编辑

Question 1

我建议采用一种非全自动方法来调整外花括号的大小。在下面的代码中，外花括号的默认大小为\big，但可以通过指定可选的大小相关参数来更改，该参数可以采用值Bigg、bigg（Big当然还有big）。我赞成最小尺寸为big（而不是正常尺寸），因为在我看来，花括号只需要“看起来”更突出一点。

在下面的屏幕截图中，第一个示例中的花括号的大小为\big（默认）；\bigg在第二个示例中，它们的大小为，其中\displaystyle对第二行有效。

\documentclass{article}
\usepackage{mathtools} % for '\DeclarePairedDelimiter' macro
\DeclarePairedDelimiter{\abs}{\lvert}{\rvert} % "absolute value" macro

\newcommand\Setst[3][big]{\csname #1l\endcsname\{ #2 : 
  \begin{array}[t]{@{}l@{}}#3\csname #1r\endcsname \}\end{array}}

\begin{document}
\[
\Setst{f}{\text{$f(t)>0$ for all $t$ with $0<t<1$}\\
          \text{and }\abs[\big]{\int_0^1 e^{-\frac{1}{2}t^2}\!f(t)\,dt}\leq1}
\]

\[ % use \displaystyle directive in second row; hence, use "bigg" curly braces
\Setst[bigg]{f}{\text{$f(t)>0$ for all $t$ with $0<t<1$}\\
  \displaystyle \text{and }\abs[\bigg]{\int_0^1 e^{-\frac{1}{2}t^2}\!f(t)\,dt}\leq1}
\] 
\end{document}

Answer

我建议采用一种非全自动方法来调整外花括号的大小。在下面的代码中，外花括号的默认大小为\big，但可以通过指定可选的大小相关参数来更改，该参数可以采用值Bigg、bigg（Big当然还有big）。我赞成最小尺寸为big（而不是正常尺寸），因为在我看来，花括号只需要“看起来”更突出一点。

在下面的屏幕截图中，第一个示例中的花括号的大小为\big（默认）；\bigg在第二个示例中，它们的大小为，其中\displaystyle对第二行有效。

\documentclass{article}
\usepackage{mathtools} % for '\DeclarePairedDelimiter' macro
\DeclarePairedDelimiter{\abs}{\lvert}{\rvert} % "absolute value" macro

\newcommand\Setst[3][big]{\csname #1l\endcsname\{ #2 : 
  \begin{array}[t]{@{}l@{}}#3\csname #1r\endcsname \}\end{array}}

\begin{document}
\[
\Setst{f}{\text{$f(t)>0$ for all $t$ with $0<t<1$}\\
          \text{and }\abs[\big]{\int_0^1 e^{-\frac{1}{2}t^2}\!f(t)\,dt}\leq1}
\]

\[ % use \displaystyle directive in second row; hence, use "bigg" curly braces
\Setst[bigg]{f}{\text{$f(t)>0$ for all $t$ with $0<t<1$}\\
  \displaystyle \text{and }\abs[\bigg]{\int_0^1 e^{-\frac{1}{2}t^2}\!f(t)\,dt}\leq1}
\] 
\end{document}

Question 2

我将以略微不同的方式设置组件，以允许\left-\right用于跨越整个构造：

\documentclass{article}

\usepackage{amsmath,amssymb}

\newcommand\setst[2]{\{#1 : 
  \begin{array}[t]{@{}l@{}}#2 \}\end{array}}

% my attempted mod
\newcommand\Setst[2]{%
  \expandafter\setsttopandbottom#2\relax
  \begin{array}{@{}l @{\ } l@{}}
    \left\{#1 :\vphantom{\setsttop}\vphantom{\setstbottom}\right.\kern-\nulldelimiterspace & 
    \begin{array}[t]{@{}l@{}}
      \setsttop \\ \kern-\nulldelimiterspace\left.\setstbottom\vphantom{\setsttop}\right\}
    \end{array}
  \end{array}
  }

\def\setsttopandbottom#1\\#2\relax{\def\setsttop{#1}\def\setstbottom{#2}}

\begin{document}

OK:
\[
  \setst{f}{f(t) > 0 \text{ for all $t$ with } 0 < t < 1 \\
      \text{and }\left|\int_{0}^{1}e^{-\frac{1}{2} t^{2}}f(t) dt\right| \leq 1 }
\]

Modified:
\[
  \Setst{f}{f(t) > 0 \text{ for all $t$ with } 0 < t < 1\\
      \text{and }\left|\int_{0}^{1}e^{-\frac{1}{2} t^{2}}f(t) dt\right| \leq 1 }
\]

\end{document}

具体来说，第二个参数中的构造的顶部和底部被提取并用于第一行（用于开头\{）以及最后一行（用于结尾\}）的测量。

上述解决方案假设您将总是在构造中使用\\，因此有两行。对于单行应用程序，必须构建单独的宏。

使用

\usepackage{array}

\newcommand\Setst[2]{%
  \expandafter\setsttopandbottom#2\relax
  \begin{array}{@{}>{\displaystyle}l @{\ } l@{}}
    \left\{#1 :\vphantom{\setsttop}\vphantom{\setstbottom}\right.\kern-\nulldelimiterspace & 
    \begin{array}[t]{@{}>{\displaystyle}l@{}}
      \setsttop \\ \kern-\nulldelimiterspace\left.\setstbottom\vphantom{\setsttop}\right\}
    \end{array}
  \end{array}
  }

array您始终可以确保使用来设置构造内的单元格\displaystyle。

Answer

我将以略微不同的方式设置组件，以允许\left-\right用于跨越整个构造：

\documentclass{article}

\usepackage{amsmath,amssymb}

\newcommand\setst[2]{\{#1 : 
  \begin{array}[t]{@{}l@{}}#2 \}\end{array}}

% my attempted mod
\newcommand\Setst[2]{%
  \expandafter\setsttopandbottom#2\relax
  \begin{array}{@{}l @{\ } l@{}}
    \left\{#1 :\vphantom{\setsttop}\vphantom{\setstbottom}\right.\kern-\nulldelimiterspace & 
    \begin{array}[t]{@{}l@{}}
      \setsttop \\ \kern-\nulldelimiterspace\left.\setstbottom\vphantom{\setsttop}\right\}
    \end{array}
  \end{array}
  }

\def\setsttopandbottom#1\\#2\relax{\def\setsttop{#1}\def\setstbottom{#2}}

\begin{document}

OK:
\[
  \setst{f}{f(t) > 0 \text{ for all $t$ with } 0 < t < 1 \\
      \text{and }\left|\int_{0}^{1}e^{-\frac{1}{2} t^{2}}f(t) dt\right| \leq 1 }
\]

Modified:
\[
  \Setst{f}{f(t) > 0 \text{ for all $t$ with } 0 < t < 1\\
      \text{and }\left|\int_{0}^{1}e^{-\frac{1}{2} t^{2}}f(t) dt\right| \leq 1 }
\]

\end{document}

具体来说，第二个参数中的构造的顶部和底部被提取并用于第一行（用于开头\{）以及最后一行（用于结尾\}）的测量。

上述解决方案假设您将总是在构造中使用\\，因此有两行。对于单行应用程序，必须构建单独的宏。

使用

\usepackage{array}

\newcommand\Setst[2]{%
  \expandafter\setsttopandbottom#2\relax
  \begin{array}{@{}>{\displaystyle}l @{\ } l@{}}
    \left\{#1 :\vphantom{\setsttop}\vphantom{\setstbottom}\right.\kern-\nulldelimiterspace & 
    \begin{array}[t]{@{}>{\displaystyle}l@{}}
      \setsttop \\ \kern-\nulldelimiterspace\left.\setstbottom\vphantom{\setsttop}\right\}
    \end{array}
  \end{array}
  }

array您始终可以确保使用来设置构造内的单元格\displaystyle。

Question 3

这个答案只是一个草稿，旨在解释我发布时的想法我的评论。

% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly 
                                 % declare the paper format.

\usepackage[T1]{fontenc}         % Not always necessary, but recommended.
% End of standard header.  What follows pertains to the problem at hand.

\usepackage{amsmath}

\makeatletter

\newcommand*{\SetSt}[2]{%
    \setbox\z@ \hbox{%
        % Inside this box, the "\\" command is `turned off'. 
        % BEWARE: WE ASSUME THAT "&" IS **NOT** USED INSIDE #2!
        \let\\\relax
        $#2$%
    }%
    \dimen@   \ht\z@
    \dimen@ii \dp\z@
    % Now typesetting begins:
    \left\{\,#1%
    % select one of the two following alternatives:
    % {}:{}% colon as separator
    \,\middle|\,% vertical bar as separator
    \vrule \@height\dimen@ \@depth\dimen@ii \@width\z@
    \right.\kern-\nulldelimiterspace
    \begin{array}[t]{@{}l@{}}%
        #2%
        \left.\kern-\nulldelimiterspace
        \vrule \@height\dimen@ \@depth\dimen@ii \@width\z@
        \right\}%
    \end{array}%
}

\makeatother



\begin{document}

With tall symbols:
\[
    \SetSt{f}{f(t) > 0 \text{ for all $t$ with } 0 < t < 1\\
        \text{and }\left|\int_{0}^{1}e^{-\frac{1}{2} t^{2}}f(t) dt\right| \leq 1 }
\]

WIthout tall symbols:
\[ \SetSt{x}{x\in A \land x\in B\\x\in C \land x\in D\\x\in E \land x\in F} \]
As you can see, no restriction is imposed on the number of lines.  Even a 
single line is OK\@:
\[ \SetSt{x}{\int_{0}^{x} f(t)\,dt > 0} \]

Another example with two lines:
\[
    \SetSt{x}{\int_{0}^{x} f(t)\,dt > 0\\\int_{0}^{x} g(t)\,dt > 0}
\]

\end{document}

至少可以从两个方面加以完善：

该\\ 命令不应该简单地重新定义为\relax，它应该吞噬一个星号和一个可选参数（如果存在）；
该\SetSt命令应该考虑当前的数学风格；这可以通过常规应用轻松实现\mathpalette。

很抱歉，但是现在我太累了，无法进行这些改进……

编辑

我现在可以发布修改后的代码版本：

% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly 
                                 % declare the paper format.

\usepackage[T1]{fontenc}         % Not always necessary, but recommended.
% End of standard header.  What follows pertains to the problem at hand.

\usepackage{amsmath}
\usepackage{array}

\makeatletter

\newcommand*\SetSt[2]{\mathpalette{\@SetSt{#1}}{#2}}
\newcommand*\@SetSt[3]{%
    % #1 := 1st argument of "\SetSt"
    % #2 := style selector (e.g., "\displaystyle")
    % #3 := 2nd argument of "\SetSt"
    \setbox\z@ \hbox{%
        % Inside this box, the "\\" command is `turned off'. 
        % BEWARE: WE ASSUME THAT "&" IS **NOT** USED INSIDE #3!
        \let\\\@Murray@NoopCR
        $#2#3$%
    }%
    \dimen@   \ht\z@
    \dimen@ii \dp\z@
    % Now typesetting begins:
    \left\{\@nonscript@thinspace #2% style selector
    % The following group protects the value of "\dimen@" and "\dimen@ii":
    \begingroup #1\endgroup
    % select one of the two following alternatives:
    % {}:{}% colon as separator
    \@nonscript@thinspace\middle|\@nonscript@thinspace % vertical bar as separator
    \@Murray@mathstrut
    \right.\kern-\nulldelimiterspace
    % The group around the cells allows this macro to be nested;
    % it is important only for the last row, but we cannot avoid specifying it 
    % for every entry.
    \begin{array}[t]{@{}>{#2\begingroup}l<{\endgroup}@{}}%
        #3\endgroup % end the group to restore the values of "\dimen@" and 
                    % "dimen@ii"
        \left.\kern-\nulldelimiterspace
        \@Murray@mathstrut
        \right\}%
        \begingroup % to balance the "\endgroup" at the end of the entry
    \end{array}%
}
\newcommand*\@Murray@NoopCR{\@ifstar\x@Murray@NoopCR\x@Murray@NoopCR}
\newcommand*\x@Murray@NoopCR{\@ifnextchar[\y@Murray@NoopCR {}}
\@ifdefinable\y@Murray@NoopCR{\def\y@Murray@NoopCR[#1]{}}
\newcommand*\@nonscript@thinspace{\nonscript\mskip\thinmuskip}
\newcommand*\@Murray@mathstrut{%
    \vrule \@height\dimen@ \@depth\dimen@ii \@width\z@
}

\makeatother



\begin{document}

With tall symbols:
\[
    \SetSt{f}{f(t) > 0 \text{ for all $t$ with } 0 < t < 1\\[\medskipamount]
        \text{and }\left|\int_{0}^{1}e^{-\frac{1}{2} t^{2}}f(t) dt\right| \leq 1 }
\]

WIthout tall symbols:
\[ \SetSt{x}{x\in A \land x\in B\\x\in C \land x\in D\\x\in E \land x\in F} \]
As you can see, no restriction is imposed on the number of lines.  Even a 
single line is OK\@:
\[ \SetSt{x}{\int_{0}^{x} f(t)\,dt > 0} \]

Another example with two lines:
\[
    \SetSt{x}{\int_{0}^{x} f(t)\,dt > 0
    \\*[\bigskipamount]% the star serves no purpose, we just want to check that 
                       % it is handled correctly
    \int_{0}^{x} g(t)\,dt > 0}
\]

An example of nesting:
\[
    \SetSt{x}{x\in A\\x\in B\\x\in\SetSt{y}{y<-1\\y>1}}
\]
This last example clearly shows the limits of the algorithm.

\end{document}

它产生的输出如下：

Answer