Eliminar el punto después de un entorno de teorema específico

Question

El siguiente código puede eliminar el punto final de todos los entornos de teoremas.

\makeatletter
\xpatchcmd{\@thm}{\thm@headpunct{.}}{\thm@headpunct{}}{}{}
\makeatother

VerEliminar punto después del teorema con amsthm e hyperref

La cuestión aquí es un poco diferente. A veces solo necesitamos eliminar el punto de ciertos entornos de teoremas, como las definiciones y el texto dentro del entorno "sta" a continuación.

\documentclass{article}
\usepackage{amsthm}
\usepackage{xpatch}
\newtheorem{theorem}{Theorem}
\newtheorem{definition}{Definition}


\newtheorem{sta}{\normalfont}
\renewcommand{\thesta}{(\arabic{sta})\unskip}

\makeatletter
\xpatchcmd{\@thm}{\thm@headpunct{.}}{\thm@headpunct{}}{}{}
\makeatother
\begin{document}

\begin{definition}
A {\it cycle} in a graph is a non-empty trail in which only the first and last vertices are equal.
\end{definition}

\begin{theorem}
An undirected graph is bipartite if and only if it does not contain an odd cycle.
\end{theorem}

First, we note that:
\begin{sta}\label{seesall}
Every bipartite graph contains no odd cycles.
\end{sta}

This prove \ref{seesall}.
\end{document} 

Answer 1

No estoy seguro de por qué no querrías el punto en las definiciones. No veo ninguna razón para romper la uniformidad.

De todos modos, la forma correcta es definir estilos de teoremas adecuados.

\documentclass{article}
\usepackage{amsthm}

% see https://tex.stackexchange.com/a/17555/4427
\newtheoremstyle{definitionnoperiod}
  {\topsep}   % ABOVESPACE
  {\topsep}   % BELOWSPACE
  {\normalfont}  % BODYFONT
  {0pt}       % INDENT (empty value is the same as 0pt)
  {\bfseries} % HEADFONT
  {}          % HEADPUNCT
  {5pt plus 1pt minus 1pt} % HEADSPACE
  {}          % CUSTOM-HEAD-SPEC
\newtheoremstyle{empty}
  {\topsep}   % ABOVESPACE
  {\topsep}   % BELOWSPACE
  {\itshape}  % BODYFONT
  {0pt}       % INDENT (empty value is the same as 0pt)
  {\normalfont} % HEADFONT
  {}         % HEADPUNCT
  {5pt plus 1pt minus 1pt} % HEADSPACE
  {\thmnumber{#2}}          % CUSTOM-HEAD-SPEC

\newtheorem{theorem}{Theorem}

\theoremstyle{definitionnoperiod}
\newtheorem{definition}{Definition}

\theoremstyle{empty}
\newtheorem{sta}{}
\renewcommand{\thesta}{(\arabic{sta})}

\begin{document}

\begin{definition}
A \emph{cycle} in a graph is a non-empty trail in which only the first 
and last vertices are equal.
\end{definition}

\begin{theorem}
An undirected graph is bipartite if and only if it does not contain an odd cycle.
\end{theorem}

First, we note that:
\begin{sta}\label{seesall}
Every bipartite graph contains no odd cycles.
\end{sta}

This proves \ref{seesall}.
\end{document} 

Tenga en cuenta que esto {\it cycle}es incorrecto por dos razones:

1. \itha estado en desuso durante unos 30 años;
2. el comando de nivel superior \emphes la opción para tales casos.

Answer 2

\documentclass[12pt]{article}
\usepackage{mathtools,amsthm,amssymb}

\newtheorem{theorem}{Theorem}

\newtheoremstyle{definitionstyle}% name
  {0pt}% space above
  {0pt}% space below
  {}% body font
  {}% indent amount
  {\bfseries}% theorem head font
  {}% punctuation after theorem head
  {0.5em}% space after theorem head
  {}% theorem head spec

\theoremstyle{definitionstyle}
\newtheorem{definition}{Definition}


\begin{document}

\begin{definition}
A function $f: A \to \mathbb{R}$ is said to be \emph{continuous at a point} $c$ in its domain $A$ if, for every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x$ in $A$, if $|x - c| < \delta$, then $|f(x) - f(c)| < \epsilon$.
\end{definition}

\begin{theorem}
If $f: [a, b] \to \mathbb{R}$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one $c$ in $(a, b)$ such that
\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]
\end{theorem}

\end{document}