Borel cross section

Question

It is known from metric space topology that a closed equivalence relation on a Polish space has either countably many or $\mathfrak{c}$ many equivalence classes.

A short elementary proof is given in Srivastava "A course on Borel sets", Theorem 2.6.7. This proof does not proceed through a Borel or analytic cross section for the equivalence relation. Is it possible that a closed equivalence relation can have no analytic cross section?

(A "cross section", or "transversal", is a set with exactly one point in each equivalence class. An equivalence relation on a space $X$ is "closed" if it is a closed subset of the product space $X\times X$.)

Several results asserting the existence of a Borel or analytic cross section assume additional conditions, such as requiring the equivalence classes to be closed, for example.

The stated theorem is a particular instance of a much deeper theorem of J. Silver: every coanalytic equivalence relation has either countably many or $\mathfrak{c}$ many equivalence classes. For a proof, see Jech "Set Theory", 3rd ed. 2002/2006, Chapter 32.

Answer 1

The answer to this question is found, for example, in Su Gao's book Invariant Descriptive Set Theory. In exercise 5.4.6 an outline is given for an example of a closed equivalence relation with no Borel selector. (A selector is a map $s\colon X\to X$ such that $s(x)\mathrel{E}x$ and $x\mathrel{E}x'\implies s(x)\mathrel{E}s(x')$. On page 87 it is explained that if $E$ is Borel then having a Borel selector is equivalent to having a Borel transversal, i.e. cross-section). For completeness, here is the statement of exercise 5.4.6.

Let $F$ be a closed subset of $\omega^\omega\times\omega^\omega$ such that the set $\{x:\exists y(x,y)\in F)\}$ is analytic and not Borel. Let $X=F$ and define the equivalence relation $E$ on $X$ by $(x,y)\mathrel{E}(x',y')\iff x=x'$. Then $E$ is closed and has no Borel selector.

Answer 2

There is a recursion theory method which is quite similar to Samuel's.

By a classical recursion theory result, there is a recursive functional $F$ so that for any $x\in \omega^{\omega}$, $F(x)$ codes an $x$-recursive tree $T\subseteq \omega^{<\omega}$ so that $[T]$ has an infinite path but no infinite path hyperarithmetic in $x$. Then let $(x_0,y_0)E(x_1,y_1)$ if $x_0=x_1$ and $y_0,y_1\in F(x)$. Then $E$ is a $\Pi^0_1$-equivalence relation and has no $\mathbf{\Sigma}^1_1$-cross section.

Note that the non-compactness plays an important role here. Otherwise, you can't find a counter example.

Answer 3

Shashi Srivastava suggests taking the equivalence relation on $\omega^\omega$ induced by a continuous $f:\omega^\omega\to \omega^\omega$ with non-Borel range. This is a closed equivalence relation with no Borel cross-section.