Use of well ordering theorem
Proving $(A\le B)\vee (B\le A)$ for sets $A$ and $B$
Set spawn point for a specific player I made a server and me and one of my friends want to do a prank on another friend, because he stole all my enchanted diamond armor.  The Prank  I want to set a player's spawn point in a pit of lava, how would I do this without affecting anyone else's spawn point? Using ArcMap 10.0, I am working on a map that shows many villages. Each village has a name attribute that I want to use for a label, but in some areas there are so many villages that not all can be labeled. I am using the Maplex label engine and have not checked the "Never Remove (allow overlap)" box.  I am quite happy that not all village names are displayed as this would be way too much. And it is ok that Maplex chooses which labels to display, as I have no way to know which village is more important than another one. This also means that I can't base the labeling on a field such as population or something else, even though that would be less random.  However, I would also like the village points that are not labeled to similarly not be displayed. I think it unlikely that someone would want to see the point without knowing the name of the village. If all points are shown, it is also difficult to tell which point a name label goes with in crowded areas. So instead of choosing just labels not to be shown, I also want the computer to chose features not to be shown.  To give you a better idea on what I mean:    Is there a way to make these features/points not draw if their label is not shown? Error with \newcommand using \multicolumn in table I want to define a new command \tableheadline that (1) formats the headline text in some way and (2) takes an optional argument for spanning multiple columns while centering the headline.  I tried the following implementation:  \documentclass{minimal}  \newcommand{\tableheadline}[2][1]{\multicolumn{#1}{c}{\textit{#2}}} % \newcommand{\tableheadline}[2][1]{\multicolumn{1}{c}{\textit{#2}}}  \begin{document}   \begin{tabular}[\textwidth]{cc}      \tableheadline[2]{C}     \tableheadline{A} &amp; \tableheadline{B} \\     0 &amp; 1 \\     2 &amp; 3 \\   \end{tabular} \end{document}   This, however, gives me an error:  ! Misplaced \omit. \multispan -&gt;\omit                    \@multispan l.8 \tableheadline[2]{C}   I also get the error when I do not even use the optional argument (see commented line in the MWE). Any ideas what I am doing wrong? So this is something I just found out about that I find odd:   Opportunity attacks—trigger off of movement that uses the creatures movement (action, move action, bonus action, reaction.)  Sentinel—Whenever a creature within 5 ft of you makes an attack at a creature other than you.   Booming blade—If the target willingly moves.   Mental Prison spell—if it moves, attacks through it, or reaches any part of body out of it.   So the “willingly” is different than other rulings for how effects related to movement works. However what exactly fits willing?   Cause fear—nope magic is having them move. Dissonant whispers—same thing. They have to immediately move.   So the creature has to want to move completely of its own accord.   Crown of Madness—Humanoid creature you choose must make a wisdom saving throw. On a fail it is charmed by you. (Next is flavor text so skipping.) While charmed in this way the target must make an attack on a creature other than itself that you mentally choose on each of its turns before moving.  It can act normally if you choose no creature or if none are within reach.   Crown of madness isn't making them move but doesnt specifically say willing.  And if the ruling for willing is movement not caused by magic what about magic causing situations where you take damage if you don't move. Would a creature shimmering with evocation energy be spared because he moved out of the magical blender that is cloud of daggers? If no, then what is the go-to rule for willing movement that I should be using?  In short, what movement counts as willing, and where can it be found? Is moving out of obvious danger caused by magic, willing? Is moving away from an ally so you don't hit them, willing?   If there isn't any actual ruling on willing, that's fine; I'll go back to running it with my interpretation, so that it has same wording as AOO. If there is an actual 5e term or explanation for willing (that isn't Sage Advice), please let me know. I just don't see why it was necessary to put a term in the spell that isn't used anywhere else. I understand it could have been overlooked since it was in a splatbook or has just been ignored since most people probably assume that there is "Forced " and "Willing" movement, with the willing being defined as: any movement from an action, move action, bonus action, or reaction. In same way AOO work. How can I check the available version of a package in the repositories? How can I check the version of the available package in the Ubuntu repositories without installing it? I'm currently the proud owner of Frost the horse. What I want to know is, what will happen to him if I ride another horse?  Obviously nothing will happen as long as I have him in sight, e.g. I'll be able to ride Frost to make him my horse again.  Scenarios  But what will happen if, before I get back on Frost, I fast travel somewhere? Will he return to where I originally stole him, will he disappear forever, or can I own two horses simultaneously?  Ownership?  If he goes back to where I originally stole him, will I have to steal him again, and therefore never be his true owner again?  Two Horses?  The reason I'm asking is because I want to get Shadowmere (the Dark Brotherhood horse) while keeping Frost. Is it possible to do this, or will riding Shadowmere make me lose Frost forever? Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$. Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.  So $(a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2$  and $(ac+bd)^2 = a^2c^2+2acbd+b^2d^2$  So the problem is reduced to proving that $a^2d^2+b^2c^2\ge2acbd$ but I am not sure how to show that Digging in deeper to Javascript Remoting and I'm a bit confused.  The  starts off by showing how to set up everything and uses code like the following:  function getRemoteAccount() {     var accountName = document.getElementById('acctSearch').value;      // This remoting call will use the page's timeout value     Visualforce.remoting.Manager.invokeAction(         '{!$RemoteAction.AccountRemoter.getAccount}',         accountName,          handleResult     ); }   However, I've also seen examples where Visualforce.remoting.Manager.invokeAction is never called.  They end up looking something like the following:  function getText() {   var text = "";   JSRemotingBasicsController.doGetText(     function(result, event) {       // do something here     }   ); }   I'm wondering what is the point to using Visualforce.remoting.Manager.invokeAction when you can just reference the RemoteAction directly like the second example?  Is there something I'm missing?  Is there a time where the second example doesn't work? Why is the change in Hamiltonian for an active infinitesimal canonical transformation defined the way it is? I'm trying to understand infinitesimal canonical transformations and conservation theorems (section 9.6 Goldstein ed3). My specific problem is with understanding eq 9.104,         $\partial H = H(B) - K(A^{'}) $, where $\partial H$ represents the change in the hamiltonian under an infinitesimal canonical transformation (from the active view point ofc). The issue only arises for generating functions that explicitly depend on time.  Here's what I've thought of so far. I understand what $K$ is from the passive viewpoint (or so I think), its the hamiltonian that would furnish Hamilton's equations for the transformed variables $Q=q+ \delta q$ and $P = p + \delta p$. In the specific case of the generating function being a time dependent hamiltonian, $K = H + \delta t \frac{\partial H}{\partial t} = H + \delta t\dot H$, which makes sense because once you transform the variables to what they'll be in time $\delta t$, you'll also need to add on a term that comes from the Hamiltonian being at a later time $t+ \delta t$.   In the book, this is explained as a transformation depending on time changes the hamiltonian, so $H(A) \rightarrow K(A^{'}) $. Which I sorta get. But why not for $H(B)$? And what about the case when the Hamiltonian is time dependent and it is the generating function? Is $H(B)$ at a later time then?   Note: $A$ represents the original variables, $B$ is the transformed variables in the active sense, so in the same phase space as $A$, and $A^{'}$ is the same point as A but in the transformed phase space (passive). tmux: shared session, one user in a pane, another in another pane, two different cursors I've decided to try tmux: have been reading the docs and googling around, trying to find a way to have two users sharing a session, each with a different cursor.  However, giving 777 permissions to the socket, or creating a group, chgrping the socket and adding both users to it, seems to let that same socket be used to share a session with only one cursor: both users can write, but always in the same cursor position.  Right now both users are in the same home server over ssh, and the idea is to be able to have:   A terminal in a, let's say, left pane, where I can type commands Another terminal in a right pane, where I can see another user typing commands  in his own terminal The same thing for the other user   What I'm doing at the moment is using two sessions (not shared) and a script -f and tail -f combination that kinda works for reading each other's key strokes, but I reckon there is probably some way of doing this using tmux sharing capabilities.  Is there a way to get this idea working with write support in each other's terminal?  What is the better way to do this? How to get a file's Media Type (MIME type)? Show that $8 \mid (a^2-b^2)$ for $a$ and $b$ both odd If $a,b \in \mathbb{Z}$ and odd, show $8 \mid (a^2-b^2)$.   Let $a=2k+1$ and $b=2j+1$. I tried to get $8\mid (a^2-b^2)$ in to some equivalent form involving congruences and I started with $$a^2\equiv b^2 \mod{8} \Rightarrow 4k^2+4k \equiv 4j^2+4j \mod{8}$$ $$\Rightarrow k^2+k-j^2-j=2m$$ for some $m \in \mathbb{Z}$ but I am not sure this is heading anywhere that I can tell.  Second attempt: Use Euler's Theorem and as $\gcd(a,8)=\gcd(b,8)=1$ and $\phi(8)=4$, $a^4 \equiv b^4 \equiv 1 \mod 8$ so $a^4-b^4\equiv 0 \mod{8}$.  I haven't gotten too much further are there any hints? If $G$ is cyclic, then $G/H$ is cyclic?   The proof I got goes like this: $G$ is cyclic, so $G=&lt;g&gt;$ for some $g\in G$. So any coset in $G/H$ would be of the form $Hg'=Hg^n$ for some $n$. So $Hg$ is an generator of $G/H$. Thus, $G/H$ is cyclic.  I might just be confusing myself, but we have only shown that $Hg'$ is in form of $(Hg)^n$. But what if we are missing some $n\in\mathbb{N}$? That is, there is no quotient group of the form $Hg^2$, for example.  To make myself a little bit clearer, I think what the above proof has done was showing that $\forall Hg'\in G/H, Hg'\in &lt;Hg&gt;$, thus $G/H \subset &lt;Hg&gt;$. I feel that this is not a complete proof. Creating all possible line segments between all points using QGIS I want to connect all points in a point feature with every possible line segment using QGIS. For example, how a pentagon has five points with ten line segments connecting them all to form a pentagram inside. Alternatively phrased: How do I make spokes when all the points are hubs? I don't care about the order, I just want every connection between two points as one line each. Is there a tool or script that can do this in QGIS? This topic () is the closest to what I want but the plugins are no longer supported. Whats the simplest way to edit and add files to "/var/www"? Having installed the web server is there a simple way to set a user able to use the graphic interface to copy files and directories to the local web server /var/www  I gave myself administrative privileges in Ubuntu but it still doesn't allow copies. Duplicate questions in feed, and questions incorrectly marked as unanswered I see duplicate questions in the feed (Why is the book still wanted if Aida has memorized it?) and the question is marked as "unanswered" even though it has an upvoted answer.  I would expect unique questions in the feed, as well as questions with answers to not be marked as unanswered.   App Version: 1.6.3.4 Device: iPhone 6 OS Version: Version 10.2 (Build 14C92) This is a question I've been asked several times by students and I tend to have a hard time phrasing it in terms they can understand. This is a natural question to ask and it is not usually well covered in textbooks, so I would like to know of various perspectives and explanations that I can use when teaching.  The question comes up naturally in what is usually students' second course in quantum physics / quantum mechanics. At that stage one is fairly comfortable with the concept of wavefunctions and with the Schrödinger equation, and has had some limited exposure to operators. One common case, for example, is to explain that some operators commute and that this means the corresponding observables are 'compatible' and that there exists a mutual eigenbasis; the commutation relation is usually expressed as $[A,B]=0$ but no more is said about that object.  This naturally leaves students wondering     what is, exactly, the physical significance of the object $[A,B]$ itself?   and this is not an easy question. I would like answers to address this directly, ideally at a variety of levels of abstraction and required background. Note also that I'm much more interested in the object $[A,B]$ itself than what the consequences and interpretations are when it is zero, as those are far easier and explored in much more depth in most resources.    One reason this is a hard question (and that commutators are such confusing objects for students) is that they serve a variety of purposes, with only thin connecting threads between them (at least as seen from the bottom-up perspective).   Commutation relations are usually expressed in the form $[A,B]=0$ even though, a priori, there appears to be little motivation for the introduction of such terminology. A lot of stock is placed behind the canonical commutation relation $[x,p]=i \hbar$, though it is not always clear what it means.  (In my view, the fundamental principle that this encodes is essentially de Broglie's relation $\lambda=h/p$; this is made rigorous by the  but that's quite a bit to expect a student to grasp at a first go.) From this there is a natural extension to the Heisenberg Uncertainty Principle, which in its general form includes a commutator (and an anticommutator, to make things worse). Canonically-conjugate pairs of observables are often introduced, and this is often aided by observations on commutators. (On the other hand, the energy-time and angle-angular momentum conjugacy relations cannot be expressed in terms of commutators, making things even fuzzier.) Commutators are used very frequently, for example, when studying the angular momentum algebra of quantum mechanics. It is clear they play a big role in encoding symmetries in quantum mechanics but it is hardly made clear how and why, and particularly why the combination $AB-BA$ should be important for symmetry considerations.  This becomes even more important in more rigorous treatments of quantum mechanics, where the specifics of the Hilbert space become less important and the algebra of observable operators takes centre stage. The commutator is the central operation of that algebra, but again it's not very clear why that combination should be special. An analogy is occasionally made to the Poisson brackets of hamiltonian mechanics, but this hardly helps - Poisson brackets are equally mysterious. This also ties the commutator in with time evolution, both on the classical side and via the Heisenberg equation of motion.   I can't think of any more at the moment but they are a huge number of opposing directions which can make everything very confusing, and there is rarely a uniting thread. So: what are commutators, exactly, and why are they so important? Nautilus context menu for current directory when there's lots of files / "Open in Terminal" Proof of A ⊆ C and B ⊆ C ⇔ A ∪ B ⊆ C I know that this is valid because by thinking about it, you can see it's true, but I'm just not sure how I would go about actually proving it mathematically. I've kind of tried proof by contradiction, since that's what I use most of the time, but it doesn't seem to work.    In my mind, A ∪ B ⊆ C kind of just implies that A ⊆ C and B ⊆ C. How can you describe someone who gives up too easily? A friend of mine said that he was a cynic. I had told him that he was giving up too easily. In this situation, the worst case outcome was rejection, which logically speaking, was equivalent to not trying in the first place.  I want to say, "No, you aren't cynical, you are X," or "No, you aren't a cynic, you are a(n) X."  The closest I can come up with is unmotivated or intimidated.  What word describes a person who gives up too easily?