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I figured that that the exam class allows me to align the two parts of the lemma using the the \part command and also I could align the two lines of equation (3.11) using \begin{align*}, \end{align*} but there are a few problems this way. First, there's more gap than normally is between all lines and also I don't know how to get the equation number to appear on the left the way it is image.

I'd like it if the answers avoided the exam class and worked in the article class instead because it seems to me like that's messing things up for me.

Edit: Here's the code:

\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}
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3  
There is really nothing special about the piece of formatted text you're showing. The items in the lemma are enumerate items, the formatting in the equation could be done with a tabular, for instance. As this is surely not the only piece of text you need to typeset, I doubt you will benefit very much from someone giving a ready-to-use solution. It's better if you show how far you got and ask for specific advice on the things you struggle with. –  Stephan Lehmke Nov 22 '12 at 16:23
1  
@StephanLehmke Hi. I think I've already specified what I'm struggling with. I need to type exactly as in the image but I'm using the exam class (for reasons that I know a bit about it) and I'd rather work in the exam class but don't know how to align stuff there. Plus the spacing between each line's a bit off for me and I need to know how to label the equation to the left. I'll edit my post and write the code so you can see. –  Bakhtiar Kasi Nov 22 '12 at 16:34
    
I meant rather work in the article class –  Bakhtiar Kasi Nov 22 '12 at 16:41

1 Answer 1

up vote 3 down vote accepted

Here is code that will give you what you want. It's not clear to me what you're having difficulty achieving. So, I put various comments in my code to clarify what I'm doing. I'd recommend you look at the documentation for amsthm. It's a very useful package for getting theorem-like environments to look as you want them to look.

\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}
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