Inclusion of orthogonal complements: if $U_1\subset\ U_2$, then $U_2^\bot\subset\ U_1^\bot$
Proving that if $V$ is an inner product space and $U_1, U_2$ are subsets, then $U_1 \subset U_2$ if and only if $U^{\perp}_2 \subset U^{\perp}_1$
How to select the most recent set of dated records from a mysql table A high school Olympiad problem How could I detect (preferably with Python) duplicate MP3 files that can be encoded with different bitrates (but they are the same song) and ID3 tags that can be incorrect?  I know I can do an  checksum of the files content but that won't work for different bitrates. And I don't know if ID3 tags have influence in generating the MD5 checksum. Should I re-encode MP3 files that have a different bitrate and then I can do the checksum? What do you recommend? It was announced on  (at 1:05:26) that Stack Exchange has its own URL shortener, http://s.tk/. Users can't create their own, they can only use the ones that are built-in. What shortened URLs are available through it? Compare using Thread.Sleep and Timer for delayed execution I have a method which should be delayed running for a specified amount of time.  Should I use  Thread thread = new Thread(() =&gt; {     Thread.Sleep(millisecond);     action(); }); thread.IsBackground = true; thread.Start();   Or  Timer timer = new Timer(o =&gt; action(), null, millisecond, -1);   I had read some  about using Thread.Sleep is bad design. But I don't really understand why.  But for using Timer, Timer has dispose method. Since the execution is delayed, I don't know how to dispose Timer. Do you have any suggestions?  Or if you have alternative codes for delayed execution are also appreciate. Examples of infinite groups such that all their respective elements are of finite order. I am in need of examples of infinite groups such that all their respective elements are of finite order. Is there a sandbox to post test questions? I have noticed a behaviour that may or may not be a  bug. It is a discrepancy between the preview and the rendered result, which is why I would need to post a question to check it out.  Is there an official place to do that? Taylor Series type inequality How will the super massive black hole affect our galaxy? indicates that the water pipes should bonded to the electrical ground even if a grounding rod is used. How would one normally tell if that is the case? Is that bonding required to be accessible, or can it be hidden behind walls or buried? Where in the water system would it normally be bonded (e.g., before the main water shutoff, after, at the hot water tank)?  I recently had my electrical panel relocated and replaced, and don't recall seeing anything running from the panel to a water pipe - in the old panel or new. The electrician put in an intersystem bonding bridge, so the new ground rod, ground from the panel, and ground from the cable service are connected to it. I suppose that's where the bond from the water pipe would go? Introduction to AdS/CFT Is it safe to use file's hash as IV? I'm encrypting some files using AES in CBC mode.    I'm also using file's digest (SHA-1) to check that data is decrypted correctly (so I need to store it with file).  Is it safe to use this digest as AES's IV and store it in the header of file? Or it has security issue? In categorical terms, why is there no canonical isomorphism from a finite dimensional vector space to its dual? I've read in several places that one motivation for category theory was to be able to give precise meaning to statements like, "finite dimensional vector spaces are canonically isomorphic to their double duals; they are isomorphic to their duals as well, but not canonically."  I've finally sat down to work through this, and -  Okay, yes, it is easy to see that the "canonical isomorphism" from $V$ to $V^{**}$ is a functor that has a natural isomorphism (in the sense of category theory) to the identity functor.  Also, I see that there is no way that the functor $V\mapsto V^*$ could have a natural isomorphism to the identity functor, because it is contravariant whereas the identity functor is covariant. My question amounts to:     Is contravariance the whole problem?   To elaborate:   I was initially disappointed by the realization that the definition of natural isomorphism doesn't apply to a pair of functors one of which is covariant and the other contravariant, because I was hoping that the lack of a canonical isomorphism $V\rightarrow V^*$ would feel more like a theorem as opposed to an artifact of the inapplicability of a definition.  Then I tried to create a definition of a natural transformation from a covariant functor $F:\mathscr{A}\rightarrow\mathscr{B}$ to a contravariant functor $G:\mathscr{A}\rightarrow\mathscr{B}$. It seems to me that this definition should be that all objects $A\in\mathscr{A}$ get a morphism $m_A:F(A)\rightarrow G(A)$ such that for all morphisms $f:A\rightarrow A'$ of $\mathscr{A}$, the following diagram (in $\mathscr{B}$) commutes:  $$\require{AMScd}\begin{CD} F(A) @&gt;m_A&gt;&gt; G(A)\\ @VF(f)VV @AAG(f)A\\ F(A') @&gt;&gt;m_{A'}&gt; G(A') \end{CD}$$  This is much more stringent a demand on the $m_A$ than the typical definition of a natural transformation.  Indeed, it is asking that $m_A=G(f)\circ m_{A'}\circ F(f)$, regardless of how $f$ or $A'$ may vary. Taking $\mathscr{A}=\mathscr{B}=\text{f.d.Vec}_k$, $F$ the identity functor and $G$ the dualizing functor, it is clear that this definition can never be satisfied unless $m_V$ is the zero map for all $V\in\text{f.d.Vec}_k$ (because take $f$ to be the zero map). In particular, it cannot be satisfied if $m_V$ is required to be an isomorphism.     Is this the right way to understand (categorically) why there is no natural isomorphism $V\rightarrow V^*$?   As an aside, are there any interesting cases of some kind of analog (the above definition or another) of natural transformations from covariant to contravariant functors?  Note: I have read a number of math.SE answers regarding why $V^*$ is not naturally isomorphic to $V$. None that I have found are addressed to what I'm asking here, which is about how categories make the question and answer precise. ( was closest.) Hence my question here. Let $E=l^p=\left\{x=\left(x_n\right): x_n\in\mathbb{R}, \sum_{1}^{\infty}\left|x_n\right|^p&lt;\infty\right\}$ with $1\le p&lt;\infty$, $\left\|x\right\|_E=\left(\sum_{1}^{\infty}\left|x_n\right|^p\right)^{1/p}.$ Let $\left(\lambda_n\right)$ is a bounded sequence in $\mathbb{R}$ and consider the operator $T$ defined by $$T(x)=\left(\lambda_1x_x,\ldots,\lambda_nx_n,\ldots\right)$$ where $x\in l^p$. Prove that $T$ is a compact operator from $E$ to $E$ iff $\lambda_n\rightarrow 0$.  Can anyone give me some hints to solve this problem? Thank you! I've noticed that, if I add \n to a pattern for substituting using sed, it does not match. Example:  $ cat &gt; alpha.txt This is a test Please do not be alarmed  $ sed -i'.original' 's/a test\nPlease do not/not a test\nBe/' alpha.txt  $ diff alpha.txt{,.original}  $ # No differences printed out   How can I get this to work? I'm trying to prove that $(ma, mb) = $|$m$|$(a, b)$ , where $(ma, mb)$ is the greatest common divisor between $ma$ and $mb$. My thoughts: If $(ma, mb) = d$ , then $d$|$ma$ and $d$|$mb$ → $d$|$max + mby$ → $d$|$m(ax+by)$. This implies that $d$|$m$ or $d$|$(ax+by)$. This is the same as  $d$|$m$ or $d$|$a$ and $d$|$b$, so $d$|$m$ or $d$|$(a,b)$. This is the same as $d$|$m|$ or $d|(a,b)$, so $d$|$|m|(a,b)$. I don't know what to do. If $a_0\in R$ is a unit, then  $\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$ How do I get $\cos{\theta} \lt \frac{\sin{\theta}}{\theta} \lt 1$? \selectcolormodel{gray} doesn't work for the duck hair, but it works for the other parts of the ducks, why?  \documentclass{ltugboat}  \usepackage{microtype}  \usepackage{graphicx}  \usepackage{ifthen} \newboolean{grayscale}    \setboolean{grayscale}{true}% for the black &amp; white version    %\setboolean{grayscale}{false}% for the colored version     \usepackage{tikzducks} \usetikzlibrary{positioning}  \newenvironment{mytikz}{%     \begin{tikzpicture}     \ifthenelse{\boolean{grayscale}}{\selectcolormodel{gray}}{}      }     {\end{tikzpicture}}  \usepackage{ifpdf}  \ifpdf  \usepackage[breaklinks,hidelinks]{hyperref}  \else \usepackage{url} \fi   \begin{document}      Why \textbackslash\texttt{selectcolormodel\string{gray\string}} does not work for the duck hair?      Black and white version:          \begin{mytikz}         \begin{scope}[xscale=-1]             \duck[longhair=pink]         \end{scope}         \begin{scope}[xshift=10pt]             \duck[crazyhair=orange]         \end{scope}     \end{mytikz}      Colored version:     \setboolean{grayscale}{false}      \begin{mytikz}         \begin{scope}[xscale=-1]             \duck[longhair=pink]         \end{scope}         \begin{scope}[xshift=10pt]             \duck[crazyhair=orange]         \end{scope}     \end{mytikz} \end{document} Imagine a black hole that is fast-approaching its final exponential throws of Hawking evaporation. Presumably, at all points in this end process there will remain a region that identifiably remains &quot;the black hole&quot; until the the very end, as opposed to huge swarm of fundamental particles that is being radiated out from it. As the mass of the black hole descends to that of individual particles, it would seem entirely feasible that the very last fermionic Hawking radiation event available to the almost-deceased black hole could leave it with an unbalanced charge, e.g. -1, and an unbalanced spin, say 1/2. It would also have some kind of mass of course, but that aspect of the final residue could be fine-tuned to any specific value by photon emissions of arbitrary frequencies. After photon emission mass trimming, the resulting black hole residuum would reach a point where it is no longer be able to evaporate into any known particle, because there is no longer any lower-mass option available to it for removing the -1 charge and 1/2 spin. The black hole residuum will at that point be stuck, so to speak, stuck with exact charge, spin, and mass features of an electron. And so my question: Is it an electron? And if so, by equivalence, is every electrons in the universe really just a particular type of black hole that cannot evaporate any further due to the constraints of charge and spin conservation? And if so, why are charge and spin so uniquely combined in such black hole remnants, so that e.g. a remnant of -1 charge and zero spin is not permitted, at least not commonly, and the mass is forced to a very specific associated level? Is there anything in the current understanding of general relativity that would explain such a curious set of restrictions on evaporation? The full generalization of this idea would of course be that all forms of black hole evaporation are ultimately constrained in ways that correspond exactly to the Standard Model, with free fundamental particles like electrons being the only stable end states of the evaporation process. The proton would be a fascinating example of an evaporation that remains incomplete in a more profound way, with the three quarks remaining incapable of isolated existence within spacetime. The strong force, from that perspective, would in some odd sense have to be a curious unbalanced remnant of those same deeper constraints on the overall gravitational evaporation process. This may all be tautological, too! That is, since Hawking radiation is guided by the particles possible, the constraints I just mentioned may be built-in and thus entirely trivial in nature. However, something deeper in the way they work together would seem... plausible, at least? If an electron is an unbalanced black hole, then the particles given off would also be black holes, and the overall process would be not one of just particle emission, but of how black holes split at low masses. Splitting with constraints imposed by the structure of spacetime itself would be a rather different way of looking at black hole evaporation, I suspect. (final note: This is just a passing thought that I've mulled over now and then through the years. Asking it was inspired by  by Ben Crowell. I should add that I doubt very seriously that that my wild speculations above have anything to do with Wheeler's concept of geons, though.)