Как вы оформляете теоремы и доказательства к ним?
Код:
\newcommand{\onto}{\xrightarrow{\text{\textup{onto}}}}
\newcommand{\ind}{\mathop{\mathrm{ind}}\nolimits}
\newcommand{\Int}{\mathop{\mathrm{Int}}\nolimits}
\newcommand{\dgmap}{\mathop{\Delta}\nolimits}
\renewcommand{\sectionname}{\S}
\renewcommand{\labelenumi}{\theenumi)}
\newtheoremstyle{myrmk}{3pt}{3pt}{\rmfamily}{\parindent}{\itshape}{.}{.5em}{}
\newtheoremstyle{mydfn}{3pt}{3pt}{\rmfamily}{\parindent}{\bfseries}{.}{.5em}{}
\newtheoremstyle{mypln}{3pt}{3pt}{\itshape}{\parindent}{\bfseries}{.}{.5em}{}
\newenvironment{prf}[1][\proofname]{\par\indent\pushQED{\qed}\itshape#1. \normalfont\ignorespaces}{\popQED}
\swapnumbers
\theoremstyle{myrmk}
\newtheorem{itm}{}[section]
\newtheorem{rmk}[itm]{Remark}
\newtheorem{con}[itm]{Construction}
\theoremstyle{mydfn}
\newtheorem{dfn}[itm]{Definition}
\newtheorem{exm}[itm]{Example}
\theoremstyle{mypln}
\newtheorem{ass}[itm]{Assertion}
\newtheorem{thm}[itm]{Theorem}
\newtheorem{prp}[itm]{Proposition}
\newtheorem{cor}[itm]{Corollary}
\newtheorem{lem}[itm]{Lemma}
\renewcommand{\thesubsection}{\Alph{subsection}}
\makeatletter
\def\subsection{\@startsection{subsection}{2}%
\z@{.5\linespacing\@plus.7\linespacing}{.5em}%
{\normalfont\bfseries\centering}}
\def\l@subsection{\@tocline{2}{0pt}{1.45pc}{5pc}{}}
\makeatother
Код:
\begin{thm}\label{S2T9}
\emph{a)} If all mappings $^{\alpha}\pi\colon Y_{\alpha}\to Y$, $\alpha \in\mathfrak A$, are separable, then the mapping $^{\mathfrak A}\pi$ is separable too.
\emph{b)} If $\mathfrak B\subseteq\mathfrak A$ and the mapping $^{\mathfrak B}\pi$ is separable then the parallel mapping $^{\phantom{\mathfrak B\setminus}\mathfrak A}_{\mathfrak A\setminus\mathfrak B}\pi$ is separable too.
\end{thm}
\begin{prf}
a) Let $x_1,x_2\in Y_{\mathfrak A}$ be points such that $x_1\ne x_2$ but $^{\mathfrak A}\pi x_1=^{\mathfrak A}\pi x_2$. Then there exists $\alpha\in\mathfrak A$ such that $^{\mathfrak A}_{\alpha}\pi x_1\ne^{\mathfrak A}_{\alpha}\pi x_2$. Since the mapping $^{\alpha}\pi$ is separable, there exist disjoint neighborhoods $U\,^{\mathfrak A}_{\alpha}\pi x_1,U\,^{\mathfrak A}_{\alpha}\pi x_2\subseteq Y_{\alpha}$. Their preimages under the mapping $^{\mathfrak A}_{\alpha}\pi$ are disjoint neighborhoods of the points $x_1$ and $x_2$.
b) Let $x_1,x_2\in Y_{\mathfrak A}$ be distinct points such that $^{\phantom{\mathfrak B\setminus}\mathfrak A}_{\mathfrak A\setminus\mathfrak B}\pi x_1=^{\phantom{\mathfrak B\setminus}\mathfrak A}_{\mathfrak A\setminus\mathfrak B}\pi x_2$. By Proposition \ref{S2P3} points $y_1=^{\mathfrak A}_{\mathfrak B}\pi x_1$ and $y_2=^{\mathfrak A}_{\mathfrak B}\pi x_2$ are distinct but $^{\mathfrak B}\pi y_1=^{\mathfrak B}\pi y_2$. Since the mapping $^{\mathfrak B}\pi$ is separable, there are disjoint neighborhoods $Uy_1,Uy_2\subseteq Y_{\mathfrak B}$. Then the sets $^{\mathfrak A}_{\mathfrak B}\pi^{-1}Uy_1$ and $^{\mathfrak A}_{\mathfrak B}\pi^{-1}Uy_2$ are disjoint neighborhoods of the point $x_1$ and~$x_2$.
\end{prf}
В таком случае я получаю пустую строку между теоремой и доказательством, что очень некрасиво.
Интервал там действительно увеличенный, но чтобы целая пустая строка…
02.01.2009, 20:00 
