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Define: $n$-skew pair of $x,y$, denoted by $\langle x,y \rangle^n$, as: $(singleton^n(x), y)$

Define: $(-n)$-skew pair of $x,y$, denoted by $\langle x,y \rangle^{-n}$, as: $(x, singleton^n(y))$

Where $(-,-)$ is the Kuratwoski ordered pair implementation, and $n$ is a natural

where: $singleton^0(x) = x$

$singleton^{i+1}(x) = \{singleton^i(x)\}$

so $\langle x,y \rangle ^0$ and $\langle x,y\rangle^{-0}$ are both level pairs.

Define: $f \text { is }n \text{-skew injection } \equiv_{df} f \text { is injection} \land \forall p \in f (p \text { is n-skew pair})$

Define: $x \leq^* y \equiv_{df} \exists n \in \mathbb Z, \exists f (f:x \to y, f \text{ is n-skew injection)}$

Define: $x =^* y \equiv_{df} \exists n \in \mathbb Z, \exists f (f:x \to y, f \text{ is n-skew bijection)}$

Where $\mathbb Z$ is the set of Integers.

Write Cantor-Bernstein-Schroeder theorem in terms of $\leq^*,=^*$ , denoted as "skew-CBS", as:

$\forall x,y [(x \leq^* y \land y \leq^* x) \to x=^* y]$

Question: Is skew-CBS consistent with NFU?

Of course both $\leq^*$ and $=^*$ are non-stratified relations, so they work externally.

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No. A counterexample to skew CBS follows from the existence of any cardinals T(kappa) < lambda < kappa [cardinals in NF(U) are Frege cardinals; T(kappa) is the cardinality of elementwise images of elements of kappa under the singleton map.]

This is a counterexample because an element of lambda will have an ordinary injective embedding into an element of kappa, then an element of kappa will have a skew embedding to a subset of size T(kappa) included in an element of lambda.

And there cannot be any skew bijection from a set of size kappa to a set of size lambda because the image of a skew bijection with domain in kappa must have range in some T^n(kappa), and lambda is clearly not any T^n(kappa) [n might be negative here, which does not affect the conclusion].

In NFU+Choice there are cardinals strictly between |V| and T(|V|).

In NF finding cardinals T(kappa) < lambda < kappa is technically tricky but I know how to do it.

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