The following code  \documentclass{scrbook} \usepackage{amsthm} \newtheorem{theo}{Theorem}  \begin{document} \begin{theo}     Rhabarber Rhabarber Rhabarber Rhabarber     \begin{enumerate}         \item Rhabarber Rhabarber         \item Rhabarber Rhabarber     \end{enumerate} \end{theo}  \begin{proof}     \begin{enumerate}         \item Rhabarber Rhabarber         \item Rhabarber Rhabarber     \end{enumerate} \end{proof} \end{document}   produces    The distance between "Proof." and "1. Rharbarber..." (marked with a red rectangle) is too great for my liking. How can I shorten it?
Consider the following example:  \documentclass{article} \newtheorem{Theorem}{Theorem} \begin{document} \begin{Theorem} On the other hand, this one has only one claim. \end{Theorem} \begin{Theorem} \begin{enumerate} \item First claim \item Second claim \end{enumerate} \end{Theorem} \end{document}   Typesetting it results in:    Notice there is extra horizontal space before the first item in the list of the second theorem.     Is there a sensible way to get rid of that?   Ideally, the solution would work with amsthm, thmtools, enumitem.  N.B.: this is related to  but not quite the same.
I have latitude and longitude as 19.0649070739746 and 73.1308670043945 respectively.   In this case both coordinates are 13 decimal places long, but sometimes I also get coordinates which are 6 decimal places long.   Do fewer decimal points affect accuracy, and what does every digit after the decimal place signify? I am looking for some (may be rarely know to nonnative English speakers) synonyms for  "immensely valuable" or "extremely valuable". Below is an attempt at a proof of the following problem. Any feedback would be greatly appreciated. Thx!  Let $E$ be a set and $O_n = \{x: d(x, E) &lt; \frac{1}{n}\}$. Show  If $E$ is compact then $m(E) = lim_{n \rightarrow \infty} m(O_n)$.  This is false for $E$ closed and unbounded or $E$ open and bounded.   Note that all sets are real and measurable refers to Lebesgue measurable.  Suppose $E$ is compact. Then $m(E) &lt; \infty$. Also, since each $O_n$ is an open set of $\mathbb{R}^d$, $m(O_n) &lt; \infty$. If either of the following is true, we have that $lim_{n \rightarrow \infty}m(O_n)=m(E)$: $O_n \nearrow E$ $O_n \searrow E$ Now  $\bigcap_{n=1}^{\infty} O_n = \{ x:d(x,E) = 0\}$. That is, $x \in \bigcap_{n=1}^{\infty}O_n \iff \exists n&gt; \frac{1}{\epsilon},\ n \in \mathbb{N} \iff x\in \{x \mid d(x,E) =0\}$. Then we have that $\bigcap_{n=1}^{\infty}O_n = E$ and $O_n \searrow E$. Thus $ \ m(E) = lim_{n \rightarrow \infty} m(O_n)$. Suppose $E$ is closed and unbounded. Let $E = \{(x,y): y = 2 \}$ and $O_n = \{(x,y): d(x,E) &lt; \frac{1}{n}\}$. Since $E$ is a line in $\mathbb{R}^2$, $m(E) = 0$. Also, since the measure of a rectangle in $\mathbb{R}^d$ is its volume, $m(O_n) = \left| O_n \right| = \infty$, since the rectangle $O_n$ is unbounded. It is therefore apparent that $m(E) \not = lim_{n \rightarrow \infty} m(O_n)$. Suppose $E$ is open and bounded. Let $E=(0,1)$ and $O_n = \{x \mid d(x,E) &lt; \frac{1}{n}\}$. $\mathbb{R}-E$ is closed and $O_n \in \mathbb{R}-E$. Since $\mathbb{R}-E$ contains all its limit points $lim_{n \rightarrow \infty}O_n = O \in \mathbb{R}-E$. Thus $lim_{n \rightarrow \infty} m(O_n) \neq m(E)$. How to disable system beep for non-privileged user I think my MacBook Pro battery was replaced by Apple's service team. How do I check that? Creating a query that uses criteria from two tables? How can I diagnose and repair missing drive space? Possible Duplicate:          The FAQ specifies that:     Use your downvotes whenever you encounter an egregiously sloppy, no-effort-expended post, or an answer that is clearly and perhaps dangerously incorrect.   But if I find a correct question, closed (or not yet) as duplicate, is downvote a correct behavior for Stack&nbsp;Overflow policy? Disable two column mode for separate part I am writing a document using the report class in two column mode, but I need a \section to spread across both columns, but how can I set that?  Something like  \documentclass[10pt,danish,a4paper,twoside,twocolumn]{report} \input{preamble.tex} \begin{document}  % This section should spread over both columns \section{Name}  \lipsum[1-3] \end{document} I have shown that if two $n\times n$ matrices $A$ and $B$ are invertible, then $AB$ is also invertible. I am having problems though with showing that $(AB)^{-1}=B^{-1} A^{-1}$. I thought of using their left-multiplication transformations but it seems that this is only associative and not commutative so I can reverse $L_A L_B$ Use a specific forwarded key from SSH-agent? Let's say I have a key for Github, along with other keys. I've added lots of keys to my ssh agent (ssh-add -L returns lots of lines) at my home computer A. In my .ssh/config I have set up which key to use with which host, so e.g.  ssh -T -vvv git@github.com 2&gt;&amp;1 | grep Offering   gives  debug1: Offering RSA public key: /Users/doxna/.ssh/id_rsa.github   Only one key is offered, as expected. But then ssh-ing to some host B with ForwardAgent yes and repeating the same command, I get  debug1: Offering RSA public key: /Users/doxna/.ssh/id_rsa.linode2 debug1: Offering RSA public key: /Users/doxna/.ssh/id_rsa.helium debug1: Offering RSA public key: /Users/doxna/.ssh/id_rsa.github   meaning it tries all my keys. This is problematic since only a limited number of keys can be tried before servers return Too many authentication failures. So I tried editing .ssh/config on host B to include  Host github.com   IdentityFile /Users/doxna/.ssh/id_rsa.github   IdentitiesOnly yes   but then I get no key offerings, but rather  debug2: key: /Users/doxna/.ssh/id_rsa.github ((nil))   which I guess means that the key was not found(?) And after all, the key is located at my home computer A, not host B, so the question is how to refer to it at host B? Hope I managed to explain the question. In my Master thesis, I would like to have the title page without any page number and an acknowledgments section and the table of contents with roman numbers. The rest of the document should be numbered arabic.  I have tried this, to have roman numbers for the Table of contents and normal numbers for the rest.  \documentclass[12pt,ngerman,openany]{report} \usepackage{titlesec} \usepackage{titletoc} \usepackage{cite} \usepackage[a4paper]{geometry} %\usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage[ngerman]{babel} %\usepackage{ngerman}{babel} \usepackage[T1]{fontenc} \usepackage{graphicx} \usepackage{float} \selectlanguage{ngerman} \usepackage{amssymb} \usepackage{amsmath} \usepackage{url} \usepackage{epstopdf} \usepackage{scrhack} \usepackage{listings} \usepackage{caption} \titleformat{\chapter}{\rmfamily\huge}{\textbf\thechapter.}{25pt}{\huge\textbf} \defcaptionname{ngerman}{\lstlistingname}{Code} \defcaptionname{ngerman}{\lstlistlistingname }{Codeverzeichnis}  \lstloadlanguages{R}  \lstset{% general command to set parameter(s)    basicstyle=\sffamily\footnotesize, % print whole listing small    keywordstyle=\sffamily\footnotesize\bfseries, % ubold black keywords    identifierstyle=, % nothing happens    commentstyle=\sffamily\footnotesize\slshape, % green comments    stringstyle=\sffamily\footnotesize, % typewriter type for strings    showstringspaces=false, % no special string spaces    numbers=left,    numberstyle=\sffamily\footnotesize,    stepnumber=1,    numbersep=10pt,    showspaces=false,    showtabs=false,    frame=lines,    morecomment=[l]{\%},    float=htbp,    numberbychapter=true  }   \begin{document} \let\cleardoublepage\clearpage   \pagenumbering{roman}  \tableofcontents  \clearpage  \vspace*{75mm} \chapter{Einleitung} \section{Motivation} bla bla bla \section{Ziel}  bla bla \chapter{Stand der Technik zur Mustererkennung} \section{Die Zeitreihen} bla bla \subsection{Beschreibung} bla bla \subsection{Erklärung}  bla bla  \subsection{Vorhersage} bla bla \subsection{Kontrolle} bla bla \section{Feature Extraktion} bla bla \subsection{Autokorrelation} \subsection{Was ist Autokorrelation?} bla bla   \subsection{Autokorrelation und Zeitreihen} bla bla  \subsubsection{Zeitreihe-Diagramm}  bla bla \subsubsection{Lagged Streudiagramm} bla bla  \subsubsection{Autocorrelation-Function (Korrelogramm)} bla bla \subsection{Spektralanalyse} \subsection{Was ist Spektral-Analyse} bla bla \end{document} Can I kill an indestructible creature with -X/-X and sufficient damage marked on it? Can I hit a 4/4 indestructible creature with a  (3 damage) and give it -1/-1 with  to kill it? (Does it matter which way around I cast those?)  I read  here and it says, damage is removed in cleanup step, so it should be on the creature still after both spells.     702.12. Indestructible   702.12a Indestructible is a static ability.   702.12b A permanent with indestructible can’t be destroyed. Such permanents aren’t destroyed by lethal damage, and they ignore the state-based action that checks for lethal damage (see rule 704.5g).   702.12c Multiple instances of indestructible on the same permanent are redundant. Matrix with labels nested in braces Proving Geometric Hahn-Banach Implies Analytic Hahn-Banach I've come across a proof online of the H.B theorem using the goemetric version.   There is a step in the proof which im not sure why is true and it is as follows:   Let $X$ be a linear space and $M \subset X \times \Bbb R$ be a maximal subspace.  Then $M = G(F)$ for $F :X \to \Bbb R$ linear.   Someone can show me why this is true?   Maybe I misunderstood the proof, its the last paragraph of Matrin's answer here -.   Thanks a lot for helping! How diodes protect H-bridge DC motor driver? I don't really understand how these diodes in this circuit and similar circuits (like driving a relay circuit) protects the controller circuit from the energy stored by the inductance of the coil. I really appreciate if someone could graphically explain it. ( I mean how the the diodes block the current and etc)  the second question about this circuit is the capacitor. what happens if it isn't there? Rejecting a PhD offer after having accepted it. Advice? Are PPAs safe to add to my system and what are some "red flags" to watch out for? How to randomly subset X% of selected points? What methods are available in ArcGIS 10.2 to randomly subset a selection of points.  For example, in the attached screenshot I am interested in keeping 20% of the selected points and deleting the rest. Inform moderators migrating a question if the migration would have been normally blocked