Long label for a tikz picture (Having multiple line label for a tikz shape)
Manual/automatic line breaks and text alignment in TikZ nodes
So, I have proven already that every open set in $\mathbb{R}$ can be written as a countable union of disjoint open intervals; i.e., that $\mathcal{O} = \bigcup_{i=1}^\infty (a_{x_i}, b_{x_i})$.  Now, consider the closed interval $\left [ a + \frac{1}{n}, b - \frac{1}{n}\right ] \subset (a, b)$.  Since the midpoint of $(a,b)$ is $\frac{b-a}{2}$, we need $\displaystyle n &gt; \frac{2}{b-a}$.  Otherwise, $\displaystyle a + \frac{b-a}{2} = a + \frac{1}{ \frac{2}{b-a}} = b - \frac{1}{ \frac{2}{b-a}} = b - \frac{b-a}{2}$, and our interval will be degenerate.  Note that $\forall (a,b) \in \mathcal{O}$,   $\displaystyle \bigcup_{n = \frac{b-a}{2}+1}^{\infty} \left[ a + \frac{1}{n}, b - \frac{1}{n} \right] = (a,b)$.  So, we have $\displaystyle \mathcal{O} = \bigcup_{i=1}^\infty \left ( \bigcup_{n = \frac{b-a}{2} + 1}^\infty \left[ a_{x_i}+\frac{1}{n}, b_{x_i}-\frac{1}{n} \right] \right)$.  But, I don't like this. It looks like I'm still just taking a union of open sets.  Please help me get this to look like the union of closed sets.  (For unbounded intervals, I know what to do - it should follow from this relatively easily). When playing as a spellcaster, is it valid the use of the Ready action to prepare Dispel Magic to counter a spell cast by an opponent? In the same vein, when declaring said action, do I have to specify which spell slot I will use when Readying the Dispel Magic spell? If that is the case, does this tactic nullify an opponent's spell? I get the impression that this is basically turning Dispel Magic into a spell with a casting time of &quot;1 reaction&quot;. Is my reading of the rules accurate? Bootstrap dropdown sub menu missing When is unbalanced data really a problem in Machine Learning? braces on the sides of an easybmat matrix I created the following matrix with , and I would like to add at the bottom (respectively right) something like an underbrace (resp. "rightbrace") under each of the two blocks, together with a number specifying the number of columns (resp. rows) of the block. This amounts to two braces on the bottom and two on the right, and each brace should have a number below (resp. at the right) of it.  I did not find anything about that on the easybmat manual. I think I see how to do the braces with the  package, but then I am not able to create the block structure of the matrix in the first place.  \documentclass{article} \usepackage{easybmat} \begin{document} \[  M(z)= \left[ \begin{BMAT}[5pt]{c|c}{c} \begin{BMAT}[5pt]{c:c:c}{ccccccc} &amp; &amp; \\ &amp; &amp; \\ &amp; &amp; \\ v_1 &amp; \dots &amp; v_m \\ &amp; &amp; \\ &amp; &amp; \\ &amp; &amp; \end{BMAT} &amp; \begin{BMAT}{c}{c|c} \begin{BMAT}[5pt]{c:c:c}{ccc} &amp; &amp; \\ w_1(a) &amp; \dots &amp; w_n(a) \\ &amp; &amp; \end{BMAT} \\ \begin{BMAT}[10pt]{c}{c} B(z) \end{BMAT} \end{BMAT} \end{BMAT}  \right] \] \end{document}     This is more or less what i wish: If you are root, and you issue   rm -rf /    Then how far can the command go? Can you recover data from this kind of an action? Even after the binaries are gone, would the running processes still be active? What would it take to make the same physical machine boot again? What files would you need to restore to make this happen?  I could try this on a VM and see, but I want to know the rationale behind what to expect if I do this. Installed 14.04 on a non-PAE (Pentium-M), but now I can't update ("This kernel does not support a non-PAE CPU.") Here is what I'm looking at. This isn't my computer. I repair PC's and I had an end user with this low powered laptop running XP so I decided to let him try out lubuntu. The install went smoothly with the non-PAE boot switch, but now I can't update anything. I'd rather not have to format and start over since he's spent time configuring his system. Any tips would be appreciated. I tried sudo apt-get clean, sudo apt-get install -f, sudo apt-get update. No go.  Also his touchpad stopped working, the wireless stopped working, the keyboard works in firefox but not chrome, and in firefox the address bar sometimes turns black and gray for some reason. I was hoping the update would help address some of this stuff.  Here's the message I get trying to update:  Get:1 http://us.archive.ubuntu.com/ubuntu/ trusty-proposed/main linux-image-3.13.0-26-generic i386 3.13.0-26.48 [14.6 MB] Get:2 http://us.archive.ubuntu.com/ubuntu/ trusty-updates/main linux-image-3.13.0-24-generic i386 3.13.0-24.47 [14.6 MB] Fetched 29.2 MB in 2min 19s (208 kB/s)                                          (Reading database ... 169286 files and directories currently installed.) Preparing to unpack .../linux-image-3.13.0-26-generic_3.13.0-26.48_i386.deb ... This kernel does not support a non-PAE CPU. dpkg: error processing archive /var/cache/apt/archives/linux-image-3.13.0-26-generic_3.13.0-26.48_i386.deb (--unpack):  subprocess new pre-installation script returned error exit status 1 Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.13.0-26-generic /boot/vmlinuz-3.13.0-26-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.13.0-26-generic /boot/vmlinuz-3.13.0-26-generic Preparing to unpack .../linux-image-3.13.0-24-generic_3.13.0-24.47_i386.deb ... This kernel does not support a non-PAE CPU. dpkg: error processing archive /var/cache/apt/archives/linux-image-3.13.0-24-generic_3.13.0-24.47_i386.deb (--unpack):  subprocess new pre-installation script returned error exit status 1 Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.13.0-24-generic /boot/vmlinuz-3.13.0-24-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.13.0-24-generic /boot/vmlinuz-3.13.0-24-generic Errors were encountered while processing:  /var/cache/apt/archives/linux-image-3.13.0-26-generic_3.13.0-26.48_i386.deb  /var/cache/apt/archives/linux-image-3.13.0-24-generic_3.13.0-24.47_i386.deb E: Sub-process /usr/bin/dpkg returned an error code (1) Code   find / -name netcdf   Output  find: `/root/.dbus': Permission denied find: `/root/.gconf': Permission denied find: `/root/.gconfd': Permission denied find: `/root/.gnome': Permission denied find: `/root/.gnome2': Permission denied find: `/root/.gnome2_private': Permission denied How to convert URL parameters to a JavaScript object? Find all functions $f\colon \mathbb R\rightarrow \mathbb R$ such that for all reals $x$ and $y$:   $$f(f(x)f(y))+f(x+y)=f(xy).$$   It was six hours ago in IMO 2017 (problem 2).  I tried  the standard way: $x=0$, $x=y$, $x=1$,etc. but without any success.   Let $f(0)=a$. Hence, for all $x\in\mathbb R$ we have $f(a(f(x))+f(x)=a$. What is the rest?   Also we have $(x-1)(y-1)-1+x+y-1=xy-1$ and  $1-(1-x)(1-y)+1-x-y=1-xy$. Should "move-to-chat" be provided on demand (as a button perhaps)? When comments get too chatty the move-to-chat feature asks if you'd like to move it out. But this often happens after the fact and thus doesn't prevent it.   I propose that:   Move-to-chat be provided as a button along side so that a user who knows their question or conversation may become drawn out may elect to preempt and take the conversation "off-line" When we canonically quantize the scalar field in QFT, we use a Lorentz invariant integration measure given by $$\widetilde{dk} \equiv \frac{d^3k}{(2\pi)^3 2\omega(\textbf{k})}.$$  How can I show that it is Lorentz invariant? Show that the polynomial $X^5 + X^3 + \bar{1}$ in $(\mathbb{Z}/2\mathbb{Z}[X])$ is irreducible Show that the polynomial $X^5 + X^3 + \bar{1}$ in $(\mathbb{Z}/2\mathbb{Z})[X]$ is irreducible. (Hint: if it were reducible, it would either have a root or be of the form $g(X) \cdot h(x)$, where deg$g(X) = 2$ and deg$h(X) = 3$) Recall that $\bar{a}$ is shorthand for the coset $a + 2\mathbb{Z} \space $in$ \space (\mathbb{Z}/2\mathbb{Z})[X]$ (for any $a \in \mathbb{Z}$)  This question is worth $7$ marks out of a possible $75$, this is my solution..  $f(X) = X^5 + X^3 + \bar{1}$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$ iff it has no roots in $\mathbb{Z}/2\mathbb{Z}$. Since $f(\bar{0}) = \bar{1}$ and $f(\bar{1}) = \bar{1}$, it follows that $f(x)$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$.  Would this be a sufficient answer? am I missing something? 90s (or earlier) Japanese anime movie: two planets at war, young man is in love with princess from the other planet, and floods his own village I would like to find out the title of a Japanese anime feature film (not a series) that I saw on Romanian TV in the early 90s.   It was about a feud between the peoples of two planets (one very technologically advanced, the other not). The two planets go to war against each other, and a young man from one of the planets falls in love with a princess from the other planet.  I recall a scene where the army of the second planet take off in hot air balloons when they go to war.  The only other thing I recall is that there is a dam next to the village where the young man lives, and for some reason he opens the dam gates and floods his own village.   I do not recall there being robots in the movie, however I remember very vaguely that the technology level was probably at the level of late 19th century Europe on one planet and pre-industrial on the other one.  What it's not:   Nobunaga the Fool Aldnoah Zero Queen Millennia Uchuu Senkan Yamato (movie) I completed the two contracts I was given (Scout and Heavy), and earned two of the 'Teufort' weapons for turning them in.  I expected to get some new contracts so I could continue to earn some more of the schmancy new weapons, but I can't seem to find out how to get more.  Do I need to buy the ticket again to unlock more contracts? If not, how are contracts assigned? ON an American street, but IN a British one. Do the twain ever meet? In the United States, we say that someone lives on a street, whereas I've noticed that British people say in. For instance:     Bubba lives on Washington Street.   Colin lives in Cavendish Avenue.   I believe we both would use at when a number is given. For instance:     Bubba lives at 16 Washington Street.   Colin lives at 7 Cavendish Avenue.   I don't think it matters if it's a road, avenue, street, circle, or lane — as far as I know, in America, we always live on a street. Does it vary in British English, or is it always in? If it varies, what are the rules? Is the C# "explicit implementation" of the interface present in Java? In C#, if you have two base interfaces with the same method (say, F()) you can use explicit implementation to perform different impl. for F(). This alloes you to differently treat the object, corresponding to the current point of view: as IMyInterface1 or IMyInterface2. Is this possible in Java? Why does an conditionally convergent infinite series, when deranged, have a different limit? $\sum u_n = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$ conditionally converges to $\log 2$; and, when deranged to $\sum u'_n = 1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\dots,$ converges to $\frac{1}{2}\log 2$. Why does a deranged conditionally convergent series have a different limit? First Course in Math Anal, p.93, by Burkill, explains that the sums $s_n, t_n$ of the first n terms of the two series are different functions of $n$. This reasoning is incomplete, because different functions of $n$ can converge to the same limit; for example, let $r_n$ be the first $2n$ terms of $\sum u_n$, then $s_n$ and $r_n$ are different functions; besides, by differently defining terms, i.e. clustering pieces of a series, e.g. \begin{align*} 1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}+\dots = \left(1-\tfrac{1}{2}-\tfrac{1}{4}\right)+\left(\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots \\ = \left(1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots,\\ 1-1+1-1+\dots = (1-1)+(1-1)+\dots \\ =(1-1+1-1)+\dots,\\ 1+1+1+1+\dots = (1+1)+(1+1)+\dots \\ = (1+1+1+1)+\dots, \end{align*} one gets the sum of first n terms as different functions, but the series's limit remains. One could view a changed limit from the perspective of the changed number of terms, Denote 'countably many', the size of natural number, as $\lvert\mathbb{N}\rvert$, then $\lvert\mathbb{N}\rvert = \lvert k\mathbb{N}\rvert = \lvert\mathbb{N}\times\mathbb{N}\rvert$. Then two series with different numbers (e.g. $\lvert \mathbb{N}\rvert$, $\lvert k\mathbb{N}\rvert$) of terms– and so having different limits– paradoxically have same number of terms–and so are possibly formed by the same terms. The paradox could be the fundamental reason. Are there other explanations? Is the use here of 'countably many' is proper? I would like to set up memoir so that when I use \footnote{foo}, foo is displayed in the margin with the proper label and preferably break over pages.  There is an option \footnotesinmargin which gives almost exactly what I need, only justified. I want it to use ragged2e and change alignment based on even/odd page so that the margin note is always flushed towards the main text area and allows hyphenation on the side of the edge.  \documentclass[12pt, draft]{memoir} \footnotesinmargin \usepackage{polyglossia} \setmainlanguage{english} \usepackage{fontspec} \defaultfontfeatures{Ligatures=TeX} \setmainfont[Ligatures=TeX,Numbers=OldStyle]{Minion Pro} \usepackage{kantlipsum}  \begin{document} \kant[1]\footnote{\kant[1]} \kant[2]\footnote{\kant[2]} \kant[3-5] \end{document}   Possible partial work-around  I have remembered a nice thesis based upon the Classic thesis template and found the  which indeed contains a possible solution:  \usepackage{marginnote} \usepackage{ragged2e} \newcommand{\sidemark}[1]{#1.{\:}}  \newcommand{\marginelement}[2][0]{%   \marginnote{%     \strictpagechecktrue     \checkoddpage     \ifoddpage%       \RaggedRight\footnotesize%     \else%       \RaggedLeft\footnotesize%     \fi%     #2%   }[#1\onelineskip]% }%  \newcommand{\sidenote}[2][0]{%   \footnotemark%   \ignorespaces%   \marginelement[#1]{%     \sidemark{\thefootnote}%     \ignorespaces#2%   }%   \unskip% }%   Nevertheless, I would still prefer a memoir-specific solution with the re-definition of footnotes when used with the option \footnotesinmargin mainly for the sake of portability. This work-around also does not work with hyperref (there is no hypertext connection) and does not break over pages automatically. I've been reading an  ( eMagazine) relating to the Raspberry Pi; "An ARM GNU/Linux box for $25."  In the article, page 17 at the bottom it shows an area on the Pi where a track zigzags next to a straight one with the explanation text:     The “wiggles” in tracks, ensure signals are matched  electrically,   reducing interference and signal delay.  This is particularly   important for high speed video  data and HDMI signals.     I have very limited knowledge of electrical engineering so perhaps this is a very simple question but why would you incorporate these 'wiggles' in a PCB design?  I realise the quote gives me an answer and I sort of understand the interference point due to problems with power cables and coaxial cables running next to each other but I'd appreciate something assuming very little knowledge that explained why you would get the problems and how wiggles help. For example, why isn't the board covered in wiggles?