I would like to use pstool with beamer to create a presentation in pdflatex with figures processed by psfrag. However, as the simple example below shows, the figures are processed using the beamer documentclass, so that they each have a full beamer "frame" around them. Is there a way around this?

\documentclass{beamer}

\usepackage{lmodern}
\usepackage{graphicx}
\usepackage[process=all,crop=pdfcrop]{pstool}

\begin{document}
\begin{frame}
    \frametitle{Circle}
    \begin{figure}
        \psfragfig[width=0.3\textwidth]{figures/circle}
        {
            \psfrag{1}{\(s\)}
        }
    \end{figure}
\end{frame}
\end{document}




I suspect that if it's possible to override the documentclass used by pstool to process the figures, that would be a solution, but I haven't found how to do this.

Thanks.


Question: If $f : [-a,a] \rightarrow \mathbb{IR}$ is Scott continuous, why are $f^-$ and $f^+$ measurable?
Answer: This answers @user3078439's question in his comment to my original answer.

First, a short argument to show that if $P,Q$ are domains and $f:P\to Q$ is Scott-continuous, then $f$ is order preserving.

Let $a\leq b \in P$. So $D:= \{a,b\}$ is a directed set with $b=\bigsqcup D$. Since $f$ is Scott-continuous, $f(D) = \{f(a), f(b)\}$ is also directed, so at least one of the relations $f(a) \leq f(b)$ or $f(b) \leq f(a)$ holds in $Q$. Now $f$ being Scott-continuous means $$f(b) = f(\bigsqcup D) = \bigsqcup f(D) = \sup\{f(a), f(b)\},$$ which directly implies $f(a)\leq f(b)$. So if $f:P\to Q$ is Scott-continuous, then it is order-preserving. (The converse of this statement does not hold.)

Since in your question the ordering on $Q:= \mathbb{IR}$ is given by $\supseteq$, I hope that I have answered your implicit question "if $f$ being Scott-continuous truly means that for all $x\leq y, f(x)\supseteq f(y)$" in the positive.

As to the question whether $f(-a)\supseteq f([-a,a])$ there is a negative answer: $f(-a)$ is just a real interval, namely $[f^-(-a), f^+(-a)]$. But $f([-a,a])$ is a set of intervals.

What is true, though, is that $f(-a)=[f^-(-a), f^+(-a)]$ is an interval that is a superset of every interval $f(x)=[f^-(x), f^+(x)]$ for all $x\in[-a,a]$. If that was the intent of your question in the comment, then there is a positive answer.