## Classification of smooth equivalence relations

The classification of smooth equivalence relations is accomplished by two exercises from Su Gao’s book, Invariant Descriptive Set Theory.

Exercise 5.4.1, Su Gao, p 132.
Show that for any equivalence relation $E$ on a Polish space $X$, $\text{id}( \omega ) \le_B E$ iff there are infinitely many $E$-equivalence classes.

Suppose $\text{id}( \omega ) \le_B E$.  So there exists Borel $f: \omega \to X$ such that $f(x) E f(y)$ iff $x = y$. Since for each $x$, the element $f(x)$ lies in a different $E$ class, there are infinitely many classes.

Suppose there are infinitely many $E$-classes; enumerate a countable subset: $\{ C_n : n \in \omega \}$.  By the Axiom of Choice, for each $n$ pick a $y_n \in C_n$, and define $f: \omega \to X$  by $f(n) = y_n$.  Since $\omega$ has the discrete topology, $f$ is cts, and so $latex \text{id}(\omega) \le_B E$.

Exercise 5.4.2, Su Gao, p 132
Show that if an equivalence relation $E$ is smooth then either $E \sim_B \text{id}(2^\omega)$, or $E \sim_B \text{id}\omega$, or for some finite $n \in \omega$, $E \sim_B \text{id}(n)$.

By Theorem 5.4.2, if $E$ is smooth, either (1) $E \sim_B \text{id}( 2^\omega )$ or (2) $E \le_B \text{id }( \omega )$.

So assume (2).  Then either $E$ has infinitely many classes or it does not.
If it does, by Exercise 5.4.1, $\text{id}(\omega) \le_B E$, hence $\text{id}(\omega) \sim_B E$.

If it does not, suppose $E$ has $n$ classes. List the classes: $C_1, \ldots, C_n$. Since $E$ is smooth, each $C_i$ is Borel.

Then define $f: X \to n$ by $f(x) = i$ iff $x \in C_i$.  Then $f$ is Borel. Also $\text{id}(n) \le E$, since $n$ has discrete topology and we can pick
a point in each equivalence class to map $i$ to $E_i$.