\section{Introduction}
 Let $f\colon {\mathbb{C}}\to{\mathbb{C}} $ be a transcendental entire function. The set 
    \[ I(f)\defeq \{z\in {\mathbb{C}}\colon \lim_{n\to\infty} f^n(z) = \infty\} \]
  is called the \emph{escaping set} of $f$, where 
     \[ f^n = \underbrace{f\circ\dots\circ f}_{n\text{ times}}\] 
    denotes the $n$-th iterate of $f$. The escaping set  was first studied by Eremenko~\cite{alexescaping} and has been the subject of intensive research in recent years;
     see e.g.~\cite{oberwolfach,rrrs,bergweilermeasure,ripponstallarderemenkopoints} and their references. The topological study of $I(f)$ turns out to be surprisingly
     intricate. For example Eremenko \cite[p.~343]{alexescaping} asked whether, for every 
     transcendental entire function, every
     connected component of $I(f)$ is unbounded. This question has become known as \emph{Eremenko's conjecture}, and is one of the most famous open problems
     in transcendental dynamics. Strengthened versions of the conjecture have since been proved false in general; compare~\cite[Theorem~1.1]{rrrs} 
     and~\cite[Theorem~1.6]{arclike}. 
     On the other hand, Rippon and Stallard~\cite{boundariesescaping} 
     have shown that $I(f)\cup\{\infty\}$ is always connected, so any counterexample to Eremenko's original conjecture would have rather
     subtle topological properties.
     
   The \emph{fast escaping set} is a certain subset of the escaping set, first introduced by Bergweiler and Hinkkanen~\cite{bergweilerhinkkanen}. This set has also received considerable attention recently; in part because it 
    appears to be more tractable than the full escaping set 
      $I(f)$. In particular, the analogue of Eremenko's conjecture holds for $A(f)$ \cite{ripponstallardfatoueremenko}. One difference between the two
      objects lies in the topological complexity of their definitions. Indeed, $A(f)$ can be written as a countable union of closed sets~\cite[Formula~(1.3)]{ripponstallardfast}; 
       so it is an $F_{\sigma}$ set. On the other hand, by definition 
      \begin{align*} I(f) &= \{z\in{\mathbb{C}}\colon \forall M\geq 0\, \exists N\geq 0 \colon \lvert f^n(z)\rvert \geq M \text{ for $n\geq N$}\} \\ 
            &=\bigcap_{M= 0}^{\infty} \bigcup_{N=0}^{\infty}  \bigcap_{n=N}^{\infty} f^{-n}({\mathbb{C}}\setminus D(0,M)), \end{align*}
    where $D(0,M)$ denotes the disc of radius $M$ around $0$.  
      So $I(f)$ is $F_{\sigma \delta}$: a countable intersection of countable unions of closed sets.       
     This raises the following question, posed by Rippon in 2009~\cite[Problem~8, p.~2960]{oberwolfach}.
     
     \begin{question}\label{qu:Fsigma}
       Is there a transcendental entire function $f$ for which $I(f)$ itself is $F_{\sigma}$? 
     \end{question}
         Since any closed subset of ${\mathbb{C}}$ is a countable union of compact sets, it is equivalent to ask whether 
         $I(f)$ is ever \emph{$\sigma$-compact} (a countable union of compact sets); compare also \cite{lipham}. It is well-known that 
        $I(f)$ cannot be a $G_{\delta}$ (a countable intersection of
        open sets); see Lemma~\ref{lemma:Gdelta} below. 
      
   As far as we are aware, there is no function $f$ for which Question~\ref{qu:Fsigma} has been previously resolved; the goal of this paper is to
     give a negative answer in general. In fact, we prove a stronger result, as follows.
     Let $\operatorname{UO}(f)$ consist of all $z\in{\mathbb{C}}$ whose orbit $\{f^n(z)\colon n\geq 0\}$ is unbounded, 
     and let $\operatorname{BU}(f) = \operatorname{UO}(f)\setminus I(f)$ denote the ``bungee set''; see \cite{bungee}. Recall that the \emph{Julia set} $J(f)$ is the set of non-normality
     of the iterates of~$f$. 
     \begin{thm}\label{thm:main}
       Let $f$ be a transcendental entire function. 
        Then every $\sigma$-compact subset of $\operatorname{UO}(f)$ omits some point of $I(f)\cap J(f)$ and some point of $\operatorname{BU}(f)\cap J(f)$. 
                 
      In particular, the sets $I(f)$, $\operatorname{UO}(f)$, $\operatorname{BU}(f)$ and their intersections with $J(f)$ are not $\sigma$-compact. 
     \end{thm}
          
     \subsection*{Acknowledgements} I thank David Lipham and Phil Rippon for interesting discussions. I also thank David Mart\'i-Pete, James Waterman and the referee for 
          helpful comments that improved
  		the presentation of the paper.

\section{Proof of the Theorem}
  We require a result on the existence of arbitrarily slowly escaping points, 
   which follows from work of Rippon and Stallard~\cite{ripponstallardannular}; see also~\cite[Theorem~1]{ripponstallardslow}. 
   \begin{thm}\label{thm:slow}
      Let $f$ be a transcendental entire function, and let $R_0\geq 0$. Then there exists $M_0>R_0$ 
       with the following property. 
        If $(a_m)_{m=0}^{\infty}$ is a sequence with $a_m\to\infty$ and $a_m\geq M_0$ for all $m$, 
        then there are $\zeta \in J(f)\cap I(f)$ and $\omega\in J(f)\cap \operatorname{BU}(f)$ with
              \[ R_0 \leq \lvert f^m(\zeta)\rvert \leq a_m\quad\text{and}\quad R_0 \leq \lvert f^m(\omega)\rvert \leq a_m \quad\text{for $m\geq 0$}. \]
   \end{thm}
    \begin{proof}
    The result is an easy consequence of~\cite{ripponstallardannular},
      which studies ``annular itineraries'' of entire functions. 
      Fix some $R>0$ such that $M(r)>r$ for $r\geq R$, where $M(r)$
      denotes the maximum modulus of $f$ on the disc $\overline{D(0,r)}$. 
      Then the \emph{annular itinerary} of a point $z$ is the sequence 
        $(s_m)_{m=0}^{\infty}$ such that
        $s_m=0$ if $\lvert f^m(z)\rvert < R$, and 
        \[ M^{s_m-1}(R) \leq \lvert f^m(z) \rvert < M^{s_m}(R) \] 
        otherwise. (Here $M^k$ is the $k$-th iterate of $r\mapsto M(r)$.)
        
      Theorem 1.2 of~\cite{ripponstallardannular} implies that for suitable $R\geq R_0$, for an entry $s_m=n\geq 0$ in an annular itinerary, 
       there is a subset $X_n\subset \{0,\dots,n+1\}$ of allowable next entries $s_{m+1}$, and that, for infinitely many $n$, all entries but at most one
       are allowable. More precisely,  
           \begin{enumerate}[(a)]
              \item any sequence $(s_m)_{m=0}^{\infty}$ with $s_{m+1}\in X_{s_m}$ for all $m$ is the annular itinerary of some $z\in J(f)$;\label{item:admissible}
              \item $n+1\in X_n$ for all $n$; 
              \item there is an increasing sequence $(n_j)_{j=1}^{\infty}$ such that $\# X_{n_j} \geq n+1$.
            \end{enumerate}
       (With the notation of~\cite{ripponstallardannular}, $X_{n_j}=\{0,\dots,n_j+1\}\setminus I_j$, where $\# I_j \leq 1$, and
           $X_n = \{n+1\}$ when $n\neq n_j$ for all $j$.) We may suppose that the sequence $(n_j)$ is chosen such that $n_1 \geq 2$.

     Set $M_0\defeq M^{n_1}(R)$ and 
      let $(a_m)_{m=0}^{\infty}$ be a sequence as in the statement of the theorem. We may assume that $(a_m)$ is non-decreasing. 
      Similarly as in the proof of \cite[Corollary~1.3~(d)]{ripponstallardannular}, we can construct 
         an annular itinerary $(s_m)_{m=0}^{\infty}$ satisfying~\ref{item:admissible} such that $s_m\to\infty$ and 
         \begin{equation}\label{eqn:slowgrowth} M^{s_m}(R) \leq a_m \qquad\text{for } m\geq 0. \end{equation}
       Indeed, for each $j$, either $n_j\in X_{n_j}$ or $n_{j}-1\in X_{n_j}$; let us denote this element of $X_{n_j}$ by $\tilde{n}_j$. Then any sequence of the form
        \[ (s_m)_{m=0}^{\infty} = \underbrace{n_1, \tilde{n}_1, n_1, \tilde{n}_1,\dots, n_1}_{\text{length }N_1},n_{1}+1, n_1+2, \dots, 
            \underbrace{n_{2}, \tilde{n}_{2}, \dots, n_{2}}_{\text{length }N_2} ,n_{2} + 1, n_{2} + 2, \dots \]
      satisfies~\ref{item:admissible}. If $N_j$ is chosen sufficiently large that 
               \begin{equation}\label{eqn:Nj} a_m \geq M^{n_{j+1}}(R) \quad\text{for } m\geq N_j, \end{equation}
        then~\eqref{eqn:slowgrowth} holds for $m\geq N_1$. It also holds for $m\leq N_1$, since $s_m\leq n_1$ for these values of $m$, and 
     $a_m \geq M_0 = M^{n_1}(R)$ by assumption. 
      
      Let $\zeta \in J(f)$ have annular itinerary $(s_m)_{m=0}^{\infty}$; then $\zeta\in I(f)$ and 
         \[
            R_0\leq R \leq M^{s_m-1}(R) \leq \lvert f^m(\zeta)\rvert  < M^{s_m}(R) \leq a_m \quad \text{for all $m$}. \]
       Hence 
       $\zeta$ has the desired properties.   
   To obtain $\omega$, we instead use an unbounded but non-escaping sequence of the form 
        \[(\tilde{s}_m)_{m=0}^{\infty} = B_1,n_{1}+1, n_1+2, \dots, n_{2}, 
            \hat{n}_2, B_2 , n_1+1,n_1+2,\dots,n_3,\hat{n}_3, B_3,\dots, \]
     where $\hat{n}_j\in X_{n_j}$ is either $n_1$ or $n_1-1$, and $B_j$ is a block alternating between entries 
      $n_1$ and $\tilde{n}_1$. To ensure that the sequence satisfies~\ref{item:admissible}, $B_j$ should end in $n_1$; it should begin
      with $n_1$ if $\hat{n}_j=n_1-1$, and with $\tilde{n}_1$ if $\hat{n}_j = n_1$. The length $N_j$ of the block $B_j$ is again chosen to satisfy~\eqref{eqn:Nj}.
    \end{proof}

\begin{proof}[{Proof of Theorem~\ref{thm:main}}]
 If $D\subset{\mathbb{C}}$ is closed and $z\in{\mathbb{C}}$, we set 
    \[ n_D(z) \defeq \min\{n\geq 0\colon f^n(z)\notin D\} \leq \infty, \]
     with the convention that $n_D(z)=\infty$ if no such $n$ exists. Observe that
       $n_D$ depends upper semicontinuously on $z$, as the infimum of
       upper semicontinuous functions. Indeed, 
        \[ n_D(z) = \min\{\chi_n(z) \colon n \geq 0\}, \qquad\text{where}\qquad
            \chi_n(z) = \begin{cases}
                             \infty & \text{if } z\in f^{-n}(D) \\
                             n & \text{if } z\notin f^{-n}(D),\end{cases} \]
      and each $\chi_n$ is upper semicontinuous since $f^{-n}(D)$ is closed. 
      Note also that 
     $z\in \operatorname{UO}(f)$ if and only if $n_D(z)<\infty$ for every compact $D\subset{\mathbb{C}}$. 

  Now let $R_0 \geq 0$ be arbitrary, and choose $M_0$ as in Theorem~\ref{thm:slow}.
   Let $X\subset \operatorname{UO}(f)$ be $\sigma$-compact; say 
   $X = \bigcup_{j=0}^{\infty} K_j$ where each $K_j$ is compact. 
     Define $M_j \defeq M_0 +j$ and $D_j \defeq \overline{D(0,M_j)}$. Then
     $n_{D_j}(z)<\infty$ for every $z\in K_j$. 
     
     Since $K_j$ is compact and $n_{D_j}$ is
     upper semicontinuous,  
     \[ n_j \defeq \max_{z\in K_j} n_{D_j}(z) \]
     exists for every $j$. Set $N_0\defeq n_0$ and $N_{j+1}\defeq \max\{n_{j+1},N_j+1\}$. 
     For $n\geq 0$, define $j(n)\defeq\min\{j\colon N_j\geq n\}$ and 
      \[ a_n \defeq M_{j(n)} = \min\{ M_j \colon  N_j\geq n \}. \]
     Then $a_n\geq M_0$ for all $n$ and $\lim_{n\to\infty} a_n = \infty$. So by Theorem~\ref{thm:slow}, there are $\zeta\in I(f)\cap J(f)$ and $\omega\in \operatorname{BU}(f)\cap J(f)$ such that
       $\lvert f^n(z)\rvert \leq a_n$ for $z\in \{\zeta,\omega\}$ and all $n\geq 0$.
       
      Let $j\geq 0$. Then, for $n\leq n_j\leq N_j$, we have
        $\lvert f^n(z)\rvert \leq a_n \leq M_j$, and hence $f^n(z) \in D_j$. So $n_{D_j}(z) > n_j$, and $z\notin K_j$ by choice of $n_j$. 
       Thus $z\notin X = \bigcup_{j=0}^{\infty} K_j$, as claimed. 
\end{proof}

By the expanding property of the Julia set, we also obtain the following answer to a question of Lipham (personal communication).

\begin{cor}\label{cor:nowheresigmacompact}
  Let $f$ be a transcendental entire function and let $Y$ be  one of the sets $I(f)\cap J(f)$, $\operatorname{UO}(f)\cap J(f)$ and $\operatorname{BU}(f)\cap J(f)$. 
   Then $Y$ is nowhere $\sigma$-compact. That is, if $X\subset Y$ is $\sigma$-compact, then
    $X$ does not contain a non-empty relatively open subset of $Y$. 
\end{cor} 
\begin{proof}
   We prove a  slightly stronger statement. Let $Y$ be as in the statement and  let  
   $X\subset \operatorname{UO}(f)\cap J(f)$ be $\sigma$-compact. If $U\subset{\mathbb{C}}$ is open with $U\cap Y \neq \emptyset$, then $X$ omits some point of 
      $U\cap Y$. 
      
    Let $R_0$ and $M_0$ be as in Theorem~\ref{thm:slow}, and set $L_0\defeq J(f)\cap \overline{D(0,M_0)}\setminus D(0,R_0)$. If $R_0$ is chosen sufficiently large, then 
      $L_0$ contains no Fatou exceptional point of $f$. (Recall that a Fatou exceptional point is a point with finite backward orbit, 
      and that $f$ has at most one finite exceptional point \cite[p.~156]{waltersurvey}.) Since $U\cap J(f)\neq \emptyset$, it follows that 
     there is some $n$ such that $f^n(U)\supset L_0$.
     
     The proof of Theorem~\ref{thm:main} shows that no $\sigma$-compact subset of $\operatorname{UO}(f)\cap J(f)$ contains $L_0\cap Y$. (Note that we may take $a_0=M_0$ in the proof
      to ensure that $\lvert \zeta \rvert,\lvert \omega\rvert \leq M_0$.) Moreover, $f^n(X)$ is $\sigma$-compact as the 
	image of a $\sigma$-compact set under a continuous function. So $f^n(X)$ omits some points of $L_0\cap Y$. Since $Y$ is backward-invariant,
	$X$ omits some points of $U\cap Y$, as claimed. 
\end{proof} 

We remark that the proof of Theorem~\ref{thm:main} establishes the following very general principle: 
\begin{prop}[Abstract version of the main theorem]\label{prop:abstract}
  Let $f\colon U\to V$ be continuous, where $U,V$are non-empty topological spaces and $U$ is
  $\sigma$-compact. Let $\operatorname{UO}(f)$ denote the set of $z\in U$ such that $f^n(z)$ is defined and in $U$ for all $n\geq 0$, but
   the orbit $\{f^n(z)\}$ is not contained in any compact subset of $U$.
   
  Suppose that $X\subset \operatorname{UO}(f)$ is $\sigma$-compact and $\Delta\subset U$ is compact. Then there is a sequence 
  $(\Delta_n)_{n=0}^{\infty}$ of compact subsets $\Delta_n\subset U$ with $\Delta\subset \Delta_0\subset \Delta_1\subset \dots$ and 
    $\bigcup_{n=0}^{\infty} \Delta_n = U$ such that $X$ contains no point $\zeta\in \operatorname{UO}(f)$ with
  $f^n(\zeta)\in \Delta_n$ for all $n\geq 0$. 
  \end{prop}
  \begin{remark}
    Note that we are not assuming any relation between the spaces $U$ and $V$. However, the statement is vacuous when $\operatorname{UO}(f)=\emptyset$, and in
      particular when $U\cap V\neq \emptyset$. 
  \end{remark}
  \begin{sketch}
     Let $(D_j)_{j=0}^{\infty}$ be an increasing sequence of compact subsets $D_j\subset U$ with $\bigcup _{j=0}^{\infty} D_j= U$. 
      Define $j(n)$ as in the proof of Theorem~\ref{thm:main}; then the sets $\Delta_n \defeq D_{j(n)}$ have the desired property. 
  \end{sketch}

  It follows that the set of escaping points is not $\sigma$-compact in any
  setting where an analogue of  Theorem~\ref{thm:slow} holds. 
  This includes:
  \begin{enumerate}[(a)]
    \item transcendental meromorphic functions ${\mathbb{C}}\to\hat{\C}$ \cite{ripponstallardslow};\label{item:mero}
    \item transcendental self-maps of the punctured plane \cite[Theorem~1.2]{martipeteCstar};\label{item:Cstar}
    \item quasiregular self-maps $f\colon {\mathbb{R}}^d\to{\mathbb{R}}^d$ of transcendental type
        \cite{nicksslow}; 
    \item continuous functions $\phi\colon [0,\infty)\to [0,\infty)$ with $\phi(t)\not\to\infty$ as $t\to\infty$,
       and such that $I(\phi)\neq \emptyset$ \cite[Theorem~2.2]{osborneripponstallard}.
  \end{enumerate}

   A more general setting than both~\ref{item:mero} and \ref{item:Cstar} is provided by
     the \emph{Ahlfors islands maps} of Epstein; see e.g.~\cite{exotic}. 
     These are maps $f\colon W\to X$, where $X$ is a compact one-dimensional
     manifold, $W\subset X$ is open, and $f$ satisfies certain transcendence 
     conditions near $\partial W$. We may define the escaping set $I(f)$ 
     as the set of points $z\in W$ with $f^n(z)\in W$ for all $n$ and
     $\operatorname{dist}(f^n(z),\partial W)\to 0$ as $n\to\infty$. It is plausible that an analogue of Theorem~\ref{thm:slow} holds for
     Ahlfors islands maps with $W\neq X$, using a similar proof
    as in~\cite{ripponstallardslow}. This would mean, by Proposition~\ref{prop:abstract}, that $I(f)$ is not 
    $\sigma$-compact for such functions.

 For completeness, we conclude by giving the simple proof that $I(f)$ is never a $G_{\delta}$ set; compare also \cite[Corollary~3.2]{liphametds}. 
\begin{lem}\label{lemma:Gdelta}
 Let $f$ be a transcendental entire function. Then $I(f)$ and $I(f)\cap J(f)$ are not $G_{\delta}$ sets. 
\end{lem}
\begin{proof}
  Every closed subset of ${\mathbb{C}}$ (or any metric space) is $G_{\delta}$; so $J(f)$ is $G_{\delta}$. 
   The intersection of two $G_{\delta}$ sets is again $G_{\delta}$, so it is enough to prove the claim for $I(f)\cap J(f)$. 
 By \cite[Theorem~2]{alexescaping}, $I(f)\cap J(f)$ is nonempty, and hence dense in $J(f)$ by Montel's theorem. 
  By Baire's theorem, any two dense $G_{\delta}$ subsets of $J(f)$ must intersect. Hence it is enough
    to observe that $\operatorname{BU}(f)\cap J(f)$ contains a dense $G_{\delta}$ by Montel's theorem, namely the set of points whose orbits are dense in $J(f)$.  
      (See \cite[Lemma~1]{bakerdominguezresidual}.) 
\end{proof}



 Lipham has pointed out the following reformulation of Corollary~\ref{cor:nowheresigmacompact}.
 \begin{cor}
    Any $G_{\delta}$ set 
    $A\subset Y\defeq J(f)\setminus I(f)$ is nowhere dense in $Y$. 

    In particular, $Y$ is $G_{\delta \sigma}$ but not \emph{strongly $\sigma$-complete}; that is, it cannot be written as a countable
    union of relatively closed $G_{\delta}$ subsets.
  \end{cor}
  \begin{proof}
    Suppose, by contradiction, that $\overline{D(\zeta,\varepsilon)}\cap Y \subset A$ for some $\zeta\in J(f)$ and $\varepsilon>0$. The complement
      of a $G_{\delta}$ set is $F_{\sigma}$, so 
    $B \defeq J(f)\cap \overline{D(\zeta,\varepsilon)} \setminus A$ is an $F_{\sigma}$ set 
     with \[ D(\zeta,\varepsilon)\cap I(f)\cap J(f)\subset B \subset I(f)\cap J(f),\] which contradicts Corollary~\ref{cor:nowheresigmacompact}. 

   The set $Y$ is $G_{\delta \sigma}$ by definition. 
     As mentioned in the proof of Lemma~\ref{lemma:Gdelta}, $Y$ contains a $G_{\delta}$ subset $U$ that is dense in $J(f)$.
     On the other hand, if $(A_k)_{k=0}^{\infty}$ is a sequence of relatively closed $G_{\delta}$ subsets of $Y$, then 
     $\overline{A_k}$ is closed and nowhere dense in $J(f)$. Hence
        \[  \bigcup_{k=0}^{\infty} A_k \subset \bigcup_{k=0}^{\infty} \overline{A_k} \not\supset U \subset Y, \]
      by Baire's theorem, and $Y$ is indeed strongly $\sigma$-complete. 
  \end{proof}

  
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