答案1
这是朝着正确的方向发展的吗?
\documentclass[fleqn]{article}
\usepackage{amsmath,amsfonts}
\usepackage{newtxtext,newtxmath}
\begin{document}
\[\tau=\left\{ G\subset\overline{\mathfrak{R}}/ \forall x\in G,
\left\{\begin{array}{@{}l@{\ /\ }l@{,\ }l@{}}
\exists a,b\in\mathfrak{R} & x\in [a,b\mathclose[\subset G&x\ne\pm\infty\\
\exists b\in\mathfrak{R} & x\in [-\infty,b[\subset G&x=-\infty\\
\exists a\in\mathfrak{R} & x\in \mathopen]a,\infty]\subset G&x=+\infty
\end{array}\right\}
\right\}
\]
\end{document}
(不过,我不确定我是否理解了这个符号。)编辑:按照@egreg 的建议,添加\mathopen
和。)\mathclose
答案2
还有另一种解决方案,有两种变体:
\documentclass{article}
\usepackage{array}
\usepackage{mathtools}
\DeclarePairedDelimiter{\set} \{ \}
\DeclareFontFamily{U}{mathx}{\hyphenchar\font45}
\DeclareFontShape{U}{mathx}{m}{n}{
<-6> mathx5 <6-7> mathx6 <7-8> mathx7
<8-9> mathx8 <9-10> mathx9
<10-12> mathx10 <12-> mathx12
}{}
\DeclareSymbolFont{mathx}{U}{mathx}{m}{n}
\DeclareFontSubstitution{U}{mathx}{m}{n}
\DeclareMathAccent{\widebar}{0}{mathx}{"73}
\begin{document}
\[
\tau = \set*{%
G \subset \widebar{\Re}\bigm / \forall X\in G, \begin{Bmatrix}
\begin{array}{@{}l!{/}l@{\:}l@{}}
\exists a, b \in\Re & x\in{]a, b[}\subset G, &x\ne \pm\infty \\
\exists b \in\Re & x\in{[-\infty, b[}\subset G, &x = -\infty \\
\exists ab \in\Re & x\in {]a, + \infty]} \subset G, &x = + \infty
\end{array}
\end{Bmatrix}}%
\]%
\[
\tau = \set*{%
G \subset \widebar{\Re}\bigm / \forall X\in G, \begin{cases}
\begin{array}{@{}l!{/}l@{\:}l@{}}
\exists a, b \in\Re & x\in{]a, b[}\subset G, &x\ne \pm\infty \\
\exists b \in\Re & x\in{[-\infty, b[}\subset G, &x = -\infty \\
\exists ab \in\Re & x\in {]a, + \infty]} \subset G, &x = + \infty
\end{array}
\end{cases}}%
\]%
\end{document}
答案3
或者也许是这样的?(请注意,我已经在间隔表达式中插入逗号作为分隔符。)
\documentclass{article}
\usepackage{amssymb} % for "\mathfrak" macro
\newcommand\R{\mathfrak{R}}
\begin{document}
\[
\tau=
\left\{
G\subset\overline{\R} \mid \forall x\in G,
\left\{
\begin{array}{@{}l@{}}
\exists a,b\in\R \mid
x\in\mathopen{]}a,b\mathclose{[} \subset G,\ x\ne\pm\infty \\
\exists b\in\R \mid
[-\infty,b\mathclose{[} \subset G,\ x=-\infty \\
\exists a\in\R \mid
\mathopen{]}a,+\infty] \subset G,\ x=+\infty
\end{array}
\right\}
\right\}
\]
\end{document}