Theorem without counter (any TeX)

I often use theorems on LaTeX and have defined various kinds of them. At times, however, it is best not to define a specific theorem which would otherwise be a once-in-a-long-time usage, as for example with the "Fundamental Theorem of Algebra" or the "Hairy Ball Theorem" and such. A pdf I found on the Internet suggests the following code:

\makeatletter
\newtheorem{@thmattr}[thm]{\theorem@attr}
\newenvironment{thmattr}[1]
{\def\theorem@attr{#1}\begin{@thmattr}}
{\end{@thmattr}}
\makeatother


The only problem is that, besides needing a definition of a counter thm (which can easily be solved by removing the [thm]), this gives such theorems a counter. So I get "Fundamental Theorem of Algebra 1", which doesn't make sense since there is only one theorem with that name. So the question is: how do I make a theorem with no counter?

• \newtheorem* is your friend, if you use amsthm ;) – yo' Mar 23 '14 at 17:00
• \usepackage{amsmath} and \newtheorem*{HBT}{Hairy Ball Theorem}. Do you have several "named theorems"? – egreg Mar 23 '14 at 17:00
• @egreg, why? Is there a trick for such situation? – Sigur Mar 23 '14 at 17:05
• @Sigur Yes, of course. – egreg Mar 23 '14 at 17:05
• @egreg, please, reference? – Sigur Mar 23 '14 at 17:07

If you have a single named theorem, the easiest way is

\usepackage{amsthm}

\newtheorem*{HBT}{Hairy Ball Theorem}


so that

\begin{HBT}
There is no nonvanishing continuous tangent vector field on
even dimensional $n$-spheres.
\end{HBT}


will produce what you want.

If you have several named theorems, then a strategy similar to what you found will work:

\newtheorem*{namedthm*}{\thistheoremname}
\newcommand{\thistheoremname}{} % initialization
\newenvironment{namedthm}[1]
{\renewcommand{\thistheoremname}{#1}\begin{namedthm*}}
{\end{namedthm*}}


and the input will be

\begin{namedthm}{Hairy Ball Theorem}
There is no nonvanishing continuous tangent vector field on
even dimensional $n$-spheres.
\end{namedthm}


You can also give the attribution in the usual way:

\begin{namedthm}{Hairy Ball Theorem}[Brouwer]
There is no nonvanishing continuous tangent vector field on
even dimensional $n$-spheres.
\end{namedthm}


Complete example; choose your preferred strategy.

\documentclass{article}
\usepackage{amsthm}

\newtheorem*{HBT}{Hairy Ball Theorem}

\newtheorem*{namedthm*}{\thistheoremname}
\newcommand{\thistheoremname}{} % initialization
\newenvironment{namedthm}[1]
{\renewcommand{\thistheoremname}{#1}\begin{namedthm*}}
{\end{namedthm*}}

\begin{document}

\begin{HBT}
There is no nonvanishing continuous tangent vector field on
even dimensional $n$-spheres.
\end{HBT}

\begin{namedthm}{Hairy Ball Theorem}
There is no nonvanishing continuous tangent vector field on
even dimensional $n$-spheres.
\end{namedthm}

\begin{namedthm}{Hairy Ball Theorem}[Brouwer]
There is no nonvanishing continuous tangent vector field on
even dimensional $n$-spheres.
\end{namedthm}

\end{document}


• Ow, I see. The theorem's name as an option. Nice! Thanks. – Sigur Mar 23 '14 at 17:22

Using ntheorem, you have the emptyand emptybreak theorem styles. The name is an optional argument. Here are 4 possibilities (I had to patch the empty style because it didn't accept a label separator):

        \documentclass[12pt,a4paper]{article}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{MinionPro}
\usepackage{amsmath}
\usepackage[svgnames, x11names]{xcolor}
\usepackage{framed}
\usepackage[framed, amsmath, thmmarks]{ntheorem}%
\newcommand*\C{\mathbf C}

\makeatletter
\renewtheoremstyle{empty}%
{\item[]}%
\makeatother

\theorembodyfont{\itshape}

\theoremstyle{empty}
\theoremseparator{.\,—}
\newtheorem{namedthm}{}
\newframedtheorem{namedfrthm}{}
\theoremstyle{emptybreak}
\theorembodyfont{\upshape\color{DarkSeaGreen4}}
\theoremseparator{\smallskip}
\newtheorem{NamedThm}{}
\newframedtheorem{NamedfrThm}{}
%\newframedtheorem{namedfrthm}}
\begin{document}

\begin{namedthm}[Fundamental Theorem of Algebra]
Every polynomial with coefficients in  $\C$ has a root in  $\C$.  In other words,  the field of complex numbers is algebraically closed.
\end{namedthm}

\begin{namedfrthm}[Fundamental Theorem of Algebra]
Every polynomial with coefficients in  $\C$ has a root in  $\C$.  In other words,  the field of complex numbers is algebraically closed.
\end{namedfrthm}

\begin{NamedThm}[Fundamental Theorem of Algebra]
Every polynomial with coefficients in  $\C$ has a root in  $\C$.  In other words,  the field of complex numbers is algebraically closed.
\end{NamedThm}

\begin{NamedfrThm}[Fundamental Theorem of Algebra]
Every polynomial with coefficients in  $\C$ has a root in  $\C$.  In other words,  the field of complex numbers is algebraically closed.
\end{NamedfrThm}

\end{document}