I need help with the following proof, which my professor added for practice (but not as homework). I am completely lost here.  Let $A$ be a nonempty subset of a metric space $X$. Define $d(\cdot ,A) : X \to [0,\infty)$ by  $$d(x,A) = \inf\{d(x,a) : a \in A\}.$$   Prove that $d(\cdot,A)$ is continuous.   Now there are some things that I do know.    The definition of continuous:Suppose X,Y are metric spaces, a ∈ X and f : X → Y. The function f is continuous at a if for every ε>0there is a δ>0such that if dX(a,x) &lt; δ, then dY (f(a),f(x)) &lt; ε.  If f is continuous at every point a ∈ X, then f is said to be continuous.  The following theorem:A function f : X → Y is continuous if and only if f−1(U) ⊂ X is open for every open set U ⊂ Y .
Let $(X,d)$ be a metric space. How to prove that for any closed $A$ a function $d(x,A)$ is continuous - I know that it is even Lipschitz continuous, but I have a problem with the proof:  $$ |d(x,a) - d(y,a)| \leq d(x,y) $$ for any $a\in A$ - but we cannot just replace it by $|d(x,A) - d(y,A)|\leq d(x,y)$ since the minimum (or infimum in general) can be attained in different points $a\in A$ for $x$ and $y$, so we only have that $$ |d(x,A)-d(y,A)|\leq d(x,y)+\sup\limits_{a,b\in A}d(a,b) $$ which does not mean continuity.
I have a .tiff and a .tfw file that I want to change into a GeoTIFF. When I try to open the file in ArcMap and ENVI 5.0, it is successful. However, in Google Earth it is not.  Is there anyway to create the GeoTiff in ArcMap or ENVI? I have FlexiPage Home_Page_Default.flexipage assigned as Default Home Lightning Page for all profiles in one sandbox. I need to migrate it to another sandbox and assign as Default Home Lightning Page for all profiles in the destination sandbox.   I am able to retrieve and deploy that flexipage but looks like I am not able to assign it automatically as Default Home Lightning Page for all profiles in the destination sandbox using Force.com ANT Migration Tool or changeset.  Is that possible to migrate default home flexi page assignment from one organization to another using ANT Migration Tool or changeset?  If this is possible then how? How long does it take for badges to be awarded? How are they generated? Write the digits of a number into an array C1/D Visa - Mexico and US We are going to change the IP address of our SharePoint Servers can any one point out the RISK and IMPACT %AX = (%AH + %AL)  So why not %EAX = (%SOME_REGISTER + %AX) for some register %SOME_REGISTER? How do I install Ubuntu 16.04 from an SD card? Could someone explain how to install Ubuntu from an SD card? I don't have any CDs or USB sticks around. My current operating system is Windows 10 64 bit. Work in ArcGlobe and I want to calculate the coordinates of the globe latitude and longitude, but the coordinates from the mouse click is read from the screen, not the globe, because when I clicked on the corner of the screen it shows me the coordinates 0,0. How can I make it so that coordinates read with globe and not with the screen? import arcpy import pythonaddins  class ToolClass2(object):     &quot;&quot;&quot;Implementation for coord_addin.tool (Tool)&quot;&quot;&quot;     def __init__(self):         self.enabled = True         self.shape = &quot;NONE&quot; # Can set to &quot;Line&quot;, &quot;Circle&quot; or &quot;Rectangle&quot; for interactive shape drawing and to activate the onLine/Polygon/Circle event sinks.     def onMouseDown(self, x, y, button, shift):         pythonaddins.MessageBox(&quot;Long&quot; + &quot; &quot; + str(x) + '\n'+ &quot;Lat&quot;+ &quot; &quot; + str(y), 'Coordinates', 0)         pass     def onMouseDownMap(self, x, y, button, shift):         #pythonaddins.MessageBox(&quot;Long&quot; + &quot; &quot; + str(x) + '\n'+ &quot;Lat&quot;+ &quot; &quot; + str(y), 'Coordinates', 0)         pass     def onMouseUp(self, x, y, button, shift):         pass     def onMouseUpMap(self, x, y, button, shift):         pass     def onMouseMove(self, x, y, button, shift):         pass     def onMouseMoveMap(self, x, y, button, shift):         pass     def onDblClick(self):         pass     def onKeyDown(self, keycode, shift):         pass     def onKeyUp(self, keycode, shift):         pass     def deactivate(self):         pass     def onCircle(self, circle_geometry):         pass     def onLine(self, line_geometry):         pass     def onRectangle(self, rectangle_geometry):         pass I want to remotely control my Android device from my PC without rooting it. Is this possible? I'm trying to include a file with a specific shape (cup) into another TikZ image, in order to use the cup as a "stamp" to draw more complex figure. The file is located in /usr/share/tikzit/shapes/cap.tikz  \documentclass{article} \usepackage[svgnames]{xcolor} \usepackage{tikz} \usetikzlibrary{decorations.markings} \usetikzlibrary{shapes.geometric} \pagestyle{empty}  \pgfdeclarelayer{edgelayer} \pgfdeclarelayer{nodelayer} \pgfsetlayers{edgelayer,nodelayer,main}  \tikzstyle{none}=[inner sep=0pt] \tikzstyle{node1}=[cap,fill=White,draw=Black]  \tikzstyle{simple}=[-,draw=Black,line width=2.000] \tikzstyle{arrow}=[-,draw=Black,postaction={decorate},decoration={markings,mark=at position .5 with {\arrow{&gt;}}},line width=2.000] \tikzstyle{tick}=[-,draw=Black,postaction={decorate},decoration={markings,mark=at position .5 with {\draw (0,-0.1) -- (0,0.1);}},line width=2.000]  \usepackage[graphics,tightpage,active]{preview} \PreviewEnvironment{tikzpicture} \newlength{\imagewidth} \newlength{\imagescale}  \begin{document} \begin{tikzpicture}     \begin{pgfonlayer}{nodelayer}         \node [style=node1] (0) at (-2.5, 1.5) {};     \end{pgfonlayer} \end{tikzpicture}  \end{document}   The code i want to include is the following:  \begin{tikzpicture}         \path [use as bounding box] (-2,-2) rectangle (2,2);         \begin{pgfonlayer}{nodelayer}                 \node [style=none] (0) at (0, 0.75) {};                 \node [style=none] (1) at (-0.5, 0) {};                 \node [style=none] (2) at (0.5, 0) {};         \end{pgfonlayer}         \begin{pgfonlayer}{edgelayer}                 \draw [in=0, out=90] (2.center) to (0.center);                 \draw [in=270, out=-90, looseness=0.75] (1.center) to (2.center);                 \draw [in=90, out=180] (0.center) to (1.center);         \end{pgfonlayer} \end{tikzpicture}   But it seems I miss how to specify the directory of cup. I get this error:     Package pgfkeys Error: The key '/tikz/cap' requires a value. I am   going to ignore this key.   It seems like I have to deal with tikzstyle or pgfkey, but I don't know how to specify an external file. Why do the measurements of this object seem erroneous? Bash-like, non-rotating filename tab completion with PowerShell Under Windows, using PowerShell, is it possible to make the filename suggestion (using the Tab key) work like with bash under Linux?  Let's assume a directory contains 3 directories: example1, example and examination.  On Windows (by default), if you type cd ex and Tab, it will auto-suggest cd .\examination. Subsequent uses of Tab will rotate the suggestion between example1, example and examination.  Using bash (typically under Linux), if you type cd ex and Tab, it will auto-complete until the character where the filenames start to differ: cd exam. Another use of Tab then suggests which directories are available:  $ cd exam examination/ example1/    example2/   Then, if you type p and Tab, it will auto-complete to example, and so on (here, offering a choice between example1 and example2).  Is there a way to make the filename suggestion work in a tree-like search like this, instead of rotating through the names?  (I'm mainly interested in file names. Bash can also refine this selection based on the file type, for example only directories for cd, which would be good, but having the file name selection itself would be enough.) How do you add CSS rules (eg strong { color: red }) by use of Javascript? How to encrypt individual folders? Let's say that I have a folder, within my Documents folder, that has files I want no one to access without a password.  Is there a way to lock down that folder so that it's password protected / encrypted?  In fact, is it possible to password-protect any individual folder? How do I show that $\sum_{cyc} \frac {a^6}{b^2 + c^2} \ge \frac {abc(a + b + c)}2?$ Let $a, b, c$ be positive real numbers, show that  $$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2.$$  I think this is likely to turn out to be proved by Hölder, but I can't see how. Any hints will be appreciated. What gear do I need for a newborn photography session? I am an amateur photographer and I have a newborn session coming up for a family friend.   My current gear is:   Canon T5i (crop sensor) Canon 18-55 IS STM Kit lens 85mm f/1.8 lens   I plan on buying a reflector for the shoot and will be using natural light as my light source.  So my question is: is this enough gear to get some nice photos? Should I get another prime lens besides the 85mm? I don't really use the kit lens.  I also have another 8 month old baby shoot coming up and an informal wedding shoot coming up so it would be nice to use the lenses in all situations. I don't need a telephoto at the moment and will probably not use one any time soon. I want to combine two ffmpeg commands to a single command. I am using ready made ffmpeg binary in my Android application. I want to concat .ts files and overlay an image.  I am using following commands:   To concat .ts files:  String[] ffmpegcommand = {"ffmpeg", "-i","concat:"+input_file_path+"|"+input_file_path1, "-c", "copy", "-bsf:a", "aac_adtstoasc", output_file_path};  Apply image overlay effect:  String[] ffmpegcommand = {"ffmpeg","-y" ,"-i", input_file_path,"-strict","experimental", "-vf","movie="+AppStaticData.BASE_FOLDER_PATH + File.separator + "watermarklogo.png"+" [watermark]; [in][watermark] overlay=main_w-overlay_w-10:10 [out]", AppStaticData.BASE_FOLDER_PATH+"/" + output_file_path}; I am trying to find the name of a direct to TV movie, possibly a 2-3 part miniseries which I believe aired on the SciFi channel in the 90s. I don't remember who starred in it, but the movie opened with an astronaut attempting reentry whose shuttle ends up exploding. He awakes in a large bubble of air in a river and after reaching the surface and swimming to shore, meets several other people, from seemingly many different times throughout history (including one alien, who died on Earth).  Over the course of the movie, we learn the world is the afterlife, where everyone goes when they die. The major plot involves launching a steamboat (built by Mark Twain, no less) to travel up a river to the mountains, where the beings responsible for bringing everyone to this world are.  Does anyone know what the title of this movie/miniseries is? I am looking for a closed form (ideally expressed as elementary functions) of the function $\exp_n(x) = \sum_{k=0}^n x^k / k!$.  I am already aware of expressing it in terms of the gamma function. Background / Motivation When counting combinations of objects with generating functions, it is useful to be able to express the partial sum $1 + x + \cdots + x^n$ as $\frac{1-x^{n+1}}{1-x}$. For example, to count the number of ways to pick 5 marbles from a bag of blue, red, and green marbles where we pick at most 3 blue marbles and at most 2 red marbles, we can consider the generating function $f(x) = (1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)$. By using the partial sum identity, we can express it as $f(x) = \left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^3}{1-x}\right)\left(\frac{1}{1-x}\right)$.  Simplify, express as simpler product of series, and find the coefficient of the $x^5$ term. I want to be able to do the same for a generating function in the form $g(x) = \exp_{n_1}(x)^{p_1} \exp_{n_2}(x)^{p_2} \cdots \exp_{n_j}(x)^{p_j}$ The easiest way to extract the coefficient of a given term $x^p / p!$ would be to use a similar closed form expression for $\exp_n(x)$ and a similar technique to $f$. Attempted Solutions Differential equation Recall that the way to prove the identity $1+x+x^2+\cdots+x^n = \frac{1-x^{n+1}}{1-x}$ is to define $S = 1 + x + x^2 + \cdots + x^n$ and notice that: $S - Sx = 1 - x^{n+1}$. Likewise, notice that $y(x) = \exp_n(x)$ satisfies $y - y' = x^n/n!$.  Via SAGE, the solution is $y(x) = \frac{c+\Gamma(n+1,x)}{n!}e^x$.  Our initial condition $y(0) = 0$ so $c=0$.  By , $\Gamma(n+1,x) = n! e^{-x} \exp_n(x)$ so we have $y(x) = \exp_n(x)$. Recurrence Relation Notice that $\exp_n(x) = \exp_{n-1}(x) + x^n/n!$. Using the unilateral Z-Transform and related properties, we find that $\mathcal{Z}[\exp_n(x)] = (z e^{x/z})/(z-1)$. Therefore, $\exp_n(x) = \mathcal{Z}^{-1}\left[(z e^{x/z})/(z-1)\right] = \frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz$. $(z^n e^{x/z})/(z-1)$ has two singularities: $z = 1$ and $z = 0$.  The point $z = 1$ is a pole of order one with residue $e^x$.  To find the residue at $z = 0$ consider the product $z^n e^{x/z} (-1/(1-z)) = -z^n \left( \sum_{m=0}^\infty x^m z^{-m} / m! \right) \left( \sum_{j=0}^\infty x^j \right)$.  The coefficient of the $z^{-1}$ term is given when $n - m + j = -1$. The residue of the point $z=0$ is then $-\sum_{m,j} x^m / m! = -\sum_{m=n+1}^\infty x^m / m!$. Let $C$ by the positively oriented unit circle centered at the origin.  By Cauchy's Residue Theorem, $\frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz = \frac{1}{2 \pi i} 2 \pi i \left(e^x - \sum_{m=n+1}^\infty x^m / m!\right) = \exp_n(x)$. Finite Calculus I've tried to evaluate the sum using finite calculus, but can't seem to make much progress.