Proof. Given a finite set $A \subset \mathbb {C}$ , $b \in \mathbb {C} \setminus A$ and $\zeta \in \mathbb {C}$ , define $P_{A, b, \zeta } \colon \mathbb {C} \rightarrow \mathbb {C}$ by

$$ \begin{align*} P_{A, b, \zeta}(z) = \zeta \prod_{a \in A} \bigg( \frac{z -a}{b -a} \bigg)^{2} \, \text{.} \end{align*} $$

Then the required conditions are satisfied.

Proof. Given a finite set $A \subset \mathbb {C}$ , $b \in \mathbb {C} \setminus A$ and $\unicode{x3bb} \in \mathbb {C}$ , define $Q_{A, b, \unicode{x3bb} } \colon \mathbb {C} \rightarrow \mathbb {C}$ by

$$ \begin{align*} Q_{A, b, \unicode{x3bb}}(z) = \unicode{x3bb} (z -b) \prod_{a \in A} \bigg( \frac{z -a}{b -a} \bigg)^{2} \, \text{.} \end{align*} $$

Then the required conditions are satisfied.

Proof. If $b \in A$ , setting $g = f$ , the required conditions are satisfied. Now, suppose that $b \in \mathbb {C} \setminus A$ . Using the notation of Lemma 2.1, for $\zeta \in \mathbb {C}$ , define

$$ \begin{align*} g_{\zeta} = f +P_{A, b, \zeta -f(b)} \in \mathcal{F}(f, A). \end{align*} $$

Then $\lim \nolimits _{\zeta \rightarrow f(b)} g_{\zeta } = f$ in $\mathcal {F}$ . Therefore, since E is dense in $\mathbb {C}$ , there exists $\zeta \in E$ such that $d_{\mathcal {F}}( f, g_{\zeta } ) < \varepsilon $ . Setting $g = g_{\zeta }$ , the required conditions are satisfied. Thus, the lemma is proved.

Proof. As f is not constant, there exists $S> R$ such that $f^{-1}(B) \cap \partial D(0, S) = \varnothing $ . Denote by $N \in \mathbb {Z}_{\geq 0}$ the number of preimages in $D(0, S)$ of the elements of B under f, counting multiplicities. As the topology of $\mathcal {F}$ is finer than the topology of local uniform convergence, reducing $\varepsilon $ if necessary, we may assume that

$$ \begin{align*} \text{ for all } g \in \mathcal{F}, \, d_{\mathcal{F}}(f, g) < \varepsilon \Longrightarrow \sup_{\partial D(0, S)} \lvert f -g \rvert < \operatorname{\mathrm{dist}}( f( \partial D(0, S) ), B ) , \end{align*} $$

so that the elements of B have together exactly N preimages in $D(0, S)$ under any $g \in \mathcal {F}$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ , counting multiplicities. Given $g \in \mathcal {F}$ , denote by:

• • $m_{g}$ the number of preimages in $D(0, R) \cap E$ of the elements of B under g, not counting multiplicities;

• • $n_{g}$ the number of preimages in $D(0, R) \cap E$ of the elements of B under g, counting multiplicities; and

• • $N_{g}$ the total number of preimages in $D(0, R)$ of the elements of B under g, counting multiplicities.

We prove that there exists $g \in \mathcal {F}(f, A)$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ and $n_{g} = N_{g}$ . Now, note that $m_{g} \leq N$ for all $g \in \mathcal {F}$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ . Therefore, it suffices to prove that, if $g \in \mathcal {F}(f, A)$ satisfies $d_{\mathcal {F}}(f, g) < \varepsilon $ and $n_{g} < N_{g}$ , then there exists $h \in \mathcal {F}(f, A)$ such that $d_{\mathcal {F}}(f, h) < \varepsilon $ and $m_{h}> m_{g}$ . Thus, suppose that $g \in \mathcal {F}(f, A)$ is such a map. Define

$$ \begin{align*} A^{\prime} = g^{-1}(B) \cap D(0, R) \cap E \, \text{.} \end{align*} $$

As $n_{g} < N_{g}$ , there exists $w \in D(0, R) \setminus E$ such that $g(w) \in B$ . Using the notation of Lemma 2.1, for $\zeta \in \mathbb {C} \setminus ( A \cup A^{\prime } )$ , define

$$ \begin{align*} h_{\zeta} = g +P_{A \cup A^{\prime}, \zeta, g(w) -g(\zeta)} \in \mathcal{F}( g, A \cup A^{\prime} ). \end{align*} $$

Then, $\lim \nolimits _{\zeta \rightarrow w} h_{\zeta } = g$ in $\mathcal {F}$ because $w \in \mathbb {C} \setminus ( A \cup A^{\prime } )$ . Therefore, since E is dense in $\mathbb {C}$ and $d_{\mathcal {F}}(f, g) < \varepsilon $ , there exists $\zeta \in ( D(0, R) \cap E ) \setminus ( A \cup A^{\prime } )$ such that $d_{\mathcal {F}}( f, h_{\zeta } ) < \varepsilon $ . Setting $h = h_{\zeta }$ , we have $m_{h}> m_{g}$ since $h(\zeta ) = g(w) \in B$ and $h \in \mathcal {F}( g, A^{\prime } )$ . Thus, we have shown that there exists $g \in \mathcal {F}(f, A)$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ and $n_{g} = N_{g}$ , which completes the proof of the lemma.

Proof. As $f \colon \mathbb {C} \rightarrow \mathbb {C}$ is not injective, we have $f^{\circ j} \neq \text {id}_{\mathbb {C}}$ for all $j \in \mathbb {Z}_{\geq 1}$ , and hence there exists $S> R$ such that $\partial D(0, S)$ contains no periodic point for f with period at most p. Denote by $N \in \mathbb {Z}_{\geq 0}$ the number of periodic points for f in $D(0, S)$ with period at most p, counting multiplicities. Since the topology of $\mathcal {F}$ is finer than the topology of local uniform convergence, reducing the number $\varepsilon $ if necessary, we may assume that any $g \in \mathcal {F}$ that satisfies $d_{\mathcal {F}}(f, g) < \varepsilon $ has exactly N periodic points in $D(0, S)$ with period at most p, counting multiplicities. Given $g \in \mathcal {F}$ , denote by:

• • $n_{g}$ the number of periodic points for g in $D(0, R)$ with period at most p whose cycle is contained in E and whose multiplier lies in $\Lambda $ , not counting multiplicities; and

• • $N_{g}$ the number of periodic points for g in $D(0, R)$ with period at most p, counting multiplicities.

We prove that there exists $g \in \mathcal {F}(f, A, C)$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ and $n_{g} = N_{g}$ . Now, note that $n_{g} \leq N$ for all $g \in \mathcal {F}$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ . Therefore, it suffices to show that, if $g \in \mathcal {F}(f, A, C)$ satisfies $d_{\mathcal {F}}(f, g) < \varepsilon $ and $n_{g} < N_{g}$ , then there exists $h \in \mathcal {F}(f, A, C)$ such that $d_{\mathcal {F}}(f, h) < \varepsilon $ and $n_{h}> n_{g}$ . Suppose that $g \in \mathcal {F}(f, A, C)$ is such a map. Define $C^{\prime }$ to be the union of the cycles for g with period at most p that intersect $D(0, R)$ , are contained in E and whose multipliers lie in $\Lambda $ . Because $\Lambda \subseteq \mathbb {C} \setminus \lbrace 1 \rbrace $ , the elements of $C^{\prime }$ are all simple periodic points for g. Therefore, as $n_{g} < N_{g}$ , there is a cycle $\Omega $ for g with period at most p that intersects $D(0, R)$ but is not contained in $C^{\prime }$ . Denoting by $\mathbb {C}^{\Omega }$ the set of all functions from $\Omega $ to $\mathbb {C}$ , define

$$ \begin{align*} \boldsymbol{Z} = \lbrace \boldsymbol{\zeta} \in \mathbb{C}^{\Omega} : \boldsymbol{\zeta} \vert_{\Omega \cap E} = \text{id}_{\Omega \cap E} \rbrace. \end{align*} $$

For $\boldsymbol {\zeta } \in \boldsymbol {Z}$ , set $\Omega _{\boldsymbol {\zeta }} = \lbrace \boldsymbol {\zeta }(\omega ) : \omega \in \Omega \rbrace $ . Since $A \subseteq E \cap g^{-1}(E)$ , $\Omega \subset \mathbb {C} \setminus ( C \cup C^{\prime } )$ and $\Omega \cap D(0, R) \neq \varnothing $ , there is a neighborhood V of $\text {id}_{\Omega }$ in $\boldsymbol {Z}$ such that, for each $\boldsymbol {\zeta } \in V$ :

• • $\boldsymbol {\zeta }(\omega ) \in \mathbb {C} \setminus A$ for all $\omega \in \Omega \setminus ( E \cap g^{-1}(E) )$ ;

• • $\Omega _{\boldsymbol {\zeta }} \subset \mathbb {C} \setminus ( C \cup C^{\prime } )$ ;

• • $\boldsymbol {\zeta }(\omega ) \neq \boldsymbol {\zeta }( \omega ^{\prime } )$ for all distinct $\omega , \omega ^{\prime } \in \Omega $ ; and

• • $\Omega _{\boldsymbol {\zeta }} \cap D(0, R) \neq \varnothing $ .

Using the notation of Lemma 2.1, for $\boldsymbol {\zeta } \in V$ , define

$$ \begin{align*} g_{\boldsymbol{\zeta}} = g +\sum_{\omega \in \Omega \setminus ( E \cap g^{-1}(E) )} \phi_{\boldsymbol{\zeta}, \omega} \in \mathcal{F}( g, A \cup C \cup C^{\prime} ) , \end{align*} $$

where, for every $\omega \in \Omega \setminus ( E \cap g^{-1}(E) )$ ,

$$ \begin{align*} \phi_{\boldsymbol{\zeta}, \omega} = P_{A \cup C \cup C^{\prime} \cup ( \Omega_{\boldsymbol{\zeta}} \setminus \lbrace \boldsymbol{\zeta}(\omega) \rbrace ), \boldsymbol{\zeta}(\omega), \boldsymbol{\zeta}( g(\omega) ) -g( \boldsymbol{\zeta}(\omega) )}. \end{align*} $$

For each $\boldsymbol {\zeta } \in V$ , we have $g_{\boldsymbol {\zeta }}( \boldsymbol {\zeta }(\omega ) ) = \boldsymbol {\zeta }( g(\omega ) )$ for all $\omega \in \Omega $ , and hence $\Omega _{\boldsymbol {\zeta }}$ forms a cycle for $g_{\boldsymbol {\zeta }}$ . Now, note that $\lim \nolimits _{\boldsymbol {\zeta } \rightarrow \text {id}_{\Omega }} g_{\boldsymbol {\zeta }} = g$ in $\mathcal {F}$ . Therefore, as E is dense in $\mathbb {C}$ and ${d_{\mathcal {F}}(f, g) < \varepsilon }$ , there exists $\boldsymbol {\zeta } \in V \cap E^{\Omega }$ such that $d_{\mathcal {F}}( f, g_{\boldsymbol {\zeta }} ) < \varepsilon $ . Using the notation of Lemma 2.2, for $\boldsymbol {\unicode{x3bb} } \in \mathbb {C}^{\Omega }$ , define

$$ \begin{align*} h_{\boldsymbol{\unicode{x3bb}}} = g_{\boldsymbol{\zeta}} +\sum_{\omega \in \Omega} \psi_{\boldsymbol{\unicode{x3bb}}, \omega} \in \mathcal{F}( g_{\boldsymbol{\zeta}}, A \cup \Omega_{\boldsymbol{\zeta}}, C \cup C^{\prime} ) , \end{align*} $$

where, for every $\omega \in \Omega $ ,

$$ \begin{align*} \psi_{\boldsymbol{\unicode{x3bb}}, \omega} = Q_{C \cup C^{\prime} \cup ( ( A \cup \Omega_{\boldsymbol{\zeta}} ) \setminus \lbrace \boldsymbol{\zeta}(\omega) \rbrace ), \boldsymbol{\zeta}(\omega), \boldsymbol{\unicode{x3bb}}(\omega)}. \end{align*} $$

Then $\Omega _{\boldsymbol {\zeta }}$ is a cycle for $h_{\boldsymbol {\unicode{x3bb} }}$ and we have $h_{\boldsymbol {\unicode{x3bb} }}^{\prime }( \boldsymbol {\zeta }(\omega ) ) = g_{\boldsymbol {\zeta }}^{\prime }( \boldsymbol {\zeta }(\omega ) ) +\boldsymbol {\unicode{x3bb} }(\omega )$ for all $\omega \in \Omega $ and all $\boldsymbol {\unicode{x3bb} } \in \mathbb {C}^{\Omega }$ . Moreover, $\lim \nolimits _{\boldsymbol {\unicode{x3bb} } \rightarrow 0} h_{\boldsymbol {\unicode{x3bb} }} = g_{\boldsymbol {\zeta }}$ in $\mathcal {F}$ . Therefore, since $\Lambda $ is dense in $\mathbb {C}$ and $d_{\mathcal {F}}( f, g_{\boldsymbol {\zeta }} ) < \varepsilon $ , there exists $\boldsymbol {\unicode{x3bb} } \in \mathbb {C}^{\Omega }$ such that $d_{\mathcal {F}}( f, h_{\boldsymbol {\unicode{x3bb} }} ) < \varepsilon $ and the multiplier of $h_{\boldsymbol {\unicode{x3bb} }}$ at $\Omega _{\boldsymbol {\zeta }}$ lies in $\Lambda $ . Setting $h = h_{\boldsymbol {\unicode{x3bb} }}$ , we have $n_{h}> n_{g}$ . Thus, we have proved that there exists $g \in \mathcal {F}(f, A, C)$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ and $n_{g} = N_{g}$ , which completes the proof of the lemma.

Proof of Theorem 1.4

Suppose that $f_{0} \in \mathcal {F}$ and that $\varepsilon \in \mathbb {R}_{> 0}$ . We show that there exists $f \in \mathcal {F}$ such that:

• • $d_{\mathcal {F}}( f_{0}, f ) < \varepsilon $ ;

• • $f^{-1}(E) = E$ ; and

• • the cycles for f are all contained in E and their multipliers all lie in $\Lambda $ .

Since $\mathcal {E}$ is a closed subset of $\mathcal {F}$ with empty interior, replacing $f_{0}$ and $\varepsilon $ if necessary, we may assume that $f_{0} \in \mathcal {F} \setminus \mathcal {E}$ and $\varepsilon \in ( 0, \operatorname {\mathrm {dist}}( f_{0}, \mathcal {E} ) )$ . Write $E = \lbrace e_{j} : j \in \mathbb {Z}_{\geq 1} \rbrace $ , with $e_{j} \neq e_{k}$ for all distinct $j, k \in \mathbb {Z}_{\geq 1}$ . For $n \in \mathbb {Z}_{\geq 1}$ , define $E_{n} = \lbrace e_{1}, \dotsc , e_{n} \rbrace $ . We show that there exists a sequence $( f_{n} )_{n \geq 0}$ of elements of $\mathcal {F}$ such that, for every $n \geq 1$ :

1. (1) $d_{\mathcal {F}}( f_{n -1}, f_{n} ) < {\varepsilon }/{2^{n}}$ ;

2. (2) $f_{n} \in \mathcal {F}( f_{n -1}, A_{n -1}, C_{n -1} )$ (see the definitions below);

3. (3) $f_{n}( e_{n} ) \in E$ ;

4. (4) $f_{n}^{-1}( E_{n} ) \cap D(0, n) \subset E$ ; and

5. (5) $f_{n}$ has the property $( \Gamma _{n, n} )$ ,

where, for every $n \geq 0$ ,

$$ \begin{align*} A_{n} = E_{n} \cup ( f_{n}^{-1}( E_{n} ) \cap D(0, n) ) \end{align*} $$

and $C_{n}$ denotes the union of the cycles for $f_{n}$ with period at most n that intersect $D(0, n)$ , with $A_{0} = C_{0} = \varnothing $ by convention. We proceed by recursion. Suppose that $n \in \mathbb {Z}_{\geq 0}$ and $f_{1}, \dotsc , f_{n} \in \mathcal {F}$ satisfy these conditions (1)–(5), and let us prove the existence of ${f_{n +1} \in \mathcal {F}}$ that satisfies the same conditions. By conditions (2), (3) and (5), we have $A_{n} \cup C_{n} \subset f_{n}^{-1}(E)$ . By Lemma 2.3, it follows that there exists $g_{n} \in \mathcal {F}$ such that

$$ \begin{align*} g_{n} \in \mathcal{F}( f_{n}, A_{n} \cup C_{n} ) , \quad d_{\mathcal{F}}( f_{n}, g_{n} ) < \frac{\varepsilon}{3 \cdot 2^{n +1}} , \quad g_{n}( e_{n +1} ) \in E. \end{align*} $$

Note that $g_{n} \in \mathcal {F} \setminus \mathcal {E}$ since $d_{\mathcal {F}}( f_{0}, g_{n} ) < \varepsilon $ by condition (1). Now, define

$$ \begin{align*} A_{n}^{\prime} = A_{n} \cup \lbrace e_{n +1} \rbrace = E_{n +1} \cup ( f_{n}^{-1}( E_{n} ) \cap D(0, n) ). \end{align*} $$

Then $A_{n}^{\prime } \subseteq E \cap g_{n}^{-1}(E)$ by conditions (2), (3) and (4). Moreover, $C_{n}$ is a union of cycles for $g_{n}$ that are all contained in E and whose multipliers all lie in $\Lambda $ by condition (5). Therefore, by Lemma 2.5, there exists $h_{n} \in \mathcal {F}$ such that

$$ \begin{align*} h_{n} \in \mathcal{F}( g_{n}, A_{n}^{\prime}, C_{n} ) , \quad d_{\mathcal{F}}( g_{n}, h_{n} ) < \frac{\varepsilon}{3 \cdot 2^{n +1}} , \quad h_{n} \text{ has the property } ( \Gamma_{n +1, n +2} ). \end{align*} $$

Note that $h_{n} \in \mathcal {F} \setminus \mathcal {E}$ since $d_{\mathcal {F}}( f_{0}, h_{n} ) < \varepsilon $ by condition (1). In particular, we have $h_{n}^{\circ j} \neq \text {id}_{\mathbb {C}}$ for all $j \in \lbrace 1, \dotsc , n +1 \rbrace $ , and hence there exists $R_{n} \in (n +1, n +2)$ such that $\partial D( 0, R_{n} )$ contains no periodic point for $h_{n}$ with period at most $n +1$ . Now, denote by $X_{n}$ the union of the cycles for $h_{n}$ with period at most $n +1$ that intersect $D( 0, R_{n} )$ . Then the cycles for $h_{n}$ in $X_{n}$ are all contained in E and their multipliers all lie in $\Lambda $ . Denote by $N_{n} \in \mathbb {Z}_{\geq 0}$ the number of periodic points for $h_{n}$ in $D( 0, R_{n} )$ with period at most $n +1$ , counting multiplicities. Since $\Lambda \subseteq \mathbb {C} \setminus \lbrace 1 \rbrace $ , the elements of $X_{n}$ are all simple periodic points for $h_{n}$ , and hence $N_{n}$ equals the cardinality of $X_{n} \cap D( 0, R_{n} )$ . Moreover, since the topology of $\mathcal {F}$ is finer than the topology of local uniform convergence, there exists $\varepsilon _{n} \in ( 0, {\varepsilon }/{(3 \cdot 2^{n +1})} )$ such that any $k_{n} \in \mathcal {F}$ such that $d_{\mathcal {F}}( h_{n}, k_{n} ) < \varepsilon _{n}$ has exactly $N_{n}$ periodic points in $D( 0, R_{n} )$ with period at most $n +1$ , counting multiplicities. Since $A_{n}^{\prime } \cup X_{n} \subset E$ , it follows from Lemma 2.4 that there exists $f_{n +1} \in \mathcal {F}$ such that

$$ \begin{align*} f_{n +1} \in \mathcal{F}( h_{n}, A_{n}^{\prime} \cup X_{n} ) , \quad d_{\mathcal{F}}( h_{n}, f_{n +1} ) < \varepsilon_{n} , \quad f_{n +1}^{-1}( E_{n +1} ) \cap D(0, n +1) \subset E. \end{align*} $$

Then $f_{n +1}$ clearly satisfies conditions (1)–(4). Moreover, $X_{n}$ is also a union of cycles for $f_{n +1}$ and $f_{n +1}$ has exactly $N_{n}$ periodic points in $D( 0, R_{n} )$ with period at most $n +1$ , counting multiplicities. Therefore, as $X_{n} \cap D( 0, R_{n} )$ has cardinality $N_{n}$ , the cycles for $f_{n +1}$ with period at most $n +1$ that intersect $D( 0, R_{n} )$ are the cycles contained in $X_{n}$ . Since these are all contained in E and their multipliers all lie in $\Lambda $ , the map $f_{n +1}$ also satisfies condition (5) because $R_{n}> n +1$ . Thus, we have proved the existence of such a sequence $( f_{n} )_{n \geq 0}$ of elements of $\mathcal {F}$ .

It follows from condition (1) that $( f_{n} )_{n \geq 0}$ is a Cauchy sequence in $( \mathcal {F}, d_{\mathcal {F}} )$ , and its limit $f \in \mathcal {F}$ satisfies $d_{\mathcal {F}}( f_{0}, f ) < \varepsilon $ . In particular, $f \in \mathcal {F} \setminus \mathcal {E}$ . We prove that f satisfies the required conditions. It suffices to show that, for each $m \in \mathbb {Z}_{\geq 1}$ :

1. (i) $f( e_{m} ) \in E$ ;

2. (ii) $f^{-1}( E_{m} ) \cap D(0, m) \subset E$ ; and

3. (iii) f has the property $( \Gamma _{m, m} )$ .

Suppose that $m \in \mathbb {Z}_{\geq 1}$ . Then $f_{n}( e_{m} ) = f_{m}( e_{m} )$ for all $n \geq m$ by condition (2), and hence $f( e_{m} ) = f_{m}( e_{m} )$ since $( f_{n} )_{n \geq 0}$ converges pointwise to f. Therefore, by condition (3), condition (i) is satisfied. By condition (2) and as $( f_{n} )_{n \geq 0}$ converges pointwise to f, for every $n \geq m$ ,

$$ \begin{align*} f_{n}^{-1}( E_{m} ) \cap D(0, m) \subseteq f_{n +1}^{-1}( E_{m} ) \cap D(0, m) \subseteq f^{-1}( E_{m} ) \cap D(0, m). \end{align*} $$

Therefore, as $f^{-1}( E_{m} ) \cap D(0, m)$ is finite since $f \in \mathcal {F} \setminus \mathcal {E}$ , there exists $N \in \mathbb {Z}_{\geq m}$ such that

$$ \begin{align*} \bigcup_{n \geq N} ( f_{n}^{-1}( E_{m} ) \cap D(0, m) ) = f_{N}^{-1}( E_{m} ) \cap D(0, m). \end{align*} $$

As $( f_{n} )_{n \geq 0}$ converges locally uniformly to f and $f \in \mathcal {F} \setminus \mathcal {E}$ , it follows that

$$ \begin{align*} f^{-1}( E_{m} ) \cap D(0, m) \subseteq \overline{\bigcup_{n \geq N} ( f_{n}^{-1}( E_{m} ) \cap D(0, m) )} = f_{N}^{-1}( E_{m} ) \cap D(0, m). \end{align*} $$

Therefore, by condition (4), condition (ii) is satisfied. For $n \in \mathbb {Z}_{\geq m}$ , define $Y_{n} \subseteq C_{n}$ to be the union of the cycles for $f_{n}$ with period at most m that intersect $D(0, m)$ . Also, define Y to be the union of the cycles for f with period at most m that intersect $D(0, m)$ . By condition (2) and since $( f_{n} )_{n \geq 0}$ converges pointwise to f, for each $n \geq m$ , we have $Y_{n} \subseteq Y_{n +1} \subseteq Y$ . Therefore, since Y is finite because $f \in \mathcal {F} \setminus \mathcal {E}$ , there exists $N \in \mathbb {Z}_{\geq m}$ such that $\bigcup \nolimits _{n \geq N} Y_{n} = Y_{N}$ . Since $( f_{n} )_{n \geq 0}$ converges locally uniformly to f and $f \in \mathcal {F} \setminus \mathcal {E}$ , it follows that

$$ \begin{align*} Y \subseteq \overline{\bigcup_{n \geq N} Y_{n}} = Y_{N}. \end{align*} $$

Moreover, $f \in \mathcal {F}( f_{N}, Y_{N} )$ by condition (2) and since $( f_{n} )_{n \geq 0}$ converges locally uniformly to f. Therefore, by condition (5), condition (iii) is also satisfied. This completes the proof of the theorem.

Proof. Suppose that $A \subset \mathbb {C}$ is a finite set such that $A^{\mathrm {sym}} = A$ . For $b \in (\mathbb {R} \cup i \mathbb {R}) \setminus A$ and $\zeta \in \mathbb {R}$ , define $P_{A, b, \zeta }^{\mathrm {axes}} \colon \mathbb {C} \rightarrow \mathbb {C}$ by

$$ \begin{align*} P_{A, b, \zeta}^{\mathrm{axes}}(z) = \zeta \prod_{a \in A} \bigg( \frac{z -a}{b -a} \bigg)^{2}. \end{align*} $$

For $b \in \mathbb {C} \setminus (\mathbb {R} \cup i \mathbb {R} \cup A)$ and $\zeta \in \mathbb {C}$ , define $P_{A, b, \zeta }^{\mathrm {away}} \colon \mathbb {C} \rightarrow \mathbb {C}$ by

$$ \begin{align*} P_{A, b, \zeta}^{\mathrm{away}}(z) = \Gamma_{A, b, \zeta}(z) \prod_{a \in A} (z -a)^{2} , \end{align*} $$

with

$$ \begin{align*} \Gamma_{A, b, \zeta}(z) = \frac{\Im(\gamma) z^{2} +\Re(\gamma) \Im( b^{2} ) -\Im(\gamma) \Re( b^{2} )}{\Im( b^{2} )} \quad \text{and} \quad \gamma = \frac{\zeta}{\prod\limits_{a \in A} (b -a)^{2}} , \end{align*} $$

where $\Re (w)$ and $\Im (w)$ denote the real and imaginary parts of any $w \in \mathbb {C}$ . Then the required conditions are satisfied.

Proof. Suppose that $A \subset \mathbb {C}$ is a finite set such that $A^{\mathrm {sym}} = A$ . For $b \in \mathbb {R}^{*} \setminus A$ and $\unicode{x3bb} \in \mathbb {R}$ , define $Q_{A, b, \unicode{x3bb} }^{\mathrm {axes}} \colon \mathbb {C} \rightarrow \mathbb {C}$ by

$$ \begin{align*} Q_{A, b, \unicode{x3bb}}^{\mathrm{axes}}(z) = \unicode{x3bb} \bigg( \frac{z^{2} -b^{2}}{2 b} \bigg) \prod_{a \in A} \bigg( \frac{z -a}{b -a} \bigg)^{2}. \end{align*} $$

For $b \in \mathbb {C} \setminus (\mathbb {R} \cup i \mathbb {R} \cup A)$ and $\unicode{x3bb} \in \mathbb {C}$ , define $Q_{A, b, \unicode{x3bb} }^{\mathrm {away}} \colon \mathbb {C} \rightarrow \mathbb {C}$ by

$$ \begin{align*} Q_{A, b, \unicode{x3bb}}^{\mathrm{away}}(z) = \Delta_{A, b, \unicode{x3bb}}(z) ( z^{4} -2 \Re( b^{2} ) z^{2} +\lvert b \rvert^{4} ) \prod_{a \in A} (z -a)^{2} , \end{align*} $$

where

$$ \begin{align*} \Delta_{A, b, \unicode{x3bb}}(z) = \frac{\Im(\delta) z^{2} +\Re(\delta) \Im( b^{2} ) -\Im(\delta) \Re( b^{2} )}{\Im( b^{2} )} \quad \text{and} \quad \delta = \frac{\unicode{x3bb}}{4 i b \Im( b^{2} ) \prod\limits_{a \in A} (b -a)^{2}}. \end{align*} $$

Then the required conditions are satisfied.

Proof. Because f is real and even and $E^{\mathrm {sym}} = E$ , replacing A by $A^{\mathrm {sym}}$ if necessary, we may assume that $A^{\mathrm {sym}} = A$ . If $A \subset f^{-1}(E)$ and $b \in A$ , then $g = f$ satisfies the required conditions. Now, suppose that $b \in \mathbb {C} \setminus A$ . Define

$$ \begin{align*} \text{locus} = \begin{cases} \text{axes} & \text{if } b \in \mathbb{R} \cup i \mathbb{R},\\ \text{away} & \text{if } b \in \mathbb{C} \setminus (\mathbb{R} \cup i \mathbb{R}), \end{cases}\! \quad \mathbb{C}^{\text{locus}} = \begin{cases} \mathbb{R} & \text{if } \text{locus} = \text{axes},\\ \mathbb{C} & \text{if } \text{locus} = \text{away}. \end{cases}\! \end{align*} $$

Note that $f(b) \in \mathbb {C}^{\text {locus}}$ . Therefore, with the notation of Lemma 3.1, for $\zeta \in \mathbb {C}^{\text {locus}}$ , we can define

$$ \begin{align*} g_{\zeta} = f +P_{A, b, \zeta -f(b)}^{\text{locus}} \in \mathcal{F}(f, A). \end{align*} $$

Then $\lim \nolimits _{\zeta \rightarrow f(b)} g_{\zeta } = f$ in $\mathcal {F}$ . Therefore, since $E \cap \mathbb {C}^{\text {locus}}$ is dense in $\mathbb {C}^{\text {locus}}$ , there exists $\zeta \in E \cap \mathbb {C}^{\text {locus}}$ such that $d_{\mathcal {F}}( f, g_{\zeta } ) < \varepsilon $ . Setting $g = g_{\zeta }$ , the required conditions are satisfied. Thus, the lemma is proved.

Proof. Since $E^{\mathrm {sym}} = E$ , replacing A by $A^{\mathrm {sym}}$ and B by $B^{\mathrm {sym}}$ if necessary, we may assume that $A^{\mathrm {sym}} = A$ and $B^{\mathrm {sym}} = B$ . Given $g \in \mathcal {F}$ , denote by:

• • $m_{g}$ the number of preimages in $D(0, R) \cap E$ of the elements of B under g, not counting multiplicities;

• • $n_{g}$ the number of preimages in $D(0, R) \cap E$ of the elements of B under g, counting multiplicities; and

• • $N_{g}$ the total number of preimages in $D(0, R)$ of the elements of B under g, counting multiplicities.

We prove that there exists $g \in \mathcal {F}(f, A)$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ and $n_{g} = N_{g}$ . Using the same preliminary arguments as in the proof of Lemma 2.4, reducing $\varepsilon $ if necessary, we may assume that $m_{g} \leq N$ for each $g \in \mathcal {F}$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ , for some bound ${N \in \mathbb {Z}_{\geq 0}}$ independent of $g \in \mathcal {F}$ . Therefore, it suffices to show that, if $g \in \mathcal {F}(f, A)$ satisfies $d_{\mathcal {F}}(f, g) < \varepsilon $ and $n_{g} < N_{g}$ , then there exists $h \in \mathcal {F}(f, A)$ such that ${d_{\mathcal {F}}(f, h) < \varepsilon }$ and $m_{h}> m_{g}$ . Suppose that $g \in \mathcal {F}(f, A)$ is such a map. Define

$$ \begin{align*} A^{\prime} = g^{-1}(B) \cap D(0, R) \cap E , \end{align*} $$

which is finite of cardinality $m_{g}$ and such that $( A^{\prime } )^{\mathrm {sym}} = A^{\prime }$ . Since $n_{g} < N_{g}$ , there exists $w \in D(0, R) \setminus E$ such that $g(w) \in B$ . Now, define

$$ \begin{align*} \text{locus} = \begin{cases} \text{axes} & \text{if } w \in \mathbb{R} \cup i \mathbb{R},\\ {\text{away}} & \text{if } w \in \mathbb{C} \setminus (\mathbb{R} \cup i \mathbb{R}), \end{cases}\! \quad \mathbb{C}^{\text{locus}} = \begin{cases} \mathbb{R} \cup i \mathbb{R} & \text{if } \text{locus} = \text{axes},\\ \mathbb{C} \setminus (\mathbb{R} \cup i \mathbb{R}) & \text{if } \text{locus} = {\text{away}}. \end{cases}\! \end{align*} $$

Using the notation of Lemma 3.1, for $\zeta \in \mathbb {C}^{\text {locus}} \setminus ( A \cup A^{\prime } )$ , define

$$ \begin{align*} h_{\zeta} = g +P_{A \cup A^{\prime}, \zeta, g(w) -g(\zeta)}^{\text{locus}} \in \mathcal{F}( g, A \cup A^{\prime} ). \end{align*} $$

Then $\lim \nolimits _{\zeta \rightarrow w} h_{\zeta } = g$ in $\mathcal {F}$ because $w \in \mathbb {C}^{\text {locus}} \setminus ( A \cup A^{\prime } )$ . Therefore, since $E \cap \mathbb {C}^{\text {locus}}$ is dense in $\mathbb {C}^{\text {locus}}$ and $d_{\mathcal {F}}(f, g) < \varepsilon $ , there exists $\zeta \in ( D(0, R) \cap E \cap \mathbb {C}^{\text {locus}} ) \setminus ( A \cup A^{\prime } )$ such that $d_{\mathcal {F}}( f, h_{\zeta } ) < \varepsilon $ . Setting $h = h_{\zeta }$ , we have $m_{h}> m_{g}$ since $h(\zeta ) = g(w) \in B$ and $h \in \mathcal {F}( g, A^{\prime } )$ . This completes the proof of the lemma.

Proof. As f is real and even and $E^{\mathrm {sym}} = E$ , replacing A by $A^{\mathrm {sym}}$ if necessary, we may assume that $A^{\mathrm {sym}} = A$ . Since the topology of $\mathcal {F}$ is finer than the topology of local uniform convergence, reducing the number $\varepsilon $ if necessary, we may also assume that $0 \in \mathbb {C} \setminus E$ or $0$ is not periodic with period at most p for any $g \in \mathcal {F}$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ . Given $g \in \mathcal {F}$ , denote by:

• • $n_{g}$ the number of periodic points for g in $D(0, R)$ with period at most p whose cycle is contained in E and whose multiplier lies in $\Lambda $ , not counting multiplicities; and

• • $N_{g}$ the number of periodic points for g in $D(0, R)$ with period at most p, counting multiplicities.

We prove that there exists $g \in \mathcal {F}(f, A, C)$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ and $n_{g} = N_{g}$ . Using the same preliminary arguments as in the proof of Lemma 2.5, reducing $\varepsilon $ if necessary, we may assume that $n_{g} \leq N$ for each $g \in \mathcal {F}$ such that $d_{\mathcal {F}}(f, g) < \varepsilon $ , for some bound ${N \in \mathbb {Z}_{\geq 0}}$ independent of $g \in \mathcal {F}$ . Therefore, it suffices to show that, if $g \in \mathcal {F}(f, A, C)$ satisfies $d_{\mathcal {F}}(f, g) < \varepsilon $ and $n_{g} < N_{g}$ , then there exists $h \in \mathcal {F}(f, A, C)$ such that $d_{\mathcal {F}}(f, h) < \varepsilon $ and $n_{h}> n_{g}$ . Thus, suppose that $g \in \mathcal {F}(f, A, C)$ is such a map. Now, denote by $C^{\prime }$ the union of the cycles for g with period at most p that intersect $D(0, R)$ , are contained in E and whose multipliers lie in $\Lambda $ . The elements of $C^{\prime }$ are all simple periodic points for g since $\Lambda \subseteq \mathbb {C} \setminus \lbrace 1 \rbrace $ . Therefore, as $n_{g} < N_{g}$ , there exists a cycle $\Omega $ for g with period at most p that intersects $D(0, R)$ but not $C^{\prime }$ . We have $g(\mathbb {R} \cup i \mathbb {R}) \subseteq \mathbb {R}$ because g is real and even, and hence either $\Omega \subset \mathbb {R}$ or $\Omega \subset \mathbb {C} \setminus (\mathbb {R} \cup i \mathbb {R})$ . Now, define

$$ \begin{align*} \text{locus} = \begin{cases} \text{axes} & \text{if } \Omega \subset \mathbb{R},\\ \text{away} & \text{if } \Omega \subset \mathbb{C} \setminus (\mathbb{R} \cup i \mathbb{R}), \end{cases}\! \quad \mathbb{C}^{\text{locus}} = \begin{cases} \mathbb{R} & \text{if } \text{locus} = \text{axes},\\ \mathbb{C} \setminus (\mathbb{R} \cup i \mathbb{R}) & \text{if } \text{locus} = \text{away}. \end{cases}\! \end{align*} $$

Denote by $\sigma \colon \mathbb {C} \rightarrow \mathbb {C}$ the complex conjugation. Since g is real, $\sigma (\Omega )$ is also a cycle for g, and in particular either $\Omega = \sigma (\Omega )$ or $\Omega \cap \sigma (\Omega ) = \varnothing $ . Define

$$ \begin{align*} \boldsymbol{Z} = \lbrace \boldsymbol{\zeta} \in ( \mathbb{C}^{\text{locus}} )^{\Omega} : \boldsymbol{\zeta} \vert_{\Omega \cap E} = \text{id}_{\Omega \cap E} \text{ and } \boldsymbol{\zeta} \circ \sigma \vert_{\Omega \cap \sigma(\Omega)} = \sigma \circ \boldsymbol{\zeta} \vert_{\Omega \cap \sigma(\Omega)} \rbrace. \end{align*} $$

For $\boldsymbol {\zeta } \in \boldsymbol {Z}$ , set $\Omega _{\boldsymbol {\zeta }} = \lbrace \boldsymbol {\zeta }(\omega ) : \omega \in \Omega \rbrace $ . As g is even and $\Omega $ and $\sigma (\Omega )$ are cycles for g, for each $\omega \in \Omega $ , the points in $\lbrace \omega \rbrace ^{\mathrm {sym}} \setminus \lbrace \omega , \sigma (\omega ) \rbrace $ are strictly preperiodic for g, and hence the set $\lbrace \omega \rbrace ^{\mathrm {sym}} \cap \Omega $ equals $\lbrace \omega , \sigma (\omega ) \rbrace $ or $\lbrace \omega \rbrace $ according to whether $\Omega = \sigma (\Omega )$ or ${\Omega \cap \sigma (\Omega ) = \varnothing }$ . Similarly, as g is real and even and E and $\Lambda $ are symmetric with respect to the real axis, each point in $( C \cup C^{\prime } )^{\mathrm {sym}}$ is either strictly preperiodic for g or it is periodic for g, its cycle is contained in E and its multiplier lies in $\Lambda $ , and hence $\Omega \subset \mathbb {C}^{\text {locus}} \setminus ( C \cup C^{\prime } )^{\mathrm {sym}}$ . Moreover, $A \subseteq E \cap g^{-1}(E)$ and $\Omega \cap D(0, R) \neq \varnothing $ . Therefore, there exists a neighborhood V of $\text {id}_{\Omega }$ in $\boldsymbol {Z}$ such that, for each $\boldsymbol {\zeta } \in V$ :

• • $\boldsymbol {\zeta }(\omega ) \in \mathbb {C}^{\text {locus}} \setminus A$ for all $\omega \in \Omega \setminus ( E \cap g^{-1}(E) )$ ;

• • $\Omega _{\boldsymbol {\zeta }} \subset \mathbb {C}^{\text {locus}} \setminus ( C \cup C^{\prime } )^{\mathrm {sym}}$ ;

• • $\lbrace \boldsymbol {\zeta }(\omega ) \rbrace ^{\mathrm {sym}} \cap \Omega _{\boldsymbol {\zeta }} = \begin {cases} \lbrace \boldsymbol {\zeta }(\omega ), \boldsymbol {\zeta }( \sigma (\omega ) ) \rbrace & \text {if } \Omega = \sigma (\Omega )\\ \lbrace \boldsymbol {\zeta }(\omega ) \rbrace & \text {if } \Omega \cap \sigma (\Omega ) = \varnothing \end {cases}$ for all $\omega \in \Omega $ ;

• • $\boldsymbol {\zeta }(\omega ) \neq \boldsymbol {\zeta }( \omega ^{\prime } )$ for all distinct $\omega , \omega ^{\prime } \in \Omega $ ; and

• • $\Omega _{\boldsymbol {\zeta }} \cap D(0, R) \neq \varnothing $ .

Using the notation of Lemma 3.1, for $\boldsymbol {\zeta } \in V$ , define

$$ \begin{align*} g_{\boldsymbol{\zeta}} = g +\alpha \cdot \sum_{\omega \in \Omega \setminus ( E \cap g^{-1}(E) )} \phi_{\boldsymbol{\zeta}, \omega} \in \mathcal{F}( g, A \cup C \cup C^{\prime} ) , \end{align*} $$

where

$$ \begin{align*} \alpha = \begin{cases} \frac{1}{2} & \text{if } \Omega \subset \mathbb{C} \setminus (\mathbb{R} \cup i \mathbb{R}) \text{ and } \Omega = \sigma(\Omega),\\ 1 & \text{if } \Omega \subset \mathbb{R} \text{ or } \Omega \cap \sigma(\Omega) = \varnothing, \end{cases} \end{align*} $$

and, for every $\omega \in \Omega \setminus ( E \cap g^{-1}(E) )$ ,

$$ \begin{align*} \phi_{\boldsymbol{\zeta}, \omega} = P_{A \cup ( C \cup C^{\prime} )^{\mathrm{sym}} \cup ( \Omega_{\boldsymbol{\zeta}}^{\mathrm{sym}} \setminus \lbrace \boldsymbol{\zeta}(\omega) \rbrace^{\mathrm{sym}} ), \boldsymbol{\zeta}(\omega), \boldsymbol{\zeta}( g(\omega) ) -g( \boldsymbol{\zeta}(\omega) )}^{\text{locus}}. \end{align*} $$

For each $\boldsymbol {\zeta } \in V$ , we have $g_{\boldsymbol {\zeta }}( \boldsymbol {\zeta }(\omega ) ) = \boldsymbol {\zeta }( g(\omega ) )$ for all $\omega \in \Omega $ , and hence $\Omega _{\boldsymbol {\zeta }}$ is a cycle for $g_{\boldsymbol {\zeta }}$ . Note that $\lim \nolimits _{\boldsymbol {\zeta } \rightarrow \text {id}_{\Omega }} g_{\boldsymbol {\zeta }} = g$ in $\mathcal {F}$ . Therefore, since $E \cap \mathbb {C}^{\text {locus}}$ is dense in $\mathbb {C}^{\text {locus}}$ and $d_{\mathcal {F}}(f, g) < \varepsilon $ , there exists $\boldsymbol {\zeta } \in V \cap E^{\Omega }$ such that $d_{\mathcal {F}}( f, g_{\boldsymbol {\zeta }} ) < \varepsilon $ . Now, define

$$ \begin{align*} \boldsymbol{L} = \lbrace \boldsymbol{\unicode{x3bb}} \in ( \mathbb{C}^{\text{locus}} )^{\Omega} : \boldsymbol{\unicode{x3bb}} \circ \sigma \vert_{\Omega \cap \sigma(\Omega)} = \sigma \circ \boldsymbol{\unicode{x3bb}} \vert_{\Omega \cap \sigma(\Omega)} \rbrace. \end{align*} $$

Note that $0 \in \mathbb {C} \setminus \Omega _{\boldsymbol {\zeta }}$ by the second sentence of the proof. Consequently, with the notation of Lemma 3.2, for $\boldsymbol {\unicode{x3bb} } \in \boldsymbol {L}$ , we can define

$$ \begin{align*} h_{\boldsymbol{\unicode{x3bb}}} = g_{\boldsymbol{\zeta}} +\alpha \cdot \sum_{\omega \in \Omega} \psi_{\boldsymbol{\unicode{x3bb}}, \omega} \in \mathcal{F}( g_{\boldsymbol{\zeta}}, A \cup \Omega_{\boldsymbol{\zeta}}, C \cup C^{\prime} ) , \end{align*} $$

where, for every $\omega \in \Omega $ ,

$$ \begin{align*} \psi_{\boldsymbol{\unicode{x3bb}}, \omega} = Q_{( C \cup C^{\prime} )^{\mathrm{sym}} \cup ( ( A \cup \Omega_{\boldsymbol{\zeta}}^{\mathrm{sym}} ) \setminus \lbrace \boldsymbol{\zeta}(\omega) \rbrace^{\mathrm{sym}} ), \boldsymbol{\zeta}(\omega), \boldsymbol{\unicode{x3bb}}(\omega)}^{\text{locus}}. \end{align*} $$

Then $\Omega _{\boldsymbol {\zeta }}$ is a cycle for $h_{\boldsymbol {\unicode{x3bb} }}$ and we have $h_{\boldsymbol {\unicode{x3bb} }}^{\prime }( \boldsymbol {\zeta }(\omega ) ) = g_{\boldsymbol {\zeta }}^{\prime }( \boldsymbol {\zeta }(\omega ) ) +\boldsymbol {\unicode{x3bb} }(\omega )$ for all $\omega \in \Omega $ and all $\boldsymbol {\unicode{x3bb} } \in \boldsymbol {L}$ . Moreover, we have $\lim \nolimits _{\boldsymbol {\unicode{x3bb} } \rightarrow 0} h_{\boldsymbol {\unicode{x3bb} }} = g_{\boldsymbol {\zeta }}$ in $\mathcal {F}$ . Therefore, since $\Lambda \cap \mathbb {C}^{\text {locus}}$ is dense in $\mathbb {C}^{\text {locus}}$ and $d_{\mathcal {F}}( f, g_{\boldsymbol {\zeta }} ) < \varepsilon $ , there exists $\boldsymbol {\unicode{x3bb} } \in \boldsymbol {L}$ such that $d_{\mathcal {F}}( f, h_{\boldsymbol {\unicode{x3bb} }} ) < \varepsilon $ and the multiplier of $h_{\boldsymbol {\unicode{x3bb} }}$ at $\Omega _{\boldsymbol {\zeta }}$ lies in $\Lambda $ . Setting $h = h_{\boldsymbol {\unicode{x3bb} }}$ , we have $n_{h}> n_{g}$ . Thus, the lemma is proved.

Proof of Theorem 1.5

In the case where $0 \in \mathbb {C} \setminus E$ , the proof is identical to that of Theorem 1.4, by using Lemmas 3.3, 3.4 and 3.5 instead of Lemmas 2.3, 2.4 and 2.5.

Thus, from now on, assume that $0 \in E$ . Suppose that $f_{0} \in \mathcal {F}$ and $\varepsilon \in \mathbb {R}_{> 0}$ . We show that there exists $f \in \mathcal {F}$ such that:

• • $d_{\mathcal {F}}( f_{0}, f ) < \varepsilon $ ;

• • $f^{-1}(E) = E$ ; and

• • the cycles for f are all contained in E and their multipliers all lie in $\Lambda $ .

Since $\mathcal {E}$ is a closed subset of $\mathcal {F}$ with empty interior, replacing $f_{0}$ and $\varepsilon $ if necessary, we may assume that $f_{0} \in \mathcal {F} \setminus \mathcal {E}$ and $\varepsilon \in ( 0, \operatorname {\mathrm {dist}}( f_{0}, \mathcal {E} ) )$ . Write $E = \lbrace e_{j} : j \in \mathbb {Z}_{\geq 1} \rbrace $ , with $e_{j} \neq e_{k}$ for all distinct $j, k \in \mathbb {Z}_{\geq 1}$ . For $n \in \mathbb {Z}_{\geq 1}$ , define $E_{n} = \lbrace e_{1}, \dotsc , e_{n} \rbrace $ . We show that there exists a sequence $( f_{n} )_{n \geq 0}$ of elements of $\mathcal {F}$ such that, for every $n \geq 1$ :

1. (1) $d_{\mathcal {F}}( f_{n -1}, f_{n} ) < {\varepsilon }/{2^{n}}$ ;

2. (2) $f_{n} \in \mathcal {F}( f_{n -1}, A_{n -1}, C_{n -1} )$ (see the definitions below);

3. (3) $f_{n}^{\circ n}(0) \in E \setminus ( A_{n} \cup C_{n} )^{\mathrm {sym}}$ ;

4. (4) $f_{n}( e_{n} ) \in E$ ;

5. (5) $f_{n}^{-1}( E_{n} ) \cap D(0, n) \subset E$ ; and

6. (6) $f_{n}$ has the property $( \Gamma _{n, n} )$ ,

where, for every $n \geq 0$ ,

$$ \begin{align*} A_{n} = E_{n} \cup ( f_{n}^{-1}( E_{n} ) \cap D(0, n) ) \cup \lbrace 0, f_{n}(0), \dotsc, f_{n}^{\circ (n -1)}(0) \rbrace \end{align*} $$

and $C_{n}$ denotes the union of the cycles for $f_{n}$ with period at most n that intersect $D(0, n)$ , with $A_{0} = C_{0} = \varnothing $ by convention. We proceed by recursion. Suppose that $n \in \mathbb {Z}_{\geq 0}$ and $f_{1}, \dotsc , f_{n} \in \mathcal {F}$ satisfy these conditions (1)–(6), and let us prove the existence of ${f_{n +1} \in \mathcal {F}}$ that satisfies the same conditions. Define

$$ \begin{align*} B_{n, 0} = A_{n} \cup C_{n} , \quad g_{n, 0} = f_{n} \quad \text{and} \quad B_{n, 1} = A_{n} \cup C_{n} \cup \lbrace e_{n +1}, f_{n}^{\circ n}(0) \rbrace. \end{align*} $$

We have $f_{n}^{\circ n}(0) \in \mathbb {C} \setminus B_{n, 0}^{\mathrm {sym}}$ by condition (3). Therefore, by Lemma 3.3 applied successively with different dense subsets of $\mathbb {C}$ , there exist $g_{n, 1}, \dotsc , g_{n, n +2} \in \mathcal {F}$ such that, for every $j \in \lbrace 1, \dotsc , n +2 \rbrace $ ,

$$ \begin{align*} g_{n, j} \in \mathcal{F}( g_{n, j -1}, B_{n, j -1} ) , \quad d_{\mathcal{F}}( g_{n, j -1}, g_{n, j} ) < \frac{\varepsilon}{(n +2) \cdot 2^{n +3}} \end{align*} $$

and

$$ \begin{align*} g_{n, j} \circ \dotsb \circ g_{n, 1}( f_{n}^{\circ n}(0) ) \in E \setminus B_{n, j}^{\mathrm{sym}} , \end{align*} $$

where, for every $j \in \lbrace 2, \dotsc , n +2 \rbrace $ ,

$$ \begin{align*} B_{n, j} = B_{n, j -1} \cup \lbrace g_{n, j -1} \circ \dotsb \circ g_{n, 1}( f_{n}^{\circ n}(0) ) \rbrace. \end{align*} $$

Set $g_{n} = g_{n, n +2}$ . Note that, for every $j \in \lbrace 1, \dotsc , n +2 \rbrace $ ,

$$ \begin{align*} g_{n}^{\circ (n +j)}(0) = g_{n, j} \circ \dotsb \circ g_{n, 1}( f_{n}^{\circ n}(0) ) \end{align*} $$

and

$$ \begin{align*} B_{n, j} = A_{n} \cup C_{n} \cup \lbrace e_{n +1} \rbrace \cup \lbrace g_{n}^{\circ n}(0), \dotsc, g_{n}^{\circ (n +j -1)}(0) \rbrace. \end{align*} $$

In particular:

1. (i) $d_{\mathcal {F}}( f_{n}, g_{n} ) < {\varepsilon }/{2^{n +3}}$ ;

2. (ii) $g_{n} \in \mathcal {F}( f_{n}, A_{n} \cup C_{n} )$ ;

3. (iii) $g_{n}^{\circ (n +j)}(0) \in E$ for all $j \in \lbrace 1, \dotsc , n +2 \rbrace $ ;

4. (iv) $g_{n}^{\circ (n +1)}(0) \in E \setminus ( A_{n} \cup C_{n} \cup \lbrace e_{n +1}, f_{n}^{\circ n}(0) \rbrace )^{\mathrm {sym}}$ ;

5. (v) $g_{n}^{\circ (n +2)}(0) \in E \setminus E_{n +1}$ ; and

6. (vi) $g_{n}^{\circ (n +j)}(0) \neq \pm g_{n}^{\circ (n +1)}(0)$ for all $j \in \lbrace 2, \dotsc , n +2 \rbrace $ .

Now, define

$$ \begin{align*} A_{n}^{\prime} = A_{n} \cup \lbrace g_{n}^{\circ n}(0), \dotsc, g_{n}^{\circ (2 n +1)}(0) \rbrace. \end{align*} $$

Then we have $A_{n}^{\prime } \cup C_{n} \subset g_{n}^{-1}(E)$ by conditions (2), (3), (4), (6), (ii) and (iii). Therefore, by Lemma 3.3, there exists $h_{n} \in \mathcal {F}$ such that

$$ \begin{align*} h_{n} \in \mathcal{F}( g_{n}, A_{n}^{\prime} \cup C_{n} ) , \quad d_{\mathcal{F}}( g_{n}, h_{n} ) < \frac{\varepsilon}{2^{n +3}} , \quad h_{n}( e_{n +1} ) \in E. \end{align*} $$

Note that $h_{n} \in \mathcal {F} \setminus \mathcal {E}$ since $d_{\mathcal {F}}( f_{0}, h_{n} ) < \varepsilon $ by conditions (1) and (i). Define

$$ \begin{align*} A_{n}^{\prime \prime} = A_{n}^{\prime} \cup \lbrace e_{n +1} \rbrace = A_{n} \cup \lbrace e_{n +1} \rbrace \cup \lbrace g_{n}^{\circ n}(0), \dotsc, g_{n}^{\circ (2 n +1)}(0) \rbrace. \end{align*} $$

Then $A_{n}^{\prime \prime } \subseteq E \cap h_{n}^{-1}(E)$ by conditions (2)–(5), (ii) and (iii). Moreover, $C_{n}$ is a union of cycles for $h_{n}$ that are all contained in E and whose multipliers all lie in $\Lambda $ by conditions (6) and (ii). Also note that $0$ is not periodic for $h_{n}$ with period at most $n +1$ since $h_{n}^{\circ (n +1)}(0) \neq h_{n}^{\circ j}(0)$ for all $j \in \lbrace 0, \dotsc , n \rbrace $ by conditions (ii) and (iv). Therefore, by Lemma 3.5, there exists $k_{n} \in \mathcal {F}$ such that

$$ \begin{align*} k_{n} \in \mathcal{F}( h_{n}, A_{n}^{\prime \prime}, C_{n} ) , \quad d_{\mathcal{F}}( h_{n}, k_{n} ) < \frac{\varepsilon}{2^{n +3}} , \quad k_{n} \text{ has the property } ( \Gamma_{n +1, n +2} ). \end{align*} $$

Note that $k_{n} \in \mathcal {F} \setminus \mathcal {E}$ since $d_{\mathcal {F}}( f_{0}, k_{n} ) < \varepsilon $ by conditions (1) and (i). It follows that $k_{n}^{\circ j} \neq \text {id}_{\mathbb {C}}$ for all $j \in \lbrace 1, \dotsc , n +1 \rbrace $ , and hence there exists $R_{n} \in (n +1, n +2)$ such that $\partial D( 0, R_{n} )$ contains no periodic point for $k_{n}$ with period at most $n +1$ . Now, denote by $X_{n}$ the union of the cycles for $k_{n}$ with period at most $n +1$ that intersect $D( 0, R_{n} )$ . Then the cycles for $k_{n}$ in $X_{n}$ are all contained in E and their multipliers all lie in $\Lambda $ . Denote by $N_{n} \in \mathbb {Z}_{\geq 0}$ the number of periodic points for $k_{n}$ in $D( 0, R_{n} )$ with period at most $n +1$ , counting multiplicities. Since $\Lambda \subseteq \mathbb {C} \setminus \lbrace 1 \rbrace $ , the elements of $X_{n}$ are all simple periodic points for $k_{n}$ , and hence $N_{n}$ equals the cardinality of $X_{n} \cap D( 0, R_{n} )$ . Moreover, since the topology of $\mathcal {F}$ is finer than the topology of local uniform convergence, there exists $\varepsilon _{n} \in ( 0, {\varepsilon }/{2^{n +3}} )$ such that every $\ell _{n} \in \mathcal {F}$ such that $d_{\mathcal {F}}( k_{n}, \ell _{n} ) < \varepsilon _{n}$ has exactly $N_{n}$ periodic points in $D( 0, R_{n} )$ with period at most $n +1$ , counting multiplicities. Since $A_{n}^{\prime \prime } \cup X_{n} \subset E$ , it follows from Lemma 3.4 that there exists $f_{n +1} \in \mathcal {F}$ such that

$$ \begin{align*} f_{n +1} \in \mathcal{F}( k_{n}, A_{n}^{\prime \prime} \cup X_{n} ) , \quad d_{\mathcal{F}}( k_{n}, f_{n +1} ) < \varepsilon_{n} , \quad f_{n +1}^{-1}( E_{n +1} ) \cap D(0, n +1) \subset E. \end{align*} $$

Then $f_{n +1}$ clearly satisfies conditions (1), (2), (4) and (5). Furthermore, $f_{n +1}$ also satisfies condition (3) by conditions (ii), (iv), (v) and (vi). Now, note that $X_{n}$ is also a union of cycles for $f_{n +1}$ and $f_{n +1}$ has exactly $N_{n}$ periodic points in $D( 0, R_{n} )$ with period at most $n +1$ , counting multiplicities. As $X_{n} \cap D( 0, R_{n} )$ has exactly $N_{n}$ elements, it follows that $X_{n}$ is the union of all the cycles for $f_{n +1}$ with period at most $n +1$ that intersect $D( 0, R_{n} )$ . As these are all contained in E and their multipliers all lie in $\Lambda $ , the map $f_{n +1}$ also satisfies condition (6) since $R_{n}> n +1$ . Thus, we have proved the existence of a sequence $( f_{n} )_{n \geq 0}$ of elements of $\mathcal {F}$ that satisfies the desired conditions. Then the rest of the proof is completely identical to that of Theorem 1.4. Thus, the theorem is proved.

Proof. As $f_{0}^{(2 j)}(0) = 10$ for all $j \in \mathbb {Z}_{\geq 1}$ and $\lVert f -f_{0} \rVert _{\mathcal {B}} < 10$ , we have $f^{(2 j)}(0)> 0$ for all $j \in \mathbb {Z}_{\geq 1}$ . Therefore, f is transcendental. Moreover, $f \colon \mathbb {R} \rightarrow \mathbb {R}$ is convex as

$$ \begin{align*} \text{ for all } x \in \mathbb{R}, \, f^{\prime \prime}(x) = \sum_{j = 0}^{+\infty} \frac{f^{(2 j +2)}(0)}{(2 j)!} x^{2 j}> 0. \end{align*} $$

Thus, the lemma is proved.

Proof. The map $\cosh $ induces a proper holomorphic map of degree two from the strip

$$ \begin{align*} S = \lbrace z \in \mathbb{C} : \Im(z) \in (-\pi, \pi) \rbrace \end{align*} $$

to the slit plane $\mathbb {C} \setminus (-\infty , -1]$ . Since $\cosh $ is even and $3 \mathbb {D} \subset S$ , there is a univalent map $\phi \colon 9 \mathbb {D} \rightarrow \mathbb {C}$ such that $\cosh (z) = \phi ( z^{2} )$ for all $z \in 3 \mathbb {D}$ . It follows that the map $f_{0} \colon 2 \mathbb {D} \rightarrow f_{0}(2 \mathbb {D})$ is proper of degree two. Moreover, we have $\phi (0) = 1$ and $\phi ^{\prime }(0) = \tfrac 12$ . Therefore, it follows from the Koebe distortion theorem that

$$ \begin{align*} D(1, r) \subset \phi(\mathbb{D}) \subset D(1, R) \quad \text{with} \quad r = \frac{{1}/{2}}{( 1 + ({1}/{9}))^{2}}> \frac{2}{5} , \quad R = \frac{{1}/{2}}{( 1 - ({1}/{9}) )^{2}} < \frac{7}{10} , \end{align*} $$

and

$$ \begin{align*} D(1, s) \subset \phi(4 \mathbb{D}) \quad \text{with} \quad s = \frac{{4}/{2}}{( 1 + ({4}/{9}) )^{2}}> \frac{9}{10}. \end{align*} $$

Thus, the desired inclusions hold, and the lemma is proved.

Proof. Set $g = f -f_{0}$ . Then, for every $z \in 2 \overline {\mathbb {D}}$ ,

$$ \begin{align*} \lvert g(z) \rvert = \bigg\lvert \sum_{j = 0}^{+\infty} \frac{g^{(2 j)}(0)}{(2 j)!} z^{2 j} \bigg\rvert \leq \lVert g \rVert_{\mathcal{B}} \cdot \sum_{j = 0}^{+\infty} \frac{2^{2 j}}{(2 j)!} < \frac{\cosh(2)}{4} < 1. \end{align*} $$

In particular,

$$ \begin{align*} f(0) = -2 +g(0) \in D(-2, 1) \cap \mathbb{R} \subset (-\infty, -1). \end{align*} $$

Moreover, by Lemma 4.3 and the argument principle, it follows that

$$ \begin{align*} \overline{D(-2, 3)} \subset f(\mathbb{D}) \subset D(-2, 8) \end{align*} $$

and every element of $D(-2, 8)$ has exactly two preimages in $2 \mathbb {D}$ under f, counting multiplicities. Therefore, $\mathbb {D} \Subset f(\mathbb {D})$ and the map $f \colon \mathbb {D} \rightarrow f(\mathbb {D})$ is proper of degree two. This completes the proof of the lemma.

Proof. Set $U = f^{-1}(\mathbb {D}) \cap \mathbb {D}$ . As $f \colon \mathbb {D} \rightarrow f(\mathbb {D})$ is a quadratic-like map, $U \Subset \mathbb {D}$ and the map $f \colon U \rightarrow \mathbb {D}$ is proper of degree two. Moreover, the unique critical point of f in $\mathbb {D}$ is $0$ since f is even, and $f(0) \in \mathbb {C} \setminus \mathbb {D}$ by assumption. Therefore, $f \colon U \rightarrow \mathbb {D}$ is an escaping quadratic-like map. Thus, U has two connected components $U_{-}$ and $U_{+}$ and these are mapped biholomorphically onto $\mathbb {D}$ by f. Denote by $g_{\pm } \colon \mathbb {D} \rightarrow U_{\pm }$ the inverse of ${f \colon U_{\pm } \rightarrow \mathbb {D}}$ . We have $U_{+} = -U_{-}$ and $g_{+} = -g_{-}$ because f is even. A standard argument using the fact that the map $g_{\pm }$ is contracting with respect to the Poincaré metric on $\mathbb {D}$ shows that, for every sign sequence $\boldsymbol {\varepsilon } = ( \epsilon _{n} )_{n \geq 0}$ , there is a point $\zeta (\boldsymbol {\varepsilon }) \in \mathbb {D}$ such that

$$ \begin{align*} \bigcap_{n \geq 0} g_{\epsilon_{0}} \circ \dotsb \circ g_{\epsilon_{n}}(\mathbb{D}) = \lbrace \zeta(\boldsymbol{\varepsilon}) \rbrace. \end{align*} $$

Now, denote by $\sigma $ the shift map, which sends a sign sequence $( \epsilon _{n} )_{n \geq 0}$ to $( \epsilon _{n +1} )_{n \geq 0}$ . Then the periodic points of $f \colon U \rightarrow \mathbb {D}$ are precisely the points $\zeta (\boldsymbol {\varepsilon })$ , where $\boldsymbol {\varepsilon }$ is a sign sequence that is periodic for $\sigma $ .

Since $f \colon \mathbb {R} \rightarrow \mathbb {R}$ is even and convex, it is increasing on $\mathbb {R}_{\geq 0}$ . As $f \colon \mathbb {D} \rightarrow f(\mathbb {D})$ is a quadratic-like map, it follows that $f(1)> 1$ . Therefore, $U_{\pm } \cap \mathbb {R} \neq \varnothing $ because we also have $f(0) < -1$ , by assumption. It follows that $U_{\pm }$ is symmetric with respect to the real axis and $g_{\pm }$ commutes with complex conjugation. Therefore, for every sign sequence $\boldsymbol {\varepsilon }$ , the set $\lbrace \zeta (\boldsymbol {\varepsilon }) \rbrace $ is also symmetric with respect to the real axis, and hence $\zeta (\boldsymbol {\varepsilon }) \in \mathbb {R}$ . In particular, every periodic point of $f \colon U \rightarrow \mathbb {D}$ is also a periodic point of $f \colon \mathbb {R} \rightarrow \mathbb {R}$ .

Finally, note that $f(x)> x$ for all $x \in [1, +\infty )$ as $f \colon \mathbb {R} \rightarrow \mathbb {R}$ is convex, $f(0) < 0$ and $f(1)> 1$ . It follows that the map $f \colon \mathbb {R} \rightarrow \mathbb {R}$ has no periodic point in $[1, +\infty )$ . Therefore, the periodic points of $f \colon \mathbb {R} \rightarrow \mathbb {R}$ all lie in $(-1, 1) \subset \mathbb {D}$ because f is even. Thus, every periodic point of $f \colon \mathbb {R} \rightarrow \mathbb {R}$ is also a periodic point of $f \colon U \rightarrow \mathbb {D}$ , and the lemma is proved.

Proof of Theorem 1.9

Define

$$ \begin{align*} \mathbb{Q}_{0} = \bigg\lbrace \frac{p}{q} \in \mathbb{Q} : p \text{ odd}, \, q \text{ even} \bigg\rbrace \quad \text{and} \quad \mathbb{Q}_{1} = \bigg\lbrace \frac{p}{q} \in \mathbb{Q} : q \text{ odd} \bigg\rbrace \end{align*} $$

to be the sets of rational numbers with even and odd denominators. Also define

$$ \begin{align*} E = \mathbb{Q}(i) , \quad \Lambda = \mathbb{Q}_{0} \cup (\mathbb{C} \setminus \mathbb{R}) \quad \text{and} \quad \Lambda^{\prime} = \mathbb{Q}_{1} \cup (\mathbb{C} \setminus \mathbb{R}). \end{align*} $$

Then E is a countable and dense subset of $\mathbb {C}$ such that $E^{\mathrm {sym}} = E$ and $E \cap (\mathbb {R} \cup i \mathbb {R})$ is dense in $\mathbb {R} \cup i \mathbb {R}$ . Moreover, $\Lambda $ and $\Lambda ^{\prime }$ are dense subsets of $\mathbb {C}$ that are symmetric with respect to the real axis, and $\Lambda \cap \mathbb {R}$ and $\Lambda ^{\prime } \cap \mathbb {R}$ are both dense in $\mathbb {R}$ . Therefore, by Theorem 1.5 and Remark 1.6, there exists $f \in \mathcal {B}$ such that:

1. (1) $\lVert f -f_{0} \rVert _{\mathcal {B}} < \tfrac 14$ ;

2. (2) $f^{-1}(E) = E$ ;

3. (3) the periodic points of f all lie in E;

4. (4) the multipliers of f at its cycles with period different from $2$ all lie in $\Lambda $ ; and

5. (5) the multipliers of f at its cycles with period $2$ all lie in $\Lambda ^{\prime }$ .

By condition (1) and Lemmas 4.2, 4.4 and 4.5, the map f is transcendental, the map $f \colon \mathbb {R} \rightarrow \mathbb {R}$ is convex and $f \colon f^{-1}(\mathbb {D}) \cap \mathbb {D} \rightarrow \mathbb {D}$ is an escaping quadratic-like map whose cycles coincide with those of $f \colon \mathbb {R} \rightarrow \mathbb {R}$ . We have $f(\mathbb {Q}) \subset E \cap \mathbb {R} = \mathbb {Q}$ by condition (2). Now, by condition (3), the periodic points of $f \colon \mathbb {R} \rightarrow \mathbb {R}$ all lie in $E \cap \mathbb {R} = \mathbb {Q}$ . Moreover, by conditions (4) and (5), the multipliers of $f \colon \mathbb {R} \rightarrow \mathbb {R}$ all lie in $\Lambda \cap \mathbb {R} = \mathbb {Q}_{0}$ , apart from that at its unique cycle with period $2$ which lies in $\Lambda ^{\prime } \cap \mathbb {R} = \mathbb {Q}_{1}$ . In particular, the multipliers of $f \colon \mathbb {R} \rightarrow \mathbb {R}$ all lie in $\mathbb {Q}$ . Finally, the product of the multipliers of $f \colon f^{-1}(\mathbb {D}) \cap \mathbb {D} \rightarrow \mathbb {D}$ at its two fixed points lies in $\mathbb {Q}_{0}$ , as both of them lie in $\mathbb {Q}_{0}$ , and hence it differs from the multiplier at the cycle with period $2$ . It follows that $f \colon f^{-1}(\mathbb {D}) \cap \mathbb {D} \rightarrow \mathbb {D}$ is not conjugate to an affine escaping quadratic-like map. Thus, the theorem is proved.