``` \documentclass[b5paper,openany]{book} \usepackage[paperwidth=185mm,paperheight=260mm,text={148mm,220mm},left=21mm,top=10.5mm]{geometry} \usepackage{ctexcap,calc,extarrows,mathrsfs,xeCJK} \usepackage{graphicx,mathtools,xcolor,titletoc,mathdots,picinpar} \usepackage{siunitx} \CTEXsetup[number={\chinese{chapter}}]{chapter} \CTEXsetup[name={习题,}]{section} \setlength\parindent{0em} \usepackage{amsmath,amssymb,bm,nccmath,xfrac} %\usepackage{txfonts}%%此行改变了全文的数学字体 \usepackage[amsmath,thmmarks,thref,hyperref]{ntheorem} %%%%%%%%%%%%%%%%%%%%%自定义一个无序号的证明环境 \theoremstyle{nonumberplain}\theoremheaderfont{\heiti} \theorembodyfont{\normalfont} \theoremsymbol{$~~\square$ } \newtheorem{Proof}{证明：} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%定义思考题的排版命令，以节为排序单位 \newcounter{Buchongti} \renewcommand{\theBuchongti}{\chinese{Buchongti}} \newcommand{\Buchongti}{\vspace*{1ex}\par{\centering \Large\heiti 补充题\refstepcounter{Buchongti} {\bf\theBuchongti}\hspace{1em}\par\vspace*{1ex}}} \newcommand{\EXP}{\mathrm{e}} \newcommand{\dx}{\ensuremath{\mathrm{d}x}} \newcommand{\dt}{\ensuremath{\mathrm{d}t}} \newcommand{\Int}[2]{\ensuremath{\medmath\int_{\hspace*{-3.5pt}#1}^{#2}}} \newcommand{\arcsinh}{\mathrm{arcsinh}} \newcommand{\arccosh}{\mathrm{arccosh}} \newcommand{\arctanh}{\mathrm{arctanh}} \newcommand{\arccoth}{\mathrm{arccoth}} \newcommand{\rank}{\ensuremath{\mathrm{rank}}} \titlecontents{section}[24mm]{\heiti}{\contentslabel[\qquad\thecontentslabel\qquad]{25mm}}{} {\titlerule*[0.5pc]{$\cdot$}\contentspage[{\makebox[0pt][r]{\thecontentspage}}]} \titlecontents{chapter}[15mm]{\heiti}{\contentslabel[\thecontentslabel]{16mm}}{} {\titlerule*[0.5pc]{$\cdot$}\contentspage[{\makebox[0pt][r]{\thecontentspage}}]} \begin{document} 14.\begin{window}[1,l,{\includegraphics[width=22mm]{14.png}},{}] {\heiti 解：}由13题(1)结论知，$(W\bigcap V_1\big)+\big(W\bigcap V_2\big)\subseteq W\bigcap\big(V_1+V_2\big)=W$，并且的确真不相等的例子，例如在几何空间中，设$V_1,V_2,W$是过原点的三个平面，且它们相交于同一条直线L，由于$V_1+V_2=V$，因此，$W\subseteq V_1+V_2. $而$(W\bigcap V_1\big)+\big(W\bigcap V_2\big)=L$，而$W\nsupseteqq L. $如果$V_1\subseteq W$，那么结论就成立了.\\理由如下：任取$\alpha\in W$，由于$W\subseteq V_1+V_2$，因此$\alpha\in V_1+V_2$，从而有$\alpha=\alpha_1+\alpha_2,\alpha_1\in V_1, \alpha_2\in V_2.$由于$V_1\subseteq W$，因此$\alpha_1\in W$，从而$\alpha_2=\alpha+\alpha_1\in W$，于是$\alpha _2\in V_2\bigcap W. $因此得出$\quad \alpha=\alpha_1+\alpha_2\in \big(W\bigcap V_1\big)+\big(W\bigcap V_2\big)$，因此 $W\subseteq\big(W\bigcap V_1\big)+\big(W\bigcap V_2\big)$，又由于$\big(W\bigcap V_1\big)+\big(W\bigcap V_2\big)\subseteq W$，所以$W=\big(W\bigcap V_1\big)+\big(W\bigcap V_2\big). $ \end{window} \end{document} ```