2 An interpolation theorem

A key part of the proof of Theorem 2 is the (ZFC) Theorem 5 below. One of the referees pointed out that Theorem 5 follows from known results; in particular, it follows from the much more powerful Theorem 3.2 of Burke [Reference Burke4] or, with modifications in the proofs, either Theorem 2 of Barth-Schneider [Reference Barth and Schneider3] or Corollary 1.9 of Burke [Reference Burke5]. Since deriving Theorem 5 from those more powerful theorems is not trivial, we choose to present our original direct proof of Theorem 5.

Recall that Cantor proved that any two countable dense subsets of $\mathbb {R}$ are order-isomorphic and that this order-isomorphism easily extends uniquely to a homeomorphism of $\mathbb {R}$ . Franklin [Reference Franklin9] considered the question of how nice this homeomorphism could be arranged to be, and showed that if D and E are countable dense subsets of $\mathbb {R}$ , then there is an order-isomorphism of D with E that extends to a real analytic function. A series of papers improved this result, culminating in Barth-Schneider [Reference Barth and Schneider3], who proved that there is an order-isomorphism of D with E that extends to an entire function $f: \mathbb {C} \to \mathbb {C}$ , answering (one interpretation of) Question 24 of Erdős [Reference Erdős8].Footnote ³ Subsequent work of Burke, mentioned above, further strengthened those results. The variant we will need for the proof of Theorem 2 follows.

Let us give a brief outline of the following proof of Theorem 5, which is inspired by the proof of Nienhuys-Thiemann [Reference Nienhuys and Thiemann14]. We will inductively define a sequence of functions $\langle f_n : n\in \mathbb {N}\rangle $ whose limit will be the desired function f. Each function $f_n$ will satisfy a version of Theorem 5(4) for finitely many points in W. When we define the next function $f_{n+1}$ , we will want it to be equal to $f_n$ on these finitely many points in W that have already been taken care of, and we will want $f_{n+1}$ to satisfy Theorem 5(4a) or Theorem 5(4b), depending on whether n is even or odd, for an additional point in W. We will write $A_n$ to denote the set of finitely many points of W that have already been taken care of at stage n with regard to Theorem 5(4a), and we will write $B_n$ to denote the set of finitely many points of W that have been taken care of in regard to Theorem 5(4b).

Suppose $\mathcal {D}$ is a partition of $\mathbb {R}$ into dense sets, W is a countable set of real numbers, and $P=(x_P,y_P)$ is a point in $\mathbb {R}^2$ . Fix a 1-1 enumeration $\{ w_n \ : \ n \in \mathbb {N} \}$ of W, and for each n, let $D_n$ be the unique member of $\mathcal {D}$ containing $w_n$ . Since $\mathcal {D}$ is a partition, we have

(*) $$ \begin{align} \forall k,n \in \mathbb{N} \ \ \big( w_k \in D_n \ \iff \ D_k=D_n \ \iff \ w_n \in D_k \big). \end{align} $$

Suppose $p:{\mathbb {R}}\to {\mathbb {R}}$ is a continuous positive function such that

(2) $$ \begin{align} \forall n \in \mathbb{N} \ \lim_{t\rightarrow\infty}\frac{p(t)}{t^n}=\infty. \end{align} $$

We will inductively define sequences $\langle f_n : n\in \mathbb {N}\rangle $ , $\langle A_n : n\in \mathbb {N}\rangle $ and $\langle B_n : n\in \mathbb {N}\rangle $ such that $A_0=\emptyset $ and $B_0=\emptyset $ and for all $n\in \mathbb {N}$ , we have

1. (I) _n $f_n:{\mathbb {C}}\to {\mathbb {C}}$ is entire and $f_n\upharpoonright {\mathbb {R}}$ is real-valued;

2. (II) _n $f_n(x_P)=y_P$ ;

3. (III) _n $\forall x\in {\mathbb {R}} \ f_n'(x)\geq \frac {1}{2}+\frac {1}{2^n}$ , and thus $f_n\upharpoonright {\mathbb {R}}$ is a bijection;

4. (IV) _n if $n>0$ , then $\forall z\in {\mathbb {C}}\ |f_n(z)-f_{n-1}(z)|<\frac {1}{2^n}p(|z|)$ ;

5. (V) _n if $n=2k+1$ is odd, then $A_n=A_{n-1}\cup \{w_k\}$ , $B_n=B_{n-1}$ and we have $w_k\neq x_P \ \implies \ f_n(w_k)\in D_k$ ;

6. (VI) _n if $n=2k+2$ is even, then $A_n=A_{n-1}$ , $B_n=B_{n-1}\cup \{w_k\}$ and we have $w_k\neq y_P \ \implies \ f_n^{-1}(w_k)\in D_k$ ; and

7. (VII) _n if $n>0$ , then $f_n\upharpoonright A_{n-1}=f_{n-1}\upharpoonright A_{n-1}$ and $f_n^{-1}\upharpoonright B_{n-1}=f_{n-1}^{-1}\upharpoonright B_{n-1}$ .

First, let us show that, assuming we have sequences $\langle f_n : n\in \mathbb {N}\rangle $ and $\langle A_n : n\in \mathbb {N}\rangle $ and $\langle B_n : n\in \mathbb {N}\rangle $ satisfying (I) $_n$ -(VII) $_n$ for all n, the pointwise limit defined by $f(z)=\lim _{n\to \infty }f_n(z)$ has all of the desired properties. Suppose D is any compact subset of ${\mathbb {C}}$ . Since $\sum _{n=1}^\infty \frac {1}{2^n}$ converges and since $p(|z|)$ is bounded on D, the fact that (IV) $_n$ holds for all n ensures that the sequence $\langle f_n : n\in \mathbb {N}\rangle $ is uniformly Cauchy on D. Hence, we can define a function $f:{\mathbb {C}}\to {\mathbb {C}}$ by letting $f(z)=\lim _{n\to \infty }f_n(z)$ . Since the sequence $\langle f_n : n\in \mathbb {N}\rangle $ is uniformly Cauchy on any compact set, it follows that the convergence of $\langle f_n : n\in \mathbb {N}\rangle $ to f is uniform on any compact set, and hence, f is an entire function.

Now let us verify that Theorem 5(1)–(4) hold for f. By (I) $_n$ and closure of ${\mathbb {R}}$ in ${\mathbb {C}}$ , we see that $f\upharpoonright {\mathbb {R}}$ is real valued, and since (III) $_n$ holds for all n, we have $f'(x)\geq \frac {1}{2}$ for all $x\in {\mathbb {R}}$ . Thus, Theorem 5(1) and Theorem 5(2) hold. Theorem 5(3) holds since the sequence $\langle f_n(x_P) : n\in \mathbb {N}\rangle $ is constantly equal to $y_P$ . To show that Theorem 5(4) holds, let us prove that for all $i\in \mathbb {N}$ , if $w_i\neq x_P$ , then $f(w_i)\in D_i$ , and if $w_i\neq y_P$ , then $f^{-1}(w_i)\in D_i$ . Fix $i\in \mathbb {N}$ . We have $w_i\in A_{2i+1}$ and $w_i\in B_{2i+2}$ , and furthermore, by (V) $_{2i+1}$ and (VI) $_{2i+2}$ , $w_i\neq x_P$ implies $f_{2i+1}(w_i)\in D_i$ and $w_i\neq y_P$ implies $f_{2i+2}^{-1}(w_i)\in D_i$ . Since (VII) $_n$ holds for all n, we see that both of the sequences $\langle f_n(w_i) : n\in \mathbb {N}\rangle $ and $\langle f_n^{-1}(w_i) : n\in \mathbb {N}\rangle $ are eventually constant, and indeed, for $n\geq 2i+2$ , we have $f_n(w_i)=f_{2i+1}(w_i)$ and $f_n^{-1}(w_i)=f_{2i+2}^{-1}(w_i)$ . Therefore, $f(w_i)=f_{2i+1}(w_i)$ and $f^{-1}(w_i)=f_{2i+2}^{-1}(w_i)$ , so (4) holds.

It remains to show that we can inductively define sequences $\langle f_n : n\in \mathbb {N}\rangle $ , $\langle A_n : n\in \mathbb {N}\rangle $ and $\langle B_n : n\in \mathbb {N}\rangle $ that satisfy (I) $_n$ –(VII) $_n$ for all $n\in \mathbb {N}$ .

Let $f_0:{\mathbb {C}}\to {\mathbb {C}}$ be $f_0(z)=\frac {3}{2}(z-x_P)+y_P$ , $A_0=\emptyset $ and $B_0=\emptyset $ . One may easily verify that (I) $_0$ –(VII) $_0$ hold. For $n> 0$ , Section 2.1 shows how $f_n$ is constructed when n is odd, and Section 2.2 shows how $f_n$ is constructed when n is even.

2.1 When n is odd

Suppose $n=2k+1>0$ is odd and that $f_i$ , $A_i$ and $B_i$ satisfying (I) $_i$ –(VII) $_i$ have already been defined for $i\leq 2k$ . If $k=0$ , we have $A_0=\emptyset $ and $B_0=\emptyset $ , whereas if $k>0$ , we have

$$\begin{align*}A_{n-1}=A_{2k}=A_{2(k-1)+2}=\{w_0,\ldots,w_{k-1}\}\end{align*}$$

and

$$\begin{align*}B_{n-1}=B_{2k}=\{w_0,\ldots,w_{k-1}\}.\end{align*}$$

In any case, we let $A_{n}=A_{n-1}\cup \{w_k\}$ and $B_{n}=B_{n-1}$ . We define $f_{n} = f_{2k+1}$ in two cases as follows.

Case 2.1. A: $\boldsymbol {w_k\notin \{x_P\}\cup A_{n-1}\cup f_{n-1}^{-1}(B_{n-1})}$ . Let us argue that there is an entire function $g_{n}$ such that

1. (i) $(\forall z\in {\mathbb {C}}) \ g_{n}(z)=0 \ \iff \ z\in \{x_P\}\cup A_{n-1}\cup f_{n-1}^{-1}(B_{n-1})$ ,

2. (ii) $(\forall z\in {\mathbb {C}}) \ |g_{n}(z)|\leq \frac {1}{2^{n}}p(|z|)$ and

3. (iii) $(\forall x\in {\mathbb {R}}) \ g_{n}'(x)\geq -\frac {1}{2^{n}}$ .

Take

$$\begin{align*}h_n(z)=(z-x_P)^{\beta_n}(z-w_0)\cdots(z-w_{k-1})(z-f_{n-1}^{-1}(w_0))\cdots(z-f_{n-1}^{-1}(w_{k-1})),\end{align*}$$

where $\beta _n\in \{1,2\}$ is such that the degree of $h_n$ is odd. We will show that for small enough positive $\alpha _n\in {\mathbb {R}}$ , the function $g_n(z)=\alpha _nh_n(z)$ satisfies (i)–(iii). Clearly, $h_n$ satisfies (i), so any such function $g_n(z)$ satisfies (i). For (ii), choose $m\in \mathbb {N}$ and some positive $c\in {\mathbb {R}}$ such that $|h_n(z)|\leq |z|^m+c$ for all $z\in {\mathbb {C}}$ . By our assumption on p, we have $\lim _{|z|\rightarrow \infty }\frac {p(|z|)}{|z|^m+c}=\infty $ , and thus we can let $D\subseteq {\mathbb {C}}$ be a large enough closed disk centered at the origin such that $z\in {\mathbb {C}}\setminus D$ implies $1\leq \frac {p(|z|)}{|z|^m+c}$ . Since p is a continuous positive function, we can choose a positive $\alpha _n\in {\mathbb {R}}$ such that $\alpha _n\leq \frac {1}{2^n}$ and $\alpha _n\leq \frac {p(|z|)}{2^n(|z|^m+c)}$ for all $z\in D$ . Then it follows that for every $z\in {\mathbb {C}}$ , we have

$$\begin{align*}|\alpha_nh_n(z)|\leq\alpha_n(|z|^m+c)\leq\frac{1}{2^n}p(|z|).\end{align*}$$

Let us verify that (iii) holds for small enough $\alpha _n$ . Since $h_n$ is odd and has a positive leading coefficient, the derivative of $h_n\upharpoonright {\mathbb {R}}$ is bounded below. So we may let $d=\inf \{h_n'(x) : x\in {\mathbb {R}}\}\in {\mathbb {R}}$ . Thus, we may choose a small enough positive $\alpha _n\in {\mathbb {R}}$ such that $\alpha _nd\geq -\frac {1}{2^n}$ , and then it follows that for all $x\in {\mathbb {R}}$ , we have $\alpha _nh_n'(x)\geq \alpha _n d\geq -\frac {1}{2^n}$ .

Using the case assumption that $w_k\notin \{x_P\}\cup A_{n-1}\cup f_{n-1}^{-1}(B_{n-1})$ , we see that $g_n(w_k)\neq 0$ , and hence it follows that the set

$$\begin{align*}\{f_{n-1}(w_k)+Mg_{n}(w_k) : M\in[0,1]\}\end{align*}$$

is a nontrivial interval of real numbers. Thus, since $D_k$ is dense in ${\mathbb {R}}$ , it follows that there is some $M_{n}\in [0,1]$ such that $f_{n-1}(w_k)+M_{n}g_{n}(w_k)\in D_k$ . We define

$$\begin{align*}f_{n}(z)=f_{n-1}(z)+M_{n}g_{n}(z).\end{align*}$$

Let us show that (I) $_n$ –(VII) $_n$ hold. It is trivial to see that (I) $_n$ and (II) $_n$ are true. For (III) $_n$ , notice that because $M_n\in [0,1]$ , and since (iii) and (III) $_{n-1}$ both hold, we have for all $x\in {\mathbb {R}}$ ,

$$\begin{align*}f_n'(x)=f_{n-1}'(x)+M_{n}g^{\prime}_{n}(x)\geq\frac{1}{2}+\frac{1}{2^{n-1}}-\frac{1}{2^{n}}=\frac{1}{2}+\frac{1}{2^{n}},\end{align*}$$

and thus $f_n:{\mathbb {R}}\to {\mathbb {R}}$ is a bijection. For (IV) $_n$ , we have for all $z\in {\mathbb {C}}$ ,

$$\begin{align*}|f_n(z)-f_{n-1}(z)|=M_{n}|g_{n}(z)|\leq \frac{1}{2^{n}}p(|z|),\end{align*}$$

where the last inequality follows since $M_n\in [0,1]$ and (ii) holds. Let us verify that (V) $_n$ holds. From the definition of $f_n=f_{2k+1}$ and the way we chose $M_{n}$ , it follows that $f_n(w_k)\in D_k$ (notice that $w_k \neq x_P$ by our case assumption). Thus, (V) $_n$ holds. (VI) $_n$ holds trivially since n is odd. To see that (VII) $_n$ holds, note that since $g_{n}(z)=0$ if $z\in \{x_P\}\cup A_{n-1}\cup f_{n-1}^{-1}(B_{n-1})$ , it follows directly from the definition of $f_n$ that $f_n\upharpoonright A_{n-1}=f_n\upharpoonright A_{n-1}$ and $f_n^{-1}\upharpoonright B_n=f_{n-1}^{-1}\upharpoonright B_{n-1}$ .

Case 2.1. B: $\boldsymbol {w_k\in \{x_P\}\cup A_{n-1}\cup f_{n-1}^{-1}(B_{n-1})}$ . Then we let $f_n=f_{n-1}$ , $A_n=A_{n-1}\cup \{w_k\}$ and $B_n=B_{n-1}$ . Let us argue that this definition of $f_n$ satisfies (V) $_n$ ; the rest of (I) $_n$ –(VII) $_n$ are easily seen to hold by the inductive hypothesis. Suppose $w_k\neq x_P$ . Since the enumeration of W is one-to-one, we have $w_k\neq w_j$ for all $j\leq k-1$ . Thus, for some $j\leq k-1$ , we have $w_k=f_{n-1}^{-1}(w_j)$ , and because $f_{n-1}(x_P)=y_P$ , $f_{n-1}$ is injective and $w_k\neq x_P$ , it follows that $w_j\neq y_P$ . Since $2j+2\leq n-1$ and since it follows by our inductive assumptions (VII) $_{\ell }$ for $\ell \leq n-1$ , that $f_{n-1}\upharpoonright A_{2j+2}=f_{2j+2}\upharpoonright A_{2j+2}$ , we see that $w_k=f_{n-1}^{-1}(w_j)=f_{2j+2}^{-1}(w_j)\in D_j$ . Then $D_j = D_k$ by (*) from page 4. So, $f_n(w_k)=f_{n-1}(w_k)=w_j\in D_j=D_k$ , and hence, (V) $_n$ holds.

2.2 When n is even

Now suppose $n=2k+2$ is even, where $k>0$ , and that $f_i$ , $A_i$ and $B_i$ satisfying (I) $_i$ –(VI) $_i$ have already been defined for $i\leq 2k+1$ . We have

$$\begin{align*}A_{2k+1}=\{w_0,\ldots,w_k\}\end{align*}$$

and

$$\begin{align*}B_{2k+1}=\{w_0,\ldots,w_{k-1}\}.\end{align*}$$

We will define $f_n$ , $A_n$ and $B_n$ in two cases as follows.

Case 2.2. A: $\boldsymbol {f_{n-1}^{-1}(w_k)\notin \{x_P\}\cup A_{n-1}\cup f_{n-1}^{-1}(B_{n-1})}$ . Then we let $g_{n}$ be an entire function such that

1. (i) $(\forall z\in {\mathbb {C}}) \ g_{n}(z)=0 \ \iff \ z\in \{x_P\}\cup A_{n-1}\cup f_{n-1}^{-1}(B_{n-1})$ ,

2. (ii) $(\forall z\in {\mathbb {C}}) \ |g_{n}(z)|\leq \frac {1}{2^{n}}p(|z|)$ and

3. (iii) $(\forall x\in {\mathbb {R}})\ g_{n}'(x)\geq -\frac {1}{2^{n}}$ .

For example, as in the case above where n was odd, we could take

$$\begin{align*}g_n(z)=\alpha_n(z-x_P)^{\beta_n}(z-w_0)\cdots(z-w_k)(z-f_{n-1}^{-1}(w_0))\cdots(z-f_{n-1}^{-1}(w_{k-1}))\end{align*}$$

satisfying (i)–(iii) by choosing $\alpha _n$ small enough and $\beta _n\in \{1,2\}$ so that the degree of $g_n$ is odd. By our inductive assumption about $f_{n-1}$ and by (iii), it follows that for any $M\in [0,1]$ and any $x\in {\mathbb {R}}$ , we have

$$\begin{align*}f_{n-1}'(x)+Mg_{n}'(x)\geq\frac{1}{2}+\frac{1}{2^{n-1}}-\frac{1}{2^{n}}=\frac{1}{2}+\frac{1}{2^{n}}>0.\end{align*}$$

Thus, the function $f_{n-1}+Mg_{n}:{\mathbb {R}}\to {\mathbb {R}}$ is a bijection. Let us argue that the set

$$\begin{align*}\{(f_{n-1}+Mg_{n})^{-1}(w_k) : M\in[0,1]\}\end{align*}$$

is a nontrivial interval of real numbers. It will suffice to show that $(f_{n-1}+g_n)^{-1}(w_k)\neq f^{-1}_{n-1}(w_k)$ . Suppose $(f_{n-1}+g_n)^{-1}(w_k)= f_{n-1}^{-1}(w_k)$ . Then $f_{n-1}(f_{n-1}^{-1}(w_k))=w_k$ and $(f_{n-1}+g_n)(f^{-1}_{n-1}(w_k))=w_k$ . This implies that the functions $f_{n-1}$ and $f_{n-1}+g_n$ are equal at the point $f_{n-1}^{-1}(w_k)$ , and hence, $g_n(f_{n-1}^{-1}(w_k))=0$ , which contradicts (i) by our case assumption that $f_{n-1}^{-1}(w_k) \notin \{x_P\}\cup A_{n-1}\cup f_{n-1}^{-1}(B_{n-1})$ .

Thus, since $D_k$ is dense in ${\mathbb {R}}$ , it follows that there is some $M_{n}\in [0,1]$ such that $(f_{n-1}+M_{n}g_{n})^{-1}(w_k)\in D_k$ . We fix such an $M_{n}$ and define

$$\begin{align*}f_{n}(z)=f_{n-1}(z)+M_{n}g_{n}(z).\end{align*}$$

We also let $A_n=A_{n-1}$ and $B_n=B_{n-1}\cup \{w_k\}$ . The verification that (I) $_n$ –(VII) $_n$ hold is straightforward and similar to the above; it is therefore left to the reader.

Case 2.2. B: $\boldsymbol {f_{n-1}^{-1}(w_k)\in \{x_P\}\cup A_{n-1}\cup f_{n-1}^{-1}(B_{n-1})}$ , or equivalently, $w_k\in \{y_P\}\cup f_{n-1}(A_{n-1})\cup B_{n-1}$ . Then we define $f_n=f_{n-1}$ . As in the odd case above, this definition of $f_n$ is easily seen to satisfy (I) $_n$ –(V) $_n$ and (VII) $_n$ . Let us check (VI) $_n$ . Suppose $w_k\neq y_P$ . Since the enumeration of W is one-to-one, we have $w_k\neq w_j$ for all $j\leq k-1$ , and hence, $w_k=f_{n-1}(w_j)$ for some $j\leq k$ , where $w_j\neq x_P$ . Since $2j+1\leq n-1$ , it follows by our inductive assumptions (V) $_{\ell }$ for $\ell \leq n-1$ that $f_{n-1}\upharpoonright A_{2j+1}=f_{2j+1}\upharpoonright A_{2j+1}$ and $w_k=f_{n-1}(w_j)=f_{2j+1}(w_j)\in D_j$ . Then by (*) from page 4, $D_j = D_k$ . So $f_n^{-1}(w_k) = f_{n-1}^{-1}(w_k) = w_j \in D_j = D_k$ .

This concludes the proof of Theorem 5.

3 Proof of Theorem 2

To prove the $\Leftarrow $ direction of Theorem 2, assume that $\big \{ f_P \ : \ P \in \mathbb {R}^2 \big \}$ is a sparse analytic system and consider the subcollection $\big \{ f_{(0,y)} \ : \ y \in \mathbb {R} \big \}$ . Since $f_{(0,y)}$ passes through the point $(0,y)$ and each $f_{(0,y)}$ is analytic, and hence continuous, it follows that for $y \ne y'$ , $f_{(0,y)} \restriction (-\infty ,0) \ne f_{(0,y')} \restriction (-\infty ,0)$ . So

$$\begin{align*}\mathcal{F}:= \Big\{ f_{(0,y)} \restriction (-\infty,0) \ : \ y \in \mathbb{R} \Big\} \end{align*}$$

is a continuum-sized collection of analytic functions on the common domain $D:=(-\infty ,0)$ . Furthermore, given any $z \in D$ , since $z \ne 0$ and the $f_P$ ’s formed a sparse analytic system, it follows that

$$\begin{align*}\{ f_{(0,y)} (z) \ : \ y \in \mathbb{R} \} \end{align*}$$

is countable. So $\mathcal {F}$ is a collection of analytic functions as in clause (2) of Erdős’ Theorem 1. So by that theorem, CH must hold.

To prove the $\Rightarrow $ direction of Theorem 2 – which is heavily inspired by Erdős’ proof of Theorem 1 – assume CH and fix an enumeration $\langle w_\alpha \ : \ \alpha < \omega _1 \rangle $ of $\mathbb {R}$ . Fix any partition $\mathcal {D}$ of the reals into countable dense subsets of $\mathbb {R}$ .Footnote ⁴ For each $\alpha < \omega _1$ , let $D_\alpha $ be the unique member of $\mathcal {D}$ containing $w_\alpha $ . Also fix an $\omega _1$ -enumeration $\langle P_\alpha =(a_\alpha ,b_\alpha ) \ : \ \alpha < \omega _1 \rangle $ of $\mathbb {R}^2$ .

Fix an $\alpha < \omega _1$ . By Theorem 5, there exists an entire $f_\alpha : \mathbb {C} \to \mathbb {C}$ such that:

1. (1) $f_\alpha \restriction \mathbb {R}$ is a real analytic bijection with strictly positive derivative;

2. (2) $f_\alpha (a_\alpha ) = b_\alpha $ (i.e., $f_\alpha \restriction \mathbb {R}$ passes through the point $P_\alpha $ );

3. (3) For each $w_\xi $ in the countable set $W_\alpha :=\{ w_\xi \ : \ \xi < \alpha \}$ ,

   1. (a) if $w_\xi \ne a_\alpha $ , then $f_\alpha (w_\xi ) \in D_\xi $ ; and

   2. (b) if $w_\xi \ne b_\alpha $ , then $f_\alpha ^{-1}(w_\xi ) \in D_\xi $ .

We claim that $\{ f_\alpha \restriction \mathbb {R} \ : \ \alpha < \omega _1 \}$ is a sparse analytic system, and the only nontrivial requirement to verify is that if $w \in \mathbb {R}$ , then both

$$\begin{align*}A_w:=\big\{ f_\alpha(w) \ : \ \alpha < \omega_1 \text{ and } w \ne a_\alpha \big\} \end{align*}$$

and

$$\begin{align*}B_w:=\big\{ f_\alpha^{-1}(w) \ : \ \alpha < \omega_1 \text{ and } w \ne b_\alpha \big\} \end{align*}$$

are countable. Say $w = w_\xi $ ; then,

$$ \begin{align*} A_w = A_{w_\xi} & \subseteq \underbrace{\{ f_\alpha(w_\xi) \ : \ \xi < \alpha < \omega_1 \text{ and } w_\xi \ne a_\alpha \}}_{\subseteq D_\xi \text{, by 3a}} \cup \underbrace{\{ f_\alpha(w_\xi) \ : \ \alpha \le \xi \}}_{\text{countable because } \xi < \omega_1}, \end{align*} $$

and hence, $A_w$ is countable. Similarly,

$$ \begin{align*} B_w =B_{w_\xi} & \subseteq \underbrace{\{ f^{-1}_\alpha(w_\xi) \ : \ \xi < \alpha < \omega_1 \text{ and } w_\xi \ne b_\alpha \}}_{\subseteq D_\xi\text{, by 3b}} \cup \underbrace{\{ f^{-1}_\alpha(w_\xi) \ : \ \alpha \le \xi \}}_{\text{countable because } \xi < \omega_1}, \end{align*} $$

and hence, $B_w$ is countable.

4 Proof of Theorem 4

The next lemma is the key connection between sparse analytic systems and predictors.

Lemma 6. Suppose $\mathcal {F}=\langle f_P \ : \ P \in \mathbb {R}^2 \rangle $ is a sparse analytic system. Let $\sim $ be the equivalence relation on $\mathbb {R}$ generated by the set

$$\begin{align*}X:=\Big\{ (u,v) \in \mathbb{R}^2 \ : \ \exists P \in \mathbb{R}^2 \ \big( u \ne x_P \ \wedge \ v \ne y_P \ \wedge \ f_P(u)=v \big) \Big\}. \end{align*}$$

Then,

1. (1) Each $\sim $ -equivalence class is countable.

2. (2) For every $P=(x_P,y_P) \in \mathbb {R}^2$ and every $z \in \mathbb {R}$ : if $z \ne x_P$ , then $z \sim f_P(z)$ .

Before proving Lemma 6, we say how the proof of Theorem 4 is finished: assuming CH, Theorem 2 yields the existence of a sparse analytic system. Let $\sim $ be the equivalence relation on $\mathbb {R}$ induced by the sparse analytic system via Lemma 6. The properties of $\sim $ listed in the conclusion of Lemma 6 satisfy the assumptions of Lemma 20 of Cox-Elpers [Reference Cox and Elpers6], and that lemma tells us that if $S:=\mathbb {R}/\sim $ and

$$\begin{align*}\mathcal{P}: {}^{\underset{\smile}{\mathbb{R}}} S \to S \end{align*}$$

is any analytic-anonymous S-predictor,Footnote ⁵ then $\mathcal {P}$ fails to predict the function $x \mapsto [x]_\sim $ for almost every $x \in \mathbb {R}$ .Footnote ⁶ In particular, there is no good analytic-anonymous S-predictor.

(Proof of Lemma 6)

Part (2) holds because, by the definition of sparse analytic system, $f_P$ is injective and $f_P(x_P)=y_P$ . So if $z \ne x_P$ , then $f_P(z) \ne y_P$ ; so not only is $z \sim f_P(z)$ , but the pair $\big (z,f_P(z)\big )$ is an element of X.

To prove part (1), since X generates $\sim $ , it suffices to prove that for every $z \in \mathbb {R}$ , both

$$\begin{align*}z^\uparrow:=\{ v \in \mathbb{R} \ : \ (z,v) \in X \} = \{ v \in \mathbb{R} \ : \ \exists P \in \mathbb{R}^2 \ ( z \ne x_P \wedge v \ne y_P \wedge f_P(z)=v ) \} \end{align*}$$

and

$$\begin{align*}z_\downarrow:=\{ u \in \mathbb{R} \ : \ (u,z) \in X \} =\{ u \in \mathbb{R} \ : \ \exists P \in \mathbb{R}^2 \ ( u \ne x_P \wedge z \ne y_P \wedge f_P(u)=z ) \} \end{align*}$$

are countable. But

$$\begin{align*}z^\uparrow \subseteq \{ f_P(z) \ : \ z \ne x_P \} \end{align*}$$

and

$$\begin{align*}z_\downarrow \subseteq \{ f_P^{-1}(z) \ : \ z \ne y_P \}, \end{align*}$$

which are both countable by definition of sparse analytic system.

5 Concluding Remarks

The notion of a sparse analytic system obviously generalizes to a sparse $\Gamma $ -system for any $\Gamma \subseteq \text {Homeo}^+(\mathbb {R})$ , and Lemma 6 easily generalizes to such systems. In fact, Section 4 of Bajpai-Velleman [Reference Bajpai and Velleman2] and Section 5 of Cox-Elpers [Reference Cox and Elpers6] can both be viewed as constructions, in ZFC alone, of sparse $\Gamma $ -systems (with $\Gamma =$ ‘increasing $C^\infty $ bijections’ in [Reference Bajpai and Velleman2] and $\Gamma =$ ‘increasing smooth diffeomorphisms’ in [Reference Cox and Elpers6]).

We have shown that CH implies a negative answer to Bajpai-Velleman’s Question 3, but it is open whether ZFC alone implies a negative solution.