使用 \begin{array}[t] 的超长括号

Question

以下是我正在键入的讲稿的片段。

\[ \mathscr F=\left\{\{B_{\a}\}_{\a\in J}: \begin{array}[t]{rl}
            i)  &   B_{\a} \text{ are open}\\
            ii) &   \{B_{\a}\} \text{ is finitely inadequate}\\
            iii)&   I \subset J \text{ and } A_{\a}=B_{\a} \text{ whenever } \a\in I
        \end{array}\right\} \]

其中\a是 的宏\alpha。这将产生以下输出。

如何控制牙套的超长顶端？

Answer 1

以下是您当前设置的两种替代方案：

1. 将构造放入另一个构造内array。

2. 而是使用cases不同的布局。

\documentclass{article}

\usepackage{amsmath,mathrsfs}

\begin{document}

\[
  \mathscr F = \left\{
    \begin{array}{@{} l @{}}
      \{B_{\alpha}\}_{\alpha\in J}:
        \begin{array}[t]{r @{~} l}
          \textit{i})   &  B_{\alpha} \text{ are open}\\
          \textit{ii})  &  \{B_{\alpha}\} \text{ is finitely inadequate}\\
          \textit{iii}) &  I \subset J \text{ and } A_{\alpha}=B_{\alpha} \text{ whenever } \alpha\in I
        \end{array}
    \end{array}
  \right\}
\]

\[
  \mathscr F =
    \{B_{\alpha}\}_{\alpha\in J}:
    \begin{cases}
      \textit{i})   &  B_{\alpha} \text{ are open}\\
      \textit{ii})  &  \{B_{\alpha}\} \text{ is finitely inadequate}\\
      \textit{iii}) &  I \subset J \text{ and } A_{\alpha}=B_{\alpha} \text{ whenever } \alpha\in I
    \end{cases}
\]

\end{document}

Answer 2

我提出了另外两种变体，基于 Bmatrix，\parbox以及排队枚举环境：

\documentclass{article}

\usepackage{amsmath,mathrsfs}
\usepackage[inline]{enumitem}

\begin{document}

\[
  \mathscr F =
    \begin{Bmatrix}
      \{B_{\alpha}\}_{\alpha\in J}:\:
        \parbox{75mm}{%
        \begin{enumerate*}[label =$ \roman*) $, itemjoin ={; \enspace}]
          \item $ B_{\alpha} $ are open
          \item $ \{B_{\alpha}\} $ is finitely inadequate
          \item$ I \subset J $ and $ A_{\alpha}=B_{\alpha} $ whenever $ \alpha\in I $.
        \end{enumerate*}}
    \end{Bmatrix}
 \]

\[
  \mathscr F =
    \begin{Bmatrix}
      \{B_{\alpha}\}_{\alpha\in J}:\;
        \parbox[t]{75mm}{%
        \begin{enumerate*}[label =$ \roman*) $, itemjoin ={; \enspace}]
          \item $ B_{\alpha} $ are open
          \item $ \{B_{\alpha}\} $ is finitely inadequate
          \item$ I \subset J $ and $ A_{\alpha}=B_{\alpha} $ whenever $\alpha\in I$.
        \end{enumerate*}}
    \end{Bmatrix}
 \]

\end{document}