O código a seguir pode remover o ponto final de todos os ambientes de teoremas.
\makeatletter
\xpatchcmd{\@thm}{\thm@headpunct{.}}{\thm@headpunct{}}{}{}
\makeatother
VerRemova o ponto após o teorema com amsthm e hyperref
A questão aqui é um pouco diferente. Às vezes, só precisamos remover o ponto de certos ambientes de teoremas, como definições e o texto dentro do ambiente "sta" abaixo.
\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}
Responder1
Não sei por que você não gostaria do ponto final nas definições. Não vejo razão para quebrar a uniformidade.
De qualquer forma, a maneira correta é definir estilos de teoremas adequados.
\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}
Por favor, observe que isso {\it cycle}
está errado por dois motivos:
\it
está obsoleto há cerca de 30 anos;- o comando de nível superior
\emph
é a escolha para tais casos.
Responder2
\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}