Let $A$ be a finite set, and let $A^\infty$ be the set of all sequences $(a_n)_{n=1}^\infty$ of elements of $A$. A set $B \subseteq A^\infty$ is a tail set if for every two sequences $\vec a, \vec b \in A$ that differ in finitely many coordinates, either both are in $A$ or both are not in $A$.

Question: How high can we find tail sets in the Borel hierarchy? That is, do tail sets can be found only in the first $\alpha$ levels of the hierarchy for some ordinal $\alpha$ (maybe even some finite ordinal $\alpha$), or can we find them as high as we look?

  • $\begingroup$ Let $A = \{0,1\}$, let $U \subset A^{\infty}$ be any Borel. Construct $t(U) \subset A^{\infty}$ by letting $x \in t(U)$ iff $\sum x = \infty$, and when $10^n1$ appears in $x$ it does infinitely many times, and the characteristic sequence of $n$ s.t. $10^n1$ appears is in $U$. Clearly $t(U)$ is tail, and is Borel because can be constructed just like $U$, replacing basic opens by $\Sigma^0_2$s stating appearance of words of form $1 0^n 1$, and recurrence. There is continuous $f : A^{\infty} \to A^{\infty}$ reducing $U$ to $t(U)$, so if $t(U)$ is $Q^0_\alpha$ then so is $f^{-1}(t(U)) = U$. $\endgroup$
    – Ville Salo
    Apr 6, 2020 at 9:48
  • $\begingroup$ Slightly wrong, since changing finitely many can break the recurrence condition. I suppose just drop the recurrence assumption. (Also I'm not sure this is a research level question.) $\endgroup$
    – Ville Salo
    Apr 6, 2020 at 9:56
  • $\begingroup$ @VilleSalo: I'm a little confused by your comment. But I do know that in $\mathcal P(\mathbb N)$, the set of all $A \subseteq \mathbb N$ with positive upper density is a tail set in $\Sigma^0_3 \setminus \Sigma^0_2$. (I can't tell if this fact contradicts your claim or not.) $\endgroup$
    – Will Brian
    Apr 6, 2020 at 12:14
  • $\begingroup$ So maybe it should be $\Sigma^0_3$ instead of $\Sigma^0_2$, I wrote $\Sigma^0_2$ and was going to think about what the correct number is but apparently didn't. We agree it is a Borel set, though, that's all we need. $\endgroup$
    – Ville Salo
    Apr 6, 2020 at 12:18
  • $\begingroup$ What I believe is correct is: encode a set $U$ into those distances between $1$s that appear infinitely many times. The resulting set is Borel, tail, and at least as high as $U$. $\endgroup$
    – Ville Salo
    Apr 6, 2020 at 12:21

1 Answer 1


Here is an example: Let $A=\{0,1\}$ and for any $x\in \{0,1\}^{\omega}$, let $n<_x m$ if $x(2^n\cdot 3^m)=1$. For any countable ordinal $\alpha$, let $x\in U_{\alpha}$ if there is some $l$ so that $<_x$ over the set $\{n\mid n>l\}$ codes a well order $\leq \alpha$.

Now for any Borel set $B$, there is real $z$ so that $B$ is $\Delta_1^1(z)$ and so there is a recursive function $\Phi$ and a $z$-recursive ordinal $\alpha$ so that $x\in B\Leftrightarrow \Phi^{x\oplus z} \mbox{codes a well order} \Leftrightarrow \Phi^{x\oplus z} \mbox{codes a well order} \leq \alpha$. Moreover $x\not\in B\Leftrightarrow \Phi^{x\oplus z} \mbox{codes a linear order with an infinite descending subsequence}$.

Now let $\Psi$ be a $z$-recursive function so that $\Phi^{x\oplus z}$ codes $n<m$ iff $\Psi^{x\oplus z}(2^n\cdot 3^m)=1$; otherwise $\Psi^{x\oplus z}(2^n\cdot 3^m)=0$.

Then it is clear that for any $x$, $x\in B$ implies $\Psi^{x\oplus z}\in U_{\alpha}$. For a contradiction, suppose that $x\not\in B$ and $\Psi^{x\oplus z} \in U_{\alpha}$ for some $x$. Then $\Phi^{x\oplus z}$ codes a linear with an infinite descending subsequence over $\{n\mid n>l\}$. Then, for arbitrarily large $l$, $\Psi^{x\oplus z}$ contains an infinite descending subsequence, a contradiction.

So $x\in B$ if and only if $\Psi^{x\oplus z}\in U_{\alpha}$.

I think that by a more complicated argument, it can be shown that this is true for any countable Borel equivalence relation.


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