Uniformization (set theory)

In set theory, the axiom of uniformization, a weak form of the axiom of choice, states that if $R$ is a subset of $X\times Y$, where $X$ and $Y$ are Polish spaces, then there is a subset $f$ of $R$ that is a partial function from $X$ to $Y$, and whose domain (in the sense of the set of all $x$ such that $f(x)$ exists) equals

$\{x\in X|\exists y\in Y(x,y)\in R\}\,$

Such a function is called a uniformizing function for $R$, or a uniformization of $R$.

Uniformization of relation R (light blue) by function f (red).

To see the relationship with the axiom of choice, observe that $R$ can be thought of as associating, to each element of $X$, a subset of $Y$. A uniformization of $R$ then picks exactly one element from each such subset, whenever the subset is nonempty.

A pointclass ${\boldsymbol {\Gamma }}$ is said to have the uniformization property if every relation $R$ in ${\boldsymbol {\Gamma }}$ can be uniformized by a partial function in ${\boldsymbol {\Gamma }}$. The uniformization property is implied by the scale property, at least for adequate pointclasses.

It follows from ZFC alone that ${\boldsymbol {\Pi }}_{1}^{1}$ and ${\boldsymbol {\Sigma }}_{2}^{1}$ have the uniformization property. It follows from the existence of sufficient large cardinals that

• ${\boldsymbol {\Pi }}_{2n+1}^{1}$ and ${\boldsymbol {\Sigma }}_{2n+2}^{1}$ have the uniformization property for every natural number $n$.
• Therefore, the collection of projective sets has the uniformization property.
• Every relation in L(R) can be uniformized, but not necessarily by a function in L(R). In fact, L(R) does not have the uniformization property (equivalently, L(R) does not satisfy the axiom of uniformization).
  • (Note: it's trivial that every relation in L(R) can be uniformized in V, assuming V satisfies AC. The point is that every such relation can be uniformized in some transitive inner model of V in which AD holds.)

References

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