New command with two definitions [duplicate]

Question

I want to define a new command that serves two slightly different purposes. Can I define a new command called \set such that it's definition is chosen according to the number of argument(s) is(are) filled? I want {1,2,3} when I use \set{1,2,3} and I want {x | x>0} when I use \set{x}{x>0}.

Answer 1

Yes, this is possible, using \NewDocumentCommand from xparse for example, making the second argument with {} behave like an optional argument then, i.e. using the g argument modifier.

Please note, that there is already a \set{} macro in the braket package, which provides for typesetting of sets, so I called the macro \myset instead and use \set{...} inside.

In principle, this could be done with an trailing optional argument with [] too, which is perhaps even better, because this enforces you to distinguish between the list version and the conditional version of the set notation.

Edit I've added the \mybetterset{}[] command as a variant to \myset, using the [] as 2nd optional argument.

\documentclass{article}

\usepackage{braket}
\usepackage{mathtools}
\usepackage{xparse}

\NewDocumentCommand{\myset}{mg}{%
  \IfValueTF{#2}{%
    \set{#1\;\vert\;#2}
  }{%
    \set{#1}%
  }%
}

%% The better command with [] as optional argument

   \NewDocumentCommand{\mybetterset}{mo}{%
  \IfValueTF{#2}{%
    \set{#1\;\vert\;#2}
  }{%
    \set{#1}%
  }%
}

\usepackage{hyperref}

\begin{document}
$\myset{1,2,3}$

$\myset{x}{x > 0}$

$\mybetterset{1,2,3}$

$\mybetterset{x}[x>0]$


\end{document}

Answer 2

You can do it, but I discourage you to. Better defining a command that distinguishes between \set{1,2,3} and \set{x|<condition>}.

You can find perhaps better methods in the documentation of mathtools.

\documentclass{article}

\usepackage{xparse}

\NewDocumentCommand{\set}{>{\SplitArgument{1}{|}}m}{\printset#1}

\NewDocumentCommand{\printset}{mm}{%
  \IfNoValueTF{#2}
   {% no |
    \{#1\}%
   }
   {% |
    \{\,#1\mid#2\,\}%
   }%
}

\begin{document}

$\set{1,2,3}=\set{x | 1\le x\le 3}$

\end{document}

Answer 3

Here is another soultion, with optional argument

\documentclass{article}
\newcommand{\set}[2][]{\left\lbrace\if\relax\detokenize{#1}\relax\else#1\mid\fi#2\right\rbrace}

\begin{document}
bla bla
\[\set{a,b,c},\set[x]{x>0}\]
$\set{a,b,c},\set[x]{x>0}$
\end{document}

Classical method

\documentclass{article}

\makeatletter
\newcommand{\set}{\@ifnextchar[{\@@set}{\@set}}
\def\@set#1{\left\lbrace#1\right\rbrace}
\def\@@set[#1]#2{\@set{#1\mid#2}}
\makeatother

\begin{document}
bla bla
\[\set{a,b,c},\set[x]{x>0}\]
$\set{a,b,c},\set[x]{x>0}$
\end{document}

Answer 4

A variant solution, also with mathtools and xparse. When used for a set defined by a property, the separator between elements and properties is a semicolon, because it is easy to type, and rarely used in maths (except for sets defined by a property). The resulting sign in the .pdf file will be a vertical bar, with a correct spacing. If for some reason you really need a semi-colon, you just enclose it between a pair of braces. Thus the syntax is very close to what one writes by hand: \set{x;P(x)}.

The size of the braces and the the vertical bar will adjust automatically to the size of the contents with the star version of the \set command, or manually for fine-tining, with optional arguments: \big, \Big, \bigg,\Bigg`.

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{fourier, erewhon}

\usepackage{xparse}
\usepackage{mathtools, nccmath}

\DeclarePairedDelimiterX{\set}[1]\{\}{\setargs{#1}}
\NewDocumentCommand{\setargs}{>{\SplitArgument{1}{;}}m}
{\setargsaux#1}
\NewDocumentCommand{\setargsaux}{mm}
{\IfNoValueTF{#2}{#1}{\nonscript\,#1\nonscript\;\delimsize\vert\nonscript\:\allowbreak #2\nonscript\,}}
%
% %% The following makes \big the default for the \set command
% \let\oldset\set
% \def\set{\futurelet\testchar\MaybeOptArgSet}
% \def\MaybeOptArgSet{\ifx[\testchar \let\next\OptArgSet
% \else \let\next\NoOptArgSet \fi \next}
% \def\OptArgSet[#1]#2{\oldset[#1]{#2}}
% \def\NoOptArgSet#1{\OptArgSet[\big]{#1}}
%
% \def\Set{\oldset*}%

%%% Syntax: \set{x ; P(x)})
\begin{document}

\begin{align*}
    & \set[\big]{x ; x > 5} & & \set*{x ; x > 5} \\[4pt]
    & \set[\bigg]{ \frac{x}{2} ; x > 5} & & \set*{ \frac{x}{2} ; x > 5} \\[4pt]
    & \set[\Bigg]{ \frac{1}{2},\frac{1}{3},\frac{1}{4},\dotsm} & & \set*{ \mfrac{1}{2},\mfrac{1}{3},\mfrac{1}{4},\dotsm}
\end{align*}

\end{document} 

In the above code I commented out some lines that result in the default implicit use of the \big optional argument (which looks better, in my opinion). The \set* command is the replaced with a new Set command.