# Align by center of symbol

I want to align my equation by the center of a symbol, not by the beginning. More precisely, I have the following:

\begin{align*}
A\subseteq B&\iff\forall a\in A:a\in B\\
A\subsetneq B&\iff\forall a\in A:a\in B\land\exists b\in B:b\notin A\\
A=B&\iff A\subseteq B\land B\subseteq A\\
A\cup B&=\{x|x\in A\lor x\in B\}\\
A\cap B&=\{x|x\in A\land x\in B\}\\
A\setminus B&=\{x|x\in A\land x\notin B\}\\
\mathcal P(A)&=\{B|B\subseteq A\}\\
\overline A&=\{x|x\notin A\}
\end{align*}


Which results in However, I want something like this: • It is easy to align-centre the = sign w.r.t. \iff, but at the cost of a wider spacing on either side. Jul 14, 2019 at 10:41
• I can't see with my eyes any difference between your two screenshots. However, there is one because David Carlisle answered you. Can you point out this subtle difference that escapes my eye? Jul 14, 2019 at 10:47
• Use \mid for the bar when specifying the bar in the set builder notation. Jul 14, 2019 at 11:05

I'm not sure this is the best way to lay out those definitions: too many symbols. Anyway, you can add half of the symbol to the left and half to the right of the alignment point.

\documentclass{article}
\usepackage{amsmath,amssymb}

\newcommand{\crel}{%
\global\setbox1=\hbox{$#1$}%
\global\dimen1=0.5\wd1
\mathrel{\hbox to\dimen1{$#1$\hss}}&\mathrel{\mspace{-\thickmuskip}\hbox to\dimen1{}}%
}

\begin{document}

\begin{aligned} A\subseteq B \crel{\Longleftrightarrow} \forall a\in A:a\in B\\ A\subsetneq B \crel{\Longleftrightarrow} \forall a\in A:a\in B\land\exists b\in B:b\notin A\\ A=B \crel{\Longleftrightarrow} A\subseteq B\land B\subseteq A\\ A\cup B \crel{=} \{x \mid x\in A\lor x\in B\}\\ A\cap B \crel{=} \{x \mid x\in A\land x\in B\}\\ A\setminus B \crel{=} \{x \mid x\in A\land x\notin B\}\\ \mathcal{P}(A) \crel{=} \{B \mid B\subseteq A\}\\ \overline{A} \crel{=} \{x \mid x\notin A\} \end{aligned}

\end{document} A different layout:

\documentclass{article}
\usepackage{amsmath,amssymb}

\begin{document}

We present some definitions of symbols that are commonly used in set theory:
\begin{itemize}
\item $A\subseteq B$ means for every $a\in A$, also $a\in B$';
\item $A\subsetneq B$ means $A\subseteq B$, but $A\ne B$';
\item $A=B$ means $A\subseteq B$ and $B\subseteq A$' (extensionality of sets);
\item $A\cup B=\{x \mid x\in A\lor x\in B\}$;
\item $A\cap B=\{x \mid x\in A\land x\in B\}$;
\item $A\setminus B=\{x \mid x\in A\land x\notin B\}$;
\item $\mathcal{P}(A)=\{B \mid B\subseteq A\}$;
\item $\overline{A}=\{x \mid x\notin A\}$.
\end{itemize}
Note that the last notation only makes sense when elements are restricted to belong to a
universe' set that is usually clear from the context.

\end{document} • Thanks, your first solution worked fine. Might I note that it's supposed to be an as short as possible layout, as I'll use it as a basis for personal notes later. Nevertheless, it's of course important that the equals signs are aligned nicely. Jul 14, 2019 at 11:36 \documentclass{article}

\usepackage{amsmath,amssymb}

\sbox0{$\iff$}
\sbox2{$=$}
\newdimen\fudge
\setlength\fudge{0.5\dimexpr\wd0-\wd2\relax}

\begin{document}

\begin{align*}
A\subseteq B&\iff\forall a\in A:a\in B\\
A\subsetneq B&\iff\forall a\in A:a\in B\land\exists b\in B:b\notin A\\
A=B&\iff A\subseteq B\land B\subseteq A\\
A\cup B\kern-\fudge&\kern\fudge=\{x|x\in A\lor x\in B\}\\
A\cap B\kern-\fudge&\kern\fudge=\{x|x\in A\land x\in B\}\\
A\setminus B\kern-\fudge&\kern\fudge=\{x|x\in A\land x\notin B\}\\
\mathcal P(A)\kern-\fudge&\kern\fudge=\{B|B\subseteq A\}\\
\overline A\kern-\fudge&\kern\fudge=\{x|x\notin A\}
\end{align*}

\end{document}

• Quite ingenious! Jul 14, 2019 at 10:59

A tabular solution: (may be useful in some cases but I suggest @egreg's solution because there the math environment is not inline)

\documentclass{article}
\usepackage{amsmath,amssymb}
\begin{document}
\begingroup
\setlength{\tabcolsep}{2pt}
\renewcommand{\arraystretch}{1.4}
\begin{tabular}{rcl}
$A \subseteq B$&$\iff$ &$\forall a\in A:a\in B$\\
$A \subsetneq B$&$\iff$& $\forall a\in A:a\in B\land\exists b\in B:b \notin A$\\
$A=B$ &$\iff$ &$A\subseteq B\land B\subseteq A$\\
$A\cup B$ &$=$ & $\{x|x\in A\lor x\in B\}$\\
$A\cap B$& $=$ &$\{x|x\in A\land x\in B\}$\\
$A\setminus B$&$=$&$\{x|x\in A\land x\notin B\}$\\
$\mathcal P(A)$ &$=$ &$\{B|B\subseteq A\}$\\
$\overline A$&$=$&$\{x|x\notin A\}$
\end{tabular}
\endgroup
\end{document} Here's a solution that's similar to the one by @koleygr. The main difference is the use of an array environment instead of a tabular environment, permitting the omission of 48 [!] $ symbols inside the environment. The solution also takes care of the spacing around the \iff symbols, and it uses \mid rather than | to denote conditioning. \documentclass{article} \usepackage{amssymb,array} \newcolumntype{C}{>{{}}c<{{}}} % column type for relational and binary operators \begin{document} We present some definitions of symbols that are commonly used in set theory:\par \begingroup % localize scope of the next two instructions \setlength{\arraycolsep}{0pt} \renewcommand{\arraystretch}{1.4}$\begin{array}{rCl}
A \subseteq B &\iff& \forall a\in A:a\in B\\
A \subsetneq B&\iff& \forall a\in A:a\in B \land
\exists b\in B:b\notin A\\
A=B           &\iff& A\subseteq B\land B\subseteq A\\
A\cup B       &=& \{x\mid x\in A\lor x\in B\}\\
A\cap B       &=& \{x\mid x\in A\land x\in B\}\\
A\setminus B  &=& \{x\mid x\in A\land x\notin B\}\\
\mathcal{P}(A)&=& \{B\mid B\subseteq A\}\\
\overline A   &=& \{x\mid x\notin A\}
\end{array}\$
\endgroup
\end{document}