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Sequences of elements like

$\{x_n\}_{n\in\mathbb{N}}$

are to awful to type. And they are very common, not only sequences of elements, but families of sets and some other sorts of mathematical structures.

I'd like to create a command in which I can type a letter, say x, and a set, say $\mathbb{N}$ such that the output be the sequence like

$\{x_n\}_{n\in\mathbb{N}}$
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4 Answers 4

up vote 5 down vote accepted

Although it’s preferred to see what you tried before, here’s an example how one can do it:

\documentclass{article}

\usepackage{amssymb}

\newcommand{\xinN}[2][N]{%
    \{#2_n\}_{n\in\mathbb{#1}}%
}

\begin{document}
\begin{equation}
    \xinN{x}
\end{equation}

\begin{equation}
    \xinN{y}
\end{equation}

\begin{equation}
    \xinN[C]{z}
\end{equation}

At inline text $\xinN{k}$.
\end{document}

The command has a mandatory argument which is what you called x, and an optional argument which defaults to N.

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4  
For the reasons I give in tex.stackexchange.com/a/34837/575, I want to object to the use of \ensuremath. Basically, particularly since the asker is clearly a beginner, I think it is best for them to maintain a strong mental separation between the idea of "math TeX" and "text TeX". –  Ryan Reich Feb 13 '12 at 16:27
1  
@RyanReich: Thank you for that comment. I removed the \ensuremath from my answer. –  Tobi Feb 14 '12 at 11:11
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I need to learn to type faster. :)

For the sake of variety, I'll add another answer. My approach is similar to Tobi's, but it uses xparse instead:

\documentclass{article}

\usepackage{amssymb}
\usepackage{xparse}

\DeclareDocumentCommand{\elementinset}{O{N}m}{\{#2_n\}_{n\in\mathbb{#1}}}

\begin{document}

$\elementinset{x}$

$\elementinset[Z]{x}$

\end{document}

Personally, I like the way it handles mandatory and optional arguments. :)

Edit: \ensuremath is gone too. :)

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if the set in the subscript is always blackboard bold, you can define a command like this:

\newcommand{\mycmd}[2]{\ensuremath
   \{x_#1\}_{#1\in\mathbb{#2}}}

and use it like

\mycmd{n}{N}

the \ensuremath means that you don't have to use $...$ when it stands alone in running text. the first argument will fill in two positions (indicated by #1) and the second will provide the letter naming the set.

and of course you can use any name you want to replace \mycmd; it's best if you give it a name that actually means something.

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Here is another solution using the xstring package so as to have a very easy way to write the sequences.

\documentclass{article}
    \usepackage{amssymb}
    \usepackage{xstring}

    \makeatletter
        \newcommand{\sequ}[1]{%
% No argument is one short cut for \sequ{u,n}.
            \ifx\relax#1\relax
                \def\@setFound{0}
                \def\@name{u}
                \def\@index{n}
            \else
% Looking for one set.
                \IfSubStr{#1}{;}{
                    \def\@setFound{1}
                    \StrBehind{#1}{;}[\@set]
                    \StrBefore{#1}{;}[\@withoutSet]
                }{
                    \def\@setFound{0}
                    \def\@withoutSet{#1}
                }
% Looking for one name and one possible index.
                \expandafter\IfSubStr\expandafter{\@withoutSet}{,}{
% We have to use a trick so as to use \@withoutSet like one string variable.
% This trick was given to me by the author of the package xstring.
                    \expandafter\StrBefore\expandafter{\@withoutSet}{,}[\@name]
                    \expandafter\StrBehind\expandafter{\@withoutSet}{,}[\@index]
                }{
                    \def\@name{\@withoutSet}
                    \def\@index{n}
                }
            \fi
% Display it !
            \ensuremath{%
                \ifnum\@setFound=1
                    \left( {\@name}_{\@index} \right)_{\@index \in \mathbb{\@set}} %
                \else
                    \left( {\@name}_{\@index} \right)%
                \fi
            }%
        }
    \makeatother


\begin{document}

\verb+$\sequ{x,k;Z}$+ : $\sequ{x,k;Z}$

\medskip

\verb+$\sequ{u;N}$+ : $\sequ{u;N}$

\medskip

\verb+$\sequ{x}$+ : $\sequ{x}$

\medskip

\verb+$\sequ{}$+ : $\sequ{}$

\end{document}
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