Compactness of maximal eventually different families

TL;DR

This paper investigates the structure and existence of effectively closed and compact maximal eventually different families of functions in certain product spaces, providing criteria for their existence.

Contribution

It introduces a characterization of when effectively compact maximal eventually different families exist in product spaces of functions.

Findings

Existence criteria for effectively compact families.

Characterization of effectively closed maximal families.

Conditions for the existence of such families.

Abstract

We show that there is an effectively closed maximal eventually different family of functions in spaces of the form $\prod_n F(n)$ for $F\colon \mathbb{N} \to \mathbb{N}\cup\{\mathbb{N}\}$ and give an exact criterion for when there exists an effectively compact such family.

Equations

$$\sum_{l\leq n}\Big{(}\Big{\lvert}\prod_{k<l}F(k)\Big{\rvert}\cdot 2^{l}\Big{)}\leq F(n),$$

$${\#}_1:{}^{<\mathbb{N}}\mathbb{N}\to \mathbb{N}$$

$$\displaystyle\operatorname{\#_{2}^{F}}\colon\bigcup_{l\in\mathbb{N}}\Big{[}\Big{(}\prod_{k<l}F(k)\Big{)}\times{}^{l}2\Big{]}\rightarrow\mathbb{N}$$

$$e(g,c)(n)={\#}_2(g\upharpoonright n,c\upharpoonright n).$$

$$I_{c^{\ast }}=\{n\in \mathbb{N}|n\equiv \sum _{i<\mathrm{lh}(c^{\ast })}{c}^{\ast }(i)\cdot 2^i(\mathrm{mod}{2}^{\mathrm{lh}(c^{\ast })})\},$$

$$n_1\in I_{c^{\ast }\upharpoonright n_0+1}.$$

$$\dot{\mathrm{e}}(g,c)(n)=\{\begin{array}{ll}e(g,c)(n)& \text{if n∈/B(g,c), or (∃n0,n1∈B(g,c)∩n)n0≺gn1,}\\ & \text{or c↾n is not good;}\\ g(n)& \text{otherwise.}\end{array}$$

$$\mathcal{E}=\{\dot{\mathrm{e}}(h,d)|h\in {\mathcal{N}}_F,d\in {}^{\mathbb{N}}2\}.$$

$$f\notin \mathcal{E}⟺(\exists f^{\prime }\in {}^{\mathbb{N}}\mathbb{N})f^{\prime }\in \mathcal{E}\wedge f{=}^{\infty }{f}^{\prime }\wedge f\ne f^{\prime }.$$

$$I_{c^{\ast }}^g=\{2\cdot {\#}_1(g\upharpoonright n)|n\in I_{c^{\ast }}\}.$$

$$X=\{n\in B(g_0,c_0)\setminus B(g_1,c_1)|\dot{\mathrm{e}}(g_0,c_0)(n)=e(g_1,c_1)(n)\}$$

$$g\upharpoonright I=e(h,d)\upharpoonright I$$

$$C=\{c^{\ast }\in {}^{<\mathbb{N}}2|\mathrm{lh}(c^{\ast })=0\vee c^{\ast }(\mathrm{lh}(c^{\ast })-1)=1\}$$

$$(\exists c^{*}\in C)(\forall n_{0}\in\operatorname{I}^{g}_{c^{*}})\text{ letting $c_{1}^{*}=c^{*}\mathbin{{}^{\frown}}0^{n_{0}-\operatorname{lh}(c^{*})}\mathbin{{}^{\frown}}1$ we have }\\ \big{[}g(n_{0})\neq\operatorname{\#_{2}}(g\mathbin{\upharpoonright}n_{0},c_{1}^{*}\mathbin{\upharpoonright}n_{0})\;\wedge(\exists n_{1}\in\operatorname{I}^{g}_{c_{1}^{*}})\;(n_{0}\mathbin{\prec^{g}}n_{1})\big{]}$$

$$g(n_0)\ne {\#}_2(g\upharpoonright n_0,c_1^{\ast }\upharpoonright n_0)$$

$$(\forall c^{*}\in C)(\exists n_{0}\in\operatorname{I}^{g}_{c^{*}})\text{ s.t.\ letting $c_{1}^{*}=c^{*}\mathbin{{}^{\frown}}0^{n_{0}-\operatorname{lh}(c^{*})}\mathbin{{}^{\frown}}1$ }\\ \big{[}g(n_{0})=\operatorname{\#_{2}}(g\mathbin{\upharpoonright}n_{0},c_{1}^{*})\;\vee(\forall n_{1}\in\operatorname{I}^{g}_{c_{1}^{*}})\;\neg(n_{0}\mathbin{\prec^{g}}n_{1})\big{]}$$

$$g(n_k)={\#}_2(g\upharpoonright n_k,c_{k+1}^{\ast }\upharpoonright n_k)\vee (\forall n_1\in I_{c_{k+1}^{\ast }}^g)\neg (n_0{\prec }^g{n}_1)$$

$$\begin{gathered} B(g^{\ast },c^{\ast })=\{2\cdot \#g^{\ast }\upharpoonright n|n<\mathrm{lh}(c^{\ast })\text{ and }{c}^{\ast }(n)=1\}\setminus \\ \{n<\mathrm{lh}(c^{\ast })|g^{\ast }(n)={\#}_2(g^{\ast }\upharpoonright n,c^{\ast }\upharpoonright n)\}. \end{gathered}$$

$$f^{\ast }\in \bigcup _{l\in \mathbb{N}}\prod _{n<l}F(n)$$

$$(g^{\ast },c^{\ast })={{\#}_2}^{-1}(f^{\ast }(m))$$

$$\sum_{l\leq m}\Big{(}\Big{\lvert}\prod_{k<l}F(e_{k})\Big{\rvert}\cdot 2^{l}\Big{)}\leq F(e_{m}).$$

$$f\upharpoonright E\in {\mathcal{E}}_0⟺(\forall n\in \mathbb{N})f\circ e\upharpoonright n\in T.$$

$$\operatorname{\ddot{e}}(f)(n)=\begin{cases}\operatorname{\#_{1}}\big{(}(f\mathbin{\upharpoonright}m)\circ e\big{)}&\text{for $n\notin E$, where $m$ is maximal such that}\\ &m\leq n\text{ and }\operatorname{\#_{1}}\big{(}(f\mathbin{\upharpoonright}m)\circ e\big{)}<F(n),\\ f(n)&\text{for $n\in E$.}\end{cases}$$

$$\{g\upharpoonright E|g\in \mathcal{E}\}={\mathcal{E}}_0,$$

$$\operatorname{\#_{1}}\big{(}(f\mathbin{\upharpoonright}m_{0}+1)\circ e\big{)}<F(n).$$

$$g(n)\in \{{\#}_1(f^{\ast }\circ e)|(\exists m\le n)f^{\ast }\in \prod _{k\in E\cap m}F(k)\}$$

$${\mathcal{C}}_{\mathcal{E}}=\{X\in P(\mathbb{N})|\{g\upharpoonright X|g\in \mathcal{E}\}\text{ is a medf in ∏n∈EF(n)}\}.$$

Full text

1. Introduction

A

In [2] Horowitz and Shelah construct in ZF a maximal eventually different family (short: medf) which is $\Delta^{1}_{1}$, i.e., effectively Borel. This was a surprise since, e.g., infinite so-called mad families cannot even be $\mathbf{\Sigma}^{1}_{1}$ (i.e., analytic; see [5, 8]). In a more recent, related result [1] they obtain a $\Delta^{1}_{1}$ maximal cofinitary group.

The present paper answers a question of Asger Törnquist [9]: Given $F\colon\mathbb{N}\to\mathbb{N}$ such that $\liminf_{n\to\infty}F(n)=\infty$ does there exist a Borel or even a compact medf in the space $\{g\in{}^{\mathbb{N}}\mathbb{N}\;|\;(\forall n\in\mathbb{N})\;g(n)<F(n)\}$? As well as answering this question, we construct a medf (which is $\Pi^{0}_{1}$, i.e., effectively closed) in Baire space in an even more elementary way than in [2] or [7].

To make the question entirely precise, we give the definition of (maximal) eventually different families a broader context:

Definition 1.1**.**

Any two functions $g_{0},g_{1}$ with domain $\mathbb{N}$ are called eventually different if and only if $\{n\in\mathbb{N}\;|\;g_{0}(n)=g_{1}(n)\}$ is finite.

Given a function $F\colon\mathbb{N}\to\mathbb{N}\cup\{\infty\}$, let $\mathcal{N}_{F}=\{g\in{}^{\mathbb{N}}\mathbb{N}\;|\;(\forall n\in\mathbb{N})\;g(n)<F(n)\}$ (with the product topology, and $\{k\in\mathbb{N}\;|\;k<F(n)\}$ discrete, as usual). A set $\mathcal{E}$ is an eventually different family in $\mathcal{N}_{F}$ if and only if $\mathcal{E}\subseteq\mathcal{N}_{F}$ and any two distinct $g_{0},g_{1}\in\mathcal{E}$ are eventually different; such a family is called maximal (or short: a medf) if and only if it is maximal among such families under inclusion.

We now state our main result, followed by a rather straightforward corollary.

Theorem 1.2**.**

Suppose $F\colon\mathbb{N}\to(\mathbb{N}\setminus\{0\})\cup\{\infty\}$. If $\lim_{n\to\infty}F(n)=\infty$ there is a perfect $\Pi^{0}_{1}(F)$ (effectively-in-$F$ closed) maximal eventually different family in $\mathcal{N}_{F}$.

Corollary 1.3**.**

Suppose $F\colon\mathbb{N}\to(\mathbb{N}\setminus\{0\})\cup\{\infty\}$. There is a compact $\Pi^{0}_{1}(F)$ medf in $\mathcal{N}_{F}$ if and only if $F(n)<\infty$ for infinitely many $n\in\mathbb{N}$. Moreover, exactly one of the following holds:

1. (1)

Every medf is finite and there is a finite medf consisting of constant functions (namely, when $\liminf_{n\to\infty}F(n)<\infty$); 2. (2)

There is a perfect $\Pi^{0}_{1}(F)$ medf but no countable medf.

We ask $(\forall n\in\mathbb{N})\;F(n)>0$ to preclude the trivial case of the empty space. As $\liminf_{n\to\infty}F(n)<\infty$ means there is $m$ such that $\{n\in\mathbb{N}\;|\;F(n)=m\}$ is infinite, the set $\{c_{k}\;|\;k<m\}$ where $c_{k}$ is the constant function with value $k$ constitutes a medf in this case. So the question posed by Törnquist is only interesting if $\lim_{n\to\infty}F(n)=\infty$ holds.

B

We fix some notation and terminology (see also [3]). For any set $X$, $|X|$ denotes its cardinality. As always in set theory, we identify $n$ with $\{k\in\mathbb{N}\;|\;k<n\}$, $\mathbb{N}$ with $\omega$ and $\mathbb{N}\cup\{\mathbb{N}\}$ with $\omega+1$, $\mathbb{N}$ is equipped with the discrete and $\omega+1$ with the order topology. We write ${}^{A}B$ to mean the set of functions from $A$ to $B$ and ${}^{<\mathbb{N}}B$ means the set of finite sequences from $B$. If $s$ is a sequence, $\operatorname{lh}(s)$ denotes its length. For a set $A$ and a function $F$ on $A$, by $\prod_{x\in A}F(x)$ we always mean the set of functions $f$ with domain $A$ such that $(\forall a\in A)\;f(a)\in F(a)$ (not a product of numbers). Both ${}^{\mathbb{N}}\mathbb{N}$ and $\prod_{x\in A}F(x)$ for $F$ as in Theorem 1.2 naturally carry the product topology.

We write $f_{0}\mathbin{=^{\infty}}f_{1}$ to mean that $f_{0}$ and $f_{1}$ are not eventually different (they are infinitely equal). Two sets $A,B\subseteq\mathbb{N}$ are called almost disjoint if and only if $A\cap B$ is finite, and an almost disjoint family is a set $\mathcal{A}\subseteq\operatorname{\mathcal{P}}(\mathbb{N})$ any two elements of which are almost disjoint. We write $A\subseteq^{*}B$ to mean $A$ is *almost contained in * $B$, i.e., $A\setminus B$ is finite.

We naturally take ‘…is recursive in …’ to apply to subsets of $\mathbf{H}(\omega)$, the set of hereditarily finite sets. Any function $F\colon\mathbb{N}\to\omega+1$ is for this purpose identified with a subset of $\mathbf{H}(\omega)$ by replacing the value $\omega$ with some fixed element $\infty$ of $\mathbf{H}(\omega)\setminus\omega$. Consult [6, 4] for more on the (effective) Borel and projective hierarchies, i.e., on $\Pi^{0}_{1}$, $\Pi^{0}_{1}(F)$, $\Delta^{1}_{1}$, … sets.

Clearly all of the results in this paper can be derived in ZF (in fact in a not so strong subsystem of second order arithmetic).

C

This note is organized as follows. In Section 2 we prove a special case of Theorem 1.2 assuming that $F$ grows quickly enough so that we can code initial segments of functions without running out of space, in Lemma 2.1. The construction we give also applies to Baire space. In Section 3 we prove Theorem 1.2 in full generality, quoting the proof of Lemma 2.1. In Section 4 we prove two simple facts which together with Theorem 1.2 imply Corollary 1.3. We close with some open questions in Section 5.

*Acknowledgements: The author gratefully acknowledges the generous support from the DNRF Niels Bohr Professorship of Lars Hesselholt. *

2. The main lemma

We start by proving a variant of Theorem 1.2 (that there is a $\Pi^{0}_{1}$ medf in $\mathcal{N}_{F}$) assuming $F$ satisfies a growth condition. The proof will at the same time make more elementary and generalize the construction of Horowitz and Shelah [2] and the present author’s version [7].

Lemma 2.1**.**

Supposing $F\colon\mathbb{N}\to\omega+1$ is such that for all $n\in\mathbb{N}$

[TABLE]

there is a perfect $\Pi^{0}_{1}(F)$ maximal eventually different family in $\mathcal{N}_{F}$.

Clearly, if $F(n)<\omega$ for every $n\in\mathbb{N}$ then this family will be compact. Note that $\prod\emptyset$ gives $\{\emptyset\}$, i.e., the set containing the empty function, whence (1) implies $F(0)>0$ and in fact $(\forall n\in\mathbb{N})\;F(n)>0$.

Before we begin with the proof, we introduce the main ingredients of the construction and define our medf $\mathcal{E}$. The definitions and proofs which follow are a further streamlined version of those in [7], where the reader will find more explanations.

Definition 2.2**.**

Suppose $F$ satisfies (1), $g\in\mathcal{N}_{F}$ and $c\in{}^{\mathbb{N}}2$.

1. (A)

Fix two $\Delta^{0}_{1}(F)$ bijective coding functions

[TABLE]

where $\operatorname{\#_{2}^{F}}$ is appropriate for $F$ in the following sense: For each $l\in\mathbb{N}$, $h^{*}\in\prod_{k<l}F(k)$ and $d^{*}\in{}^{l}2$, $\operatorname{\#_{2}^{F}}(h^{*},d^{*})<F(l).$ This is possible by (1). When $F$ is clear from the context write $\operatorname{\#_{2}}(h^{*},d^{*})$ for $\operatorname{\#_{2}^{F}}(h^{*},d^{*})$. 2. (B)

Let $\operatorname{e}(g,c)$ be the function in $\mathcal{N}_{F}$ given by

[TABLE] 3. (C)

Suppose $c^{*}\in{}^{<\mathbb{N}}2$. Let

[TABLE]

i.e., the set of $n$ with binary expansion of the form $i_{k}\ldots i_{l}c^{*}(l-1)\ldots c^{*}(0)$, where $l=\operatorname{lh}(c^{*})$ (for some $i_{l},\ldots i_{k}\in\{0,1\}$ and $k\in\mathbb{N}\setminus l$). 4. (D)

Suppose $c^{*}\in{}^{<\mathbb{N}}2$. We say $c^{*}$ is good if and only if whenever $n_{0}<n_{1}$ are two consecutive elements of ${(c^{*})^{-1}}[\{1\}]$ (i.e., $c^{*}(n)=0$ when $n_{0}<n<n_{1}$) then

[TABLE]

We also say $c\in{}^{\mathbb{N}}2$ is good if and only if the same as above holds, i.e., if for every $n\in\mathbb{N}$, $c\mathbin{\upharpoonright}n$ is good. 5. (E)

Let $\operatorname{B}(g,c)=\{2\cdot\#g\mathbin{\upharpoonright}n\;|\;c(n)=1\}\setminus\{n\in\mathbb{N}\;|\;g(n)=\operatorname{e}(g,c)(n)\}.$ 6. (F)

We define a strict partial order $\mathbin{\prec^{g}}$ on $\mathbb{N}$, letting $n_{0}\mathbin{\prec^{g}}n_{1}$ if and only if the following hold:

1. (i)

$n_{0}<n_{1}$, 2. (ii)

for each $i\in\{0,1\}$ letting $(h_{i}^{*},d_{i}^{*})=\operatorname{\#_{2}}^{-1}(g(n_{i}))$ it holds that $h_{i}^{*}$ has length $n_{i}$ (which implies that also $d_{i}^{*}$ has length $n_{i}$), 3. (iii)

$h_{0}^{*}\subseteq h_{1}^{*}$, 4. (iv)

$d_{0}^{*}\subseteq d_{1}^{*}$. 7. (G)

Define $\operatorname{\dot{e}}(g,c)\in\mathcal{N}_{F}$ by

[TABLE] 8. (H)

We define our medf $\mathcal{E}$ as follows:

[TABLE]

Remark 2.3**.**

It is very easy to see that $\mathcal{E}$ is $\Sigma^{1}_{1}(F)$; thus if $\mathcal{E}$ is an medf, $\mathcal{E}$ is $\Delta^{1}_{1}(F)$ because

[TABLE] 2. 2.

If $(g_{0},c_{0})\neq(g_{1},c_{1})$, $e(g_{0},c_{0})$ and $e(g_{1},c_{1})$ are eventually different, where for each $i\in\{0,1\}$, $(g_{i},c_{i})\in\mathcal{N}_{F}\times{}^{\mathbb{N}}2$. 3. 3.

For $g\in\mathcal{N}_{F}$ and $I\subseteq\mathbb{N}$, $I$ is a set of $\mathbin{\prec^{g}}$-comparable points if and only if $g\mathbin{\upharpoonright}I=\operatorname{e}(h,d)\mathbin{\upharpoonright}I$ for some $h\in\mathcal{N}_{F}$ and $d\in{}^{\mathbb{N}}2$. 4. 4.

Note that $\operatorname{B}(g,c)$ is by definition a subset of $2\mathbb{N}$; this ensures that $g$ and $c$ be recovered from $\operatorname{\dot{e}}(g,c)$ in a simple fashion. This is only a matter of convenience; we could delete “$2\cdot$” in 2.2(E) (the only slight change necessary would be in the proof of Claim 2.9 below). 5. 5.

If $c_{0}^{*},c_{1}^{*}\in{}^{<\mathbb{N}}2$ and $c_{0}^{*}\subseteq c_{1}^{*}$ then $\operatorname{I}_{c_{1}^{*}}\subseteq\operatorname{I}_{c_{0}^{*}}$. 6. 6.

If $c_{0}^{*},c_{1}^{*}\in{}^{l}2$ for some $l\in\mathbb{N}$ and $c_{0}^{*}\neq c_{1}^{*}$, clearly $\operatorname{I}_{c_{0}^{*}}\cap\operatorname{I}_{c_{1}^{*}}=\emptyset$. 7. 7.

The set \{{c^{-1}}[\{1\}]\;|\;c\in{}^{\mathbb{N}}2,\text{ c is good}\} is an almost disjoint family: For assume $c_{0}\neq c_{1}$. Find $n\in\mathbb{N}$ such that $c_{0}\mathbin{\upharpoonright}n\neq c_{1}\mathbin{\upharpoonright}n$ and note that for each $i\in\{0,1\}$, ${c_{i}^{-1}}[\{1\}]$ is almost contained in $\operatorname{I}_{c_{i}\mathbin{\upharpoonright}n}$ since $c_{i}$ is good.

We now prove our main lemma.

Proof of Lemma 2.1. Fix $F$ satisfying (1) and let $\mathcal{E}$ etc. be as in Definition 2.2. The proof is split up into several claims. We first show:

Claim 2.4**.**

*The set \{\operatorname{B}(g,c)\;|\;g\in\mathcal{N}_{F},c\in{}^{\mathbb{N}}2,\text{ c is good}\} is an almost disjoint family. *

It will facilitate our argument to introduce the following notation:

Definition 2.5**.**

Suppose $c^{*}\in{}^{<\mathbb{N}}2$ and $g\in\mathcal{N}_{F}$. Let

[TABLE]

Note for $g_{0},g_{1}\in\mathcal{N}_{F}$, and $c_{0}^{*},c_{1}^{*}\in{}^{<\mathbb{N}}2$ we have that $\operatorname{I}^{g_{0}}_{c_{0}^{*}}\cap\operatorname{I}^{g_{1}}_{c_{1}^{*}}$ is finite whenever $g_{0}\neq g_{1}$; and $\operatorname{I}^{g_{0}}_{c_{0}^{*}}\cap\operatorname{I}^{g_{1}}_{c_{1}^{*}}=\emptyset$ whenever $c_{0}^{*}$ and $c_{1}^{*}$ are incomparable w.r.t. $\subseteq$. Moreover note for the proof of Claim 2.7 below that $\operatorname{I}^{g_{0}}_{c_{1}^{*}}\subseteq\operatorname{I}^{g_{0}}_{c_{0}^{*}}$ if $c_{0}^{*}\subseteq c_{1}^{*}$.

We now prove the claim.

Proof of Claim 2.4. Let $(g_{i},c_{i})\in\mathcal{N}_{F}\times{}^{\mathbb{N}}2$ for each $i\in\{0,1\}$ and suppose $(g_{0},c_{0})\neq(g_{1},c_{1})$. Find $n^{*}$ such that $(g_{0}\mathbin{\upharpoonright}n^{*},c_{0}\mathbin{\upharpoonright}n^{*})\neq(g_{1}\mathbin{\upharpoonright}n^{*},c_{1}\mathbin{\upharpoonright}n^{*})$. As for each $i\in\{0,1\}$ we have $\operatorname{B}(g_{i},c_{i})\mathbin{\subseteq^{*}}\operatorname{I}^{g_{i}}_{c_{i}\mathbin{\upharpoonright}n^{*}}$ and $\operatorname{I}^{g_{0}}_{c_{0}\mathbin{\upharpoonright}n^{*}}\cap\operatorname{I}^{g_{1}}_{c_{1}\mathbin{\upharpoonright}n^{*}}$ is finite, we are done. ∎Claim (2.4).

Now it is easy to show:

Claim 2.6**.**

The set $\mathcal{E}$ is an eventually different family.

Proof of claim. Let $f_{i}\in\mathcal{E}$ for each $i\in\{0,1\}$ and assume $f_{0}\neq f_{1}$. Find $g_{i}\in\mathcal{N}_{F}$ and $c_{i}\in{}^{\mathbb{N}}2$ such that $f_{i}=\operatorname{\dot{e}}(g_{i},c_{i})$ for each $i\in\{0,1\}$.

Clearly $\operatorname{\dot{e}}(g_{0},c_{0})$ and $\operatorname{\dot{e}}(g_{1},c_{1})$ can only agree on finitely many points outside of $\operatorname{B}(g_{0},c_{0})\cup\operatorname{B}(g_{1},c_{1})$. By the previous claim and by symmetry, it therefore suffices to show that the set $X$ defined by

[TABLE]

is finite. Assume that $n_{0},n_{1}\in X$ and $n_{0}\neq n_{1}$; then $\operatorname{\dot{e}}(g_{0},c_{0})(n)=\operatorname{e}(g_{0},c_{0})(n)$ whenever $n>n_{1}$ by the definition of $\operatorname{\dot{e}}(g_{0},c_{0})$ and we are done. ∎Claim 2.6.

The next claim is the combinatorial heart of the entire construction and the basis of the following proof that $\mathcal{E}$ is maximal.

Claim 2.7**.**

For any $g\in\mathcal{N}_{F}$ one of the following holds:

1. (1)

There exists an infinite set $I$ together with functions $h\in\mathcal{N}_{F}$ and $d\in{}^{\mathbb{N}}2$ such that

1. (a)

$g\mathbin{\upharpoonright}I=\operatorname{e}(h,d)\mathbin{\upharpoonright}I$, and 2. (b)

$I\cap\operatorname{B}(h,d)$* is finite.* 2. (2)

There exists a good function $c\in{}^{\mathbb{N}}2$ such that no two $n_{0},n_{1}\in\operatorname{B}(g,c)$ are comparable with respect to $\mathbin{\prec^{g}}$.

We postpone the proof of the claim and first show assuming this claim that $\mathcal{E}$ is maximal.

Claim 2.8**.**

Assuming Claim 2.7, the eventually different family $\mathcal{E}$ is maximal.

Proof of Claim 2.8. Let $g\in\mathcal{N}_{F}$ be given. If Case 1 in Claim 2.7 holds find an infinite set $I$, $h\in\mathcal{N}_{F}$ and $d\in{}^{\mathbb{N}}2$ such that

[TABLE]

and $I\cap\operatorname{B}(h,d)$ is finite. As $\operatorname{\dot{e}}(h,d)\in\mathcal{E}$ and $\operatorname{\dot{e}}(h,d)$ agrees with $\operatorname{e}(h,d)$ and thus with $g$ on all but finitely many points in $I$, we are done.

If on the other hand Case 2 in Claim 2.7 holds, we may find a good function $c\in{}^{\mathbb{N}}2$ such that $g$ agrees with $\operatorname{\dot{e}}(g,c)$ on an infinite set, and we are also done, proving maximality. ∎Claim 2.8.

Now it is high time we prove Claim 2.7.

Proof of Claim 2.7. Write

[TABLE]

i.e., let $C$ denote set of finite sequences from $\{0,1\}$ which end in $1$ together with the empty sequence. Let $g\in\mathcal{N}_{F}$ be given.

Assume first that

[TABLE]

Fix $c^{*}$ witnessing the existential quantifier above and let $n_{0}=\min\operatorname{I}^{g}_{c^{*}}$ noting that

[TABLE]

where $c_{1}^{*}=c^{*}\mathbin{{}^{\frown}}0^{n_{0}-\operatorname{lh}(c^{*})}\mathbin{{}^{\frown}}1$.

By (2) and as $\operatorname{I}^{g}_{d^{*}}\subseteq\operatorname{I}^{g}_{c^{*}}$ for every $d^{*}\in C$ which extends $c^{*}$, we can for each $k\in\mathbb{N}$ recursively find $n_{k+1}$ so that $n_{k+1}\in\operatorname{I}^{g}_{c^{*}}$ and $n_{k}\mathbin{\prec^{g}}n_{k+1}$. Thus we can find $h\in\mathcal{N}_{F}$ and $d\in{}^{\mathbb{N}}2$ such that $g\mathbin{\upharpoonright}I=\operatorname{e}(h,d)\mathbin{\upharpoonright}I$, where $I=\{n_{k}\;|\;k\in\mathbb{N}\}$.

Moreover $I\cap\operatorname{B}(h,d)$ is finite: Since $g(n_{0})\neq\operatorname{\#_{2}}(g\mathbin{\upharpoonright}n_{0},c_{1}^{*}\mathbin{\upharpoonright}n_{0})$, $(h\mathbin{\upharpoonright}n_{0},d\mathbin{\upharpoonright}n_{0})\neq(g\mathbin{\upharpoonright}n_{0},c_{1}^{*}\mathbin{\upharpoonright}n_{0})$ by (3), and so $\operatorname{I}^{h}_{d\mathbin{\upharpoonright}n_{0}}\cap\operatorname{I}^{g}_{c_{1}^{*}\mathbin{\upharpoonright}n_{0}}=\emptyset$ while $\operatorname{B}(h,d)\subseteq^{*}\operatorname{I}^{h}_{d\mathbin{\upharpoonright}n_{0}}$ and $I\subseteq^{*}\operatorname{I}^{g}_{c_{1}^{*}\mathbin{\upharpoonright}n_{0}}$ (the latter holds since $n_{k}\geq n_{0}$ for each $k\in\mathbb{N}$). Thus we have that Case 1 of the claim holds.

Now assume to the contrary that (2) fails, i.e., it holds that

[TABLE]

Let $c_{0}^{*}=\emptyset$ and recursively chose $n_{k}$ and $c_{k+1}^{*}$ for each $k\in\mathbb{N}$ such that $c_{k+1}^{*}=c_{k}^{*}\mathbin{{}^{\frown}}0^{n_{k}-\operatorname{lh}(c^{*})}\mathbin{{}^{\frown}}1$ and

[TABLE]

holds. Letting $c=\bigcup_{k\in\mathbb{N}}c_{k}^{*}$, we have that $c\in{}^{\mathbb{N}}2$ is good, and no $n_{0},n_{1}\in\operatorname{B}(g,c)$ are comparable w.r.t. $\mathbin{\prec^{g}}$. Thus, Case 2 of the claim holds. ∎Claim (2.7).

Finally, we have:

Claim 2.9**.**

The medf $\mathcal{E}$ is $\Pi^{0}_{1}(F)$.

This is fairly obvious. To be able to formulate a concise proof we extend Definitions 2.2(E) and 2.2(F) in a straightforward manner:

Definition 2.10**.**

Suppose for some $l\in\mathbb{N}$, $c^{*}\in{}^{l}2$ and $g^{*}\in\prod_{n<l}F(n)$. Define

[TABLE]

Moreover let $n_{0}\mathbin{\prec^{g^{*}}}n_{1}$ if and only if the following hold:

1. (i)

$n_{0}<n_{1}<\operatorname{lh}(g^{*})$, 2. (ii)

for each $i\in\{0,1\}$ letting $(h_{i}^{*},d_{i}^{*})=\operatorname{\#_{2}}^{-1}(g^{*}(n_{i}))$ it holds that $h_{i}^{*}$ has length $n_{i}$ (which implies that also $d_{i}^{*}$ has length $n_{i}$), 3. (iii)

$h_{0}^{*}\subseteq h_{1}^{*}$, 4. (iv)

$d_{0}^{*}\subseteq d_{1}^{*}$.

Proof of Claim 2.9. Clearly, $\mathcal{E}=[T]$ where $T$ is the tree consisting of those

[TABLE]

such that for any odd $m<\operatorname{lh}(f^{*})$ letting

[TABLE]

we have $\operatorname{lh}(g^{*})=\operatorname{lh}(c^{*})=m$, and for every $n<m$,

1. (I)

$f^{*}(n)=g^{*}(n)$ if all of the following holds:

1. (a)

$c^{*}$ is good, 2. (b)

no two $n_{0},n_{1}\in\operatorname{B}(g^{*},c^{*})\cap n$ are comparable w.r.t. $\mathbin{\prec^{g^{*}}}$, 3. (c)

$n\in\operatorname{B}(g^{*},c^{*})$; 2. (II)

if any of (Ia)–(Ic) above fails, $f^{*}(n)=\operatorname{\#_{2}}(g^{*}\mathbin{\upharpoonright}n,c^{*}\mathbin{\upharpoonright}n)$.

Lastly, clearly $T$ is $\Delta^{0}_{1}(F)$. ∎Claim 2.9 and Lemma 2.1

3. Proof of the main theorem

Before we give the proof of Theorem 1.2, i.e., that there is a $\Pi^{0}_{1}$ medf in $\mathcal{N}_{F}$, it will be convenient to give a yet broader definition of ‘maximal eventually different family’:

Definition 3.1**.**

Any two functions $g_{0},g_{1}$ with countable domain $X$ are called eventually different if and only if $\{x\in X\;|\;g_{0}(x)=g_{1}(x)\}$ is finite.

Suppose $E\subseteq\omega$ and $F\colon E\to\omega+1$. A set $\mathcal{E}$ is an eventually different family in $\prod_{n\in E}F(n)$ if and only if $\mathcal{E}\subseteq\mathcal{\prod}_{n\in E}F(n)$ and any two distinct $g_{0},g_{1}\in\mathcal{E}$ are eventually different; such a family is called maximal (or short: a medf) if and only if it is maximal among such families under inclusion.

We now have the prerequisites to give a transparent proof of our main result. This proof has a precursor in [7] where we also enlarged a medf defined on a factor space to a medf in the entire (product) space.

Proof of Theorem 1.2.

The proof is slightly easier if we assume that $\operatorname{\#_{1}}$ was chosen so that $\operatorname{\#_{1}}\emptyset=0$, so let us make this assumption from now on. As $\lim_{n\to\infty}F(n)=\omega$ we may find a sequence $\langle e_{m}\;|\;m\in\mathbb{N}\rangle$ which is $\Delta^{0}_{1}(F)$ such that $e_{0}=0$ and for each $m\in\mathbb{N}$ we have

[TABLE]

Let $E=\{e_{m}\;|\;m\in\mathbb{N}\}$ and let $e\colon\mathbb{N}\to\mathbb{N}$ denote the map $m\mapsto e_{m}$.

As by (5), $F\circ e$ satisfies the growth condition in Theorem 2.1 the proof of said theorem gives us a $\Pi^{0}_{1}(F)$ medf $\mathcal{E}_{0}$ in the space $\prod_{n\in E}F(n)$. In fact, the proof gives us a $\Delta^{0}_{1}(F)$ tree $T$ such that for any $f\in\mathcal{N}_{F}$,

[TABLE]

For $f\in\mathcal{E}_{0}$ define $\operatorname{\ddot{e}}(f)\in\mathcal{N}_{F}$ as follows:

[TABLE]

This is well defined as $\operatorname{\#_{1}}\emptyset<F(n)$ for all $n\in\mathbb{N}$. (It may be interesting to note that one can delete the requirement $m\leq n$ above in case $F[\mathbb{N}\setminus E]\subseteq\mathbb{N}$ for the purposes of the present proof.)

We show that $\mathcal{E}=\{\operatorname{\ddot{e}}(f)\;|\;f\in\mathcal{E}_{0}\}$ is a $\Pi^{0}_{1}(F)$ medf. It is maximal as

[TABLE]

and $\mathcal{E}_{0}$ is maximal in $\prod_{n\in E}F(n)$: Whenever $h\in\mathcal{N}_{F}$ there is $f\in\mathcal{E}_{0}$ such that $h\mathbin{\upharpoonright}E$ and $f$ agree on infinitely many points from $E$, so $\operatorname{\ddot{e}}(f)\mathbin{=^{\infty}}h$.

Clearly, $\mathcal{E}$ is also an eventually different family, as $\mathcal{E}_{0}$ is: For two distinct functions $f_{0}$ and $f_{1}$ from $\mathcal{E}_{0}$, find $m_{0}\in E$ is such for all $m\in E\setminus m_{0}$, $f_{0}(m)\neq f_{1}(m)$. Further, find $n_{0}>m_{0}$ such that for all $n\geq n_{0}$

[TABLE]

Then for all $n\in\mathbb{N}\setminus n_{0}$ we have $\operatorname{\ddot{e}}(f_{0})(n)\neq\operatorname{\ddot{e}}(f_{1})(n)$.

We show $\mathcal{E}$ is $\Pi^{0}_{1}(F)$. Obviously $g\in\mathcal{E}$ if and only if for every $n,m\in\mathbb{N}$

• •

$(g\mathbin{\upharpoonright}n)\circ e\in T$, and

• •

whenever $n\notin E$, and $m$ is maximal such that $m\leq n$ and \operatorname{\#_{1}}\big{(}(g\mathbin{\upharpoonright}m)\circ e\big{)}<F(n), we have g(n)=\operatorname{\#_{1}}\big{(}(g\mathbin{\upharpoonright}m)\circ e\big{)}.

Clearly we can compute $(g\mathbin{\upharpoonright}n)\circ e$ from $g\mathbin{\upharpoonright}n$ (relative to $F$). Thus all the requirements in the above definition of $\mathcal{E}$ past the universal quantifier over $n$ and $m$ are $\Delta^{0}_{1}(F)$. ∎

4. Proof of the corollary

Corollary 1.3 now immediately follows using the following rather trivial facts. We include proofs for the convenience of the reader. For the remainder, fix $F\colon\mathbb{N}\to(\omega+1)\setminus\{0\}$.

Fact 4.1**.**

There is a compact $\Pi^{0}_{1}(F)$ medf on $\mathcal{N}_{F}$ if and only if $F(n)<\omega$ for infinitely many $n\in\mathbb{N}$.

Proof.

Let $D=\{n\in\mathbb{N}\;|\;F(n)<\omega\}$. If $D$ is finite, clearly there cannot be a compact medf, as for every compact $\mathcal{E}\subseteq\mathcal{N}_{F}$ there is $f\in\mathcal{N}_{F}$ such that $f$ eventually dominates every $g\in\mathcal{E}$, i.e., $\{n\in\mathbb{N}\;|\;f(n)\leq g(n)\}$ is finite.

Conversely, suppose $D$ is infinite and show there is a compact $\Pi^{0}_{1}(F)$ medf. Clearly we can assume $\lim_{n\to\infty}F(n)=\omega$ as otherwise there is a recursive finite medf consisting of constant functions.

The proof is almost the same as that of Theorem 1.2, so we only point out the necessary changes. When defining $\langle e_{n}\;|\;n\in\mathbb{N}\rangle$ as in the proof of Theorem 1.2, let $e_{0}$ be the least element of $D$ and when choosing $e_{m}$ for $m>0$ demand in addition that $e_{m}\in D$ as well.

When defining $\operatorname{\ddot{e}}(f)$ from $f\in\mathcal{E}_{0}$, demand that $\operatorname{\ddot{e}}(f)(n)=0$ for $n<e_{0}$. The $\Pi^{1}_{0}$ condition for membership in $\mathcal{E}$ is easily adapted from the one in the proof of Theorem 1.2. The rest of the proof can be followed verbatim; we produce a compact medf $\mathcal{E}$ as for every $g\in\mathcal{E}$ and every $n\in\mathbb{N}$ we have $g(n)<F(n)<\omega$ if $n\in E$ and for $n\notin E$ we have

[TABLE]

where the right-hand side is a finite set. ∎

Finally we have:

Fact 4.2**.**

If $\lim_{n\to\omega}F(n)=\omega$ every medf is uncountable; otherwise, every medf is finite.

Proof.

If $\liminf_{n\to\omega}F(n)<\omega$, every eventually different family is finite: towards a contradiction find $m^{*}\in\mathbb{N}$ such that $\{n\in\mathbb{N}\;|\;F(n)<m^{*}\}$ is infinite. By the pigeonhole principle, there is no eventually different family of size $m^{*}$.

If on the other hand $\lim_{n\to\omega}F(n)=\omega$, a simple diagonalization argument shows that there is no countable medf. ∎

5. Questions

Is it the case that for some $F\colon\mathbb{N}\to\mathbb{N}$ there is a compact $\Pi^{0}_{1}(F)$ maximal cofinitary group in $\mathcal{N}_{F}$? 2. 2.

For which $F$ is the answer to the previous question ‘yes’ (if any)? It is easy to see that it is necessary that $F(n)>n$ for all but finitely many $n$. 3. 3.

Is there a natural, minimal fragment of second order arithmetic which proves there is a $\Pi^{0}_{1}$ eventually different family? 4. 4.

For any set $X$ let $X^{[\infty]}$ denote the set of infinite subsets of $X$. Given any $F\colon\mathbb{N}\to\{\mathbb{N}\}\cup\mathbb{N}$ and a medf $\mathcal{E}$ on $\mathcal{N}_{F}$ consider the co-ideal

[TABLE]

Is there a closed medf $\mathcal{E}$ in $\mathcal{N}$ or $\mathcal{N}_{F}$ (under some assumption on $F$) such that $\mathcal{C}_{\mathcal{E}}=\mathbb{N}^{[\infty]}$?