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.

This entry was posted in Uncategorized. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s