\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amssymb,cases}
\usepackage[margin=2cm]{geometry}
\usepackage{mathrsfs}
\begin{document}
\noindent
$\displaystyle \mathcal{M}'= \bigcup_{ \mathcal{F} \subset \mathcal{E} ,\, \mathcal{F} \text{is countable}}\mathcal{F}(\mathcal{F}).$
\end{document}
答案1
使用\substack{}
\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amssymb,cases}
\usepackage[margin=2cm]{geometry}
\usepackage{mathrsfs}
\begin{document}
\noindent
$\displaystyle \mathcal{M}'= \bigcup_{\substack{\mathcal{F} \subset \mathcal{E} \\ \mathcal{F} \text{ is countable}}}\mathcal{F}(\mathcal{F}).$
\end{document}
答案2
\smashoperator
使用from mathtools
(无需加载)进行间距改进amsmath
:
\documentclass[a4paper,12pt]{article}
\usepackage{mathtools,amssymb,cases}
\usepackage[margin=2cm]{geometry}
\usepackage{mathrsfs}
\begin{document}
\noindent
$\displaystyle \mathcal{M}'=\smashoperator{ \bigcup_{\substack{\mathcal{F} \subset \mathcal{E} \\ \mathcal{F} \text{ countable}}}}\mathcal{M}(\mathcal{F}).$
\end{document}