我如何排版这样的内容（数学 - 下图）？

Question

我发现考试课允许我使用命令对齐引理的两个部分\part，我也可以使用对齐方程 (3.11) 的两行\begin{align*}，\end{align*}但这种方式存在一些问题。首先，所有行之间的间隙比正常情况下要大，而且我不知道如何让方程编号像图像一样显示在左侧。

如果答案避开考试类别而是放在文章类别中，那就太好了，因为在我看来，这对我来说很乱。

编辑：这是代码：

\documentclass[12pt, a4paper]{exam}
\usepackage{lipsum}
\usepackage{amsmath, amsthm}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{xcolor}
\usepackage{enumitem}
\usepackage{multicol}
\usepackage{lipsum}
\usepackage{xfrac}% http://ctan.org/pkg/xfrac
\usepackage{multicol}

\begin{document}
\renewcommand{\thepartno}{\roman{partno}}
\noindent
\textbf{Lemma 3.4.}
\begin{parts}
\part\textit{For every $\alpha$ there is a cardinal number greater than $\alpha$.}
\part \textit{If $X$ is a set of cardinals, then $\sup X$ is a cardinal} \\
\end{parts}
\noindent\textit{Proof.} (i)
For any set $X$, let
\begin{align*} 
        h(X) = & \text{the least} \ \alpha \ \text{such that there is no one-to-one} \\
               & \text{function of} \ \alpha \ \text{into} \ X 
\end{align*}
\end{document}

Answer 1

以下代码可以满足您的需求。我不清楚您在实现上遇到了什么困难。因此，我在代码中添加了各种注释，以阐明我在做什么。我建议您查看 的文档amsthm。这是一个非常有用的包，可以让定理类环境看起来像您希望的那样。

\documentclass[leqno]{article} % leqno sets equation numbers on the left hand side
\usepackage{amsmath,amssymb,amsthm}
\usepackage{enumitem}
%..%
\numberwithin{equation}{section} % gets equation numbers to show current section: (3.1)
\newcounter{cntlemma}
\numberwithin{cntlemma}{section} % gets your new theorem to number by section and theorem: Lemma 3.4.
\newtheorem{mylemma}[cntlemma]{Lemma} % ties in your new theorem environment with counter 'cntlemma'
\setcounter{section}{3}   % this is completely unnecessary
\setcounter{cntlemma}{3}  % this is completely unnecessary
\setcounter{equation}{10} % this is completely unnecessary 
\begin{document}

\begin{mylemma} ~
\begin{enumerate}[label={\normalfont(\roman*)}]
\item For every $\alpha$ there is a cardinal number greater than $\alpha$.\label{point-1}
\item If $X$ is a seet of cardinals, then $\sup{X}$ is a cardinal.
\end{enumerate}
\end{mylemma}
    For every $\alpha$, let $\alpha^+$ be the least cardinal number greater
    than $\alpha$, the \emph{caridinal successor}  of $\alpha$.\\[0.5\baselineskip]

\noindent\emph{Proof.} \ref{point-1} For any set $X$, let
    \begin{align}
        h(X) = \text{\parbox[t]{3in}{\raggedright{}the least $\alpha$ such that there is
        no one-to-one function of $\alpha$ into $X$.}}
    \end{align}
There is only a set of possible well-orderings of subsets of $X$.  Hence
there is only a set of ordinals for which a one-to-one function of $\alpha$
into $X$ exists.  Thus $h(X)$ exists.

\end{document}