\documentclass{article} \usepackage{amssymb} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Created=Monday, June 25, 2001 17:45:57} %TCIDATA{LastRevised=Saturday, June 01, 2002 20:37:16} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038
\QTR{small}{Matematick\U{e1} anal\U{fd}za I online - D\U{f4}kazy\dotfill \thepage }} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \author{A. U. Thor} \title{Lab Report} \date{The Date } \maketitle \begin{abstract} A Laboratory report created with Scientific Notebook \end{abstract} \section{Postupnosti a rady re\'{a}lnych \v{c}\'{\i}sel} \subsection{D\^{o}kaz vety} \textbf{D\^{o}kaz: }Nech $\left\{ a_{n}\right\} _{n=1}^{\infty }$ je ohrani% \v{c}en\'{a}. Potom existuje $M\in \mathbf{R}$, tak\'{e} \v{z}e $a_{n}\in \left\langle -M,M\right\rangle ,\,\forall n\in \mathbf{N}.$ 1) Vyberieme jeden bod $P\in \left( -M,M\right) $ a rozde\v{l}me interval $% \left\langle -M,M\right\rangle $ na dva uzavret\'{e} intervaly $\left\langle -M,P\right\rangle $ a $\left\langle P,M\right\rangle $. Vyberme ten z uzavret% \'{y}ch intervalov $\left\langle -M,P\right\rangle ,\left\langle P,M\right\rangle ,$ v ktorom sa nach\'{a}dza nekone\v{c}ne mnoho \v{c}lenov postupnosti $\left\{ a_{n}\right\} _{n=1}^{\infty }$ a z neho vyberieme prvok $a_{n_{1}}.$ 2) Zopakujme na intervale vybratom v prvom kroku ten ist\'{y} postup. Tak vyberieme prvok $a_{n_{2}}.$ 3) T\'{y}m ist\'{y}m postupom vyber\'{a}me \v{c}lenov vybranej postupnosti $% \left\{ a_{n_{k}}\right\} _{k=1}^{\infty }.$ Postupnos\v{t} prav\'{y}ch koncov\'{y}ch bodov vybrat\'{y}ch uzavret\'{y}ch intervalov ako aj postupnos% \v{t} ich \v{l}av\'{y}ch koncov\'{y}ch bodov bude pod\v{l}a princ\'{\i}pu do seba zapadaj\'{u}cich intervalov konvergova\v{t} k nejak\'{e}mu \v{c}\'{\i}% slu $L\in \left\langle -M,M\right\rangle ,$ potom aj $\lim_{k\longrightarrow \infty }a_{n_{k}}=L.\blacksquare $ \begin{center} \begin{tabular}{|c|} \hline \hyperref{{\small Sp\"{a}\v{t}}}{}{}{M101.tex#8} \\ \hline \end{tabular} \end{center} \end{document}