Is it possible to skip the first line in a theorem environment?

Question

I'm currently writing up some solutions, and I like to rewrite the problem before answering it.

I've set exercise to be a theorem environment, and the code I'm using is

\begin{exercise} 
\begin{enumerate} 
\item Show that $R$ is symmetric iff $R^{-1}\subseteq R$.
\item Show that $R$ is transitive iff $R\circ R\subseteq R$.
\end{enumerate}
\begin{proof}
\end{proof}
\end{exercise}

However, when I build, the first item (a) appears immediately to the right of the Exercise heading, and the second item (b) appears below and indented:

I've tried using \newline and \linebreak, and even skipping a line before typing \begin{enumerate} after the \begin{exercise} command but those don't work. Is there a way to push (a) down below Exercise 3.32. so that it's lined up with (b)? Thank you.

As requested, my preamble is:

\documentclass[11pt]{article}
\input{other/packages.tex}
\input{other/theoremdef.tex}

where packages.tex is

\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amscd}
\usepackage{url}
\usepackage[top=1.3in, bottom=1.3in, left=1.3in, right=1.3in]{geometry}

% header and footer
\pagestyle{headings}

\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{stackrel}
\usepackage{mathrsfs}
\usepackage{verbatim}
\usepackage{enumerate}
\usepackage{hyperref}
\usepackage{xy}
\input xy
\xyoption{all}

And theoremdef.tex is:

\newtheorem{exercise}{\bf Exercise}
\newcommand{\dom}{\text{dom}\ }
\newcommand{\ran}{\text{ran}\ }
\newcommand{\fld}{\text{fld}\ }
\newcommand{\op}[2]{\langle #1,#2\rangle}
\newcommand{\ot}[3]{\langle #1,#2,#3\rangle}
\newcommand{\ms}[1]{\mathscr{#1}}

Answer 1

Insert

\leavevmode

after the \begin{exercise}. That puts you into horizontal mode, which is where you want to be when you encounter the \begin{enumerate}.

Answer 2

I believe this is what you are looking for:

\begin{exercise} 
\begin{enumerate} 
\item[]
\item Show that $R$ is symmetric iff $R^{-1}\subseteq R$.
\item Show that $R$ is transitive iff $R\circ R\subseteq R$.
\end{enumerate}

Answer 3

See theorem or ntheorem package. Here is an example:

\documentclass{article}
\usepackage{theorem} % or ntheorem
\theoremstyle{break}
\newtheorem{exercise}{Exercise}
\begin{document}
\begin{exercise}
\begin{enumerate}
  \item foo
  \item bar
\end{enumerate}
\end{exercise}
\end{document}

It may be a little more complex to do that using amsthm. In amsthm we can define:

\usepackage{amsthm}
\newtheoremstyle{break}{3pt}{3pt}{\itshape}{}{\bfseries}{.}{\newline}{}
\theoremstyle{break}
\newtheorem{exercise}{Exercise}

It works fine without a enumerate environment at the beginning, but will fail for your code. This is a solution:

\begin{exercise}\mbox{}
\begin{enumerate}
  \item foo
  \item bar
\end{enumerate}
\end{exercise}

It seems no elegant solution. Therefore I prefer ntheorem to modify theorem styles.

Answer 4

I've extracted the following from thmtools' handling of the listhack key:

\def\thmt@amsthmlistbreakhack{%
  \leavevmode
  \vspace{-\baselineskip}%
  \par
  \everypar{\setbox\z@\lastbox\everypar{}}%
}

I find that without the \vspace, the vertical space before the list is excessive. I don't remember right now why I put the \everypar stanza there, in particular, why I didn't use \noindent. I'll check that again.

Answer 5

I (sort of accidentally) discovered that simply using \item immediately after \begin{theorem} can achieve the desired effect. For example, try

\begin{theorem}[Name of Theorem]
\item This is a theorem with multiple parts.
    \begin{enumerate}
    \item stuff here.
    \item and here.
    \item and here, as well.
    \end{enumerate}
\end{theorem}

I haven't yet found an instance that this fails for. Hope this helps.

Answer 6

As Leo Liu suggested, I do this:

\begin{theorem}
  \mbox{}
  \begin{enumerate}
  \item foo
  \item bar
  \end{enumerate}
\end{theorem}

Answer 7

You can use \newline as a quick fix. As newline needs a line to break, you can give an empty one by ~.

\begin{exercise} 
~\newline
\begin{enumerate} 
\item Show that $R$ is symmetric iff $R^{-1}\subseteq R$.
\item Show that $R$ is transitive iff $R\circ R\subseteq R$.
\end{enumerate}
\begin{proof}
\end{proof}
\end{exercise}