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\begin{document}

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\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\begin{proof}
  which implies that the sequence $S[\phi_{k}]^{1/k}$
  converges. (It is close to a decreasing sequence.)
  For surfaces,
  $S$ does not behave as well under covers, but we still have
  that if $\widetilde \phi$ is a cover of $\phi$, then $S[\phi] \le S[\widetilde  \phi] \le \max(1,S[\phi])$
\end{proof}

\end{document}

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