Sequence contained on a set converges to the supreme of the set
Prove there is an increasing sequence ($s_n$) of points in $S$ such that $\lim s_n = \sup S$.
Consider a complete metric space $(X,d)$ and $T\colon X\to X$. Suppose there exists $n\in\mathbb{N}$ such that the n-th power of $T$ is $q$-contractive. Show that then $T$ has exactly one fixed point $\overline{x}\in X$.      The n-th power is defined inductively: $$ T^{n+1}(x):=T(T^{n}(x)), n\in\mathbb{N}. $$ And for a function $T\colon X\to X$, $q$-contractive means  $$ \exists 0\leq q&lt;1~\forall~x,y\in X: d(T(x),T(y))\leq q d(x,y). $$    Now to the proof.  I think I have to show, that $T$ is q-contractive, because then it follows with Banach, that $T$ does have exactly one fixed point. So I have to show, that there exists a $0\leq q&lt;1$ so that for all $x,y\in X$ it is $$ d(T(x),T(y))\leq q\cdot d(x,y). $$ And most likely I have to use the q-contractivity of $T^n$, i.e. that there exists a $0\leq q &lt;1$ so that for all $x,y\in X$ it is $$ d(T^n(x),T^n(y))\leq q\cdot d(x,y). $$  Can you help me? Let $E$ be measurable and define $f:E\rightarrow\mathbb{R}$ such that $\{x\in E : f(x)&gt;c\}$ is measurable for all $c\in\mathbb{Q}$, is $f$ measurable?  There are a number of equivalent definitions for the measurability of a function and the most obvious one would be to show that $\{x\in E : f(x)&gt;c\}$ is measurable for all $c\in\mathbb{R}$.  Thus my strategy has been to consider an arbitrary irrational $y$ and use the density of the rationals in the reals to show that there exists some open set $O$ such that $m^*(O-\{x\in E : f(x)&gt;y\})&lt; \varepsilon$.  I would do this by choosing some rational $q&lt;y$, close enough to $y$, such that the set of values in $E$ which get sent under $f$ to values strictly greater than $q$ and less than or equal to $y$ is small enough.  But that doesn't in general seem possible since I could keep sending intervals of some fixed positive length to smaller and smaller regions just beneath $y$.  However conversely, such a function would not be enough for a counter example, since finding such a function only rules out this strategy, but doesn't necessarily entail non-measurability.  Is there a better definition of the measurability of a set I should use? or does the statement actually not imply measurability?  Thanks. Solving equation involving Euler's totient function Given $m\in\mathbb{N}$, is there a general way to solve the equation \begin{equation} \varphi(n)=m \end{equation} where $\varphi$ is the Euler's totient function? For example, I want to solve for all the $n$'s such that $\varphi(n)=4$. Is there any way to solve it that are also applicable to many (if not general) cases? I know that the compiler will sometimes initialize memory with certain patterns such as 0xCD and 0xDD.  What I want to know is when and why this happens.  When  Is this specific to the compiler used?  Do malloc/new and free/delete work in the same way with regard to this?  Is it platform specific?  Will it occur on other operating systems, such as Linux or VxWorks?  Why  My understanding is this only occurs in Win32 debug configuration, and it is used to detect memory overruns and to help the compiler catch exceptions.  Can you give any practical examples as to how this initialization is useful?  I remember reading something (maybe in Code Complete 2) saying that it is good to initialize memory to a known pattern when allocating it, and certain patterns will trigger interrupts in Win32 which will result in exceptions showing in the debugger.  How portable is this? How to replace a video on YouTube I need to make a small correction in a video I uploaded to YouTube. Googling shows that this is not possible. However, I remember seeing it done before. I posted a comment on a video saying that a certain part of the video was missing. I got a reply back that it was fixed, and when I went to see the video again, it was changed. How was this done? Computing the $n^{\textrm{th}}$ permutation of bits. I've seen  post about the $n^{\textrm{th}}$ permutation of a set but that is not what I need. If you have a bit string (ones and zeros only) there are  to quickly permute the NEXT lexicographically ordered bit permutation. For example take $$000111 \rightarrow 001011 \rightarrow 001101\rightarrow\cdots$$ etc. If the string is long then there are going to be approximately one Bajillion of these things. And I want to know how to compute the $n^{\textrm{th}}$ guy without doing the exhaust.  Backgound: This is for a parallel computing job where I need to farm out the search space of elements of a set.  Foreground: I own a copy of Knuth's 4th volume of "The Art of Computer Programming" and I think the answer is in there but I can't seem to find it. (It's like 900 pages).  I'm posting here in the hopes that someone has knowledge of this (obviously). Even if you can point me to a source, say the part in Knuth's book where he describes this problem I would be most grateful. If a nonempty set of real numbers is open and closed, is it $\mathbb{R}$? Why/Why not? In other words, are $\emptyset$ and $\mathbb{R}$ the only open and closed sets in $\mathbb{R}$? Why/Why not?  I tried by assuming a set is equal to its interior points and contains its limit points.  A bounded set will not do since stuff like $[1,4]$ and $\{5\}$ will not work, though that is not really proof. Help please?  Anyway, it must then be unbounded.  If $a$ is a real number then $(a,\infty)$, $(-\infty,a)$, $[a,\infty)$ and $(-\infty,a]$ don't seem to cut it so it must be $\mathbb{R}$. Just curious whether the process takes a while, and whether it scrapes through history to determine if it was once valid or not. Currently, I have 6 0-vote accepted answers and 20 1+-vote answers (), and that makes it > 5 and > 20% but not for long if voting goes up or down from here. ;-P  Update:  Special thanks to Grace Note for pointing out the  to follow. I upgraded this to a bug because it's been 48 hours during which the criteria supposedly were met. 6 0-votes, 20-23 1+ votes during the whole period. The criteria only last so long, so it may go out of scope soon. If it only checks every 3 days, I'm not sure if that's intentional or a bug. What is the reason for the top secret two space newline markdown weirdness? Consider the following Gaussian Integral $$I = \int_{-\infty}^{\infty} e^{-x^2} \ dx$$  The usual trick to calculate this is to consider $$I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} \ dx \right) \left(\int_{-\infty}^{\infty} e^{-y^{2}} \ dy \right)$$  and convert to polar coordinates. We get $\sqrt{\pi}$ as the answer.   Is it possible to get the same answer by considering $I^{3}, I^{4}, \dots, I^{n}$? Proof of this fairly obscure differentiation trick? Suppose we're tying to differentiate the function $f(x)=x^x$. Now the textbook method would be to notice that $f(x)=e^{x \log{x}}$ and use the chain rule to find $$f'(x)=(1+\log{x})\ e^{x \log{x}}=(1+\log{x})\ x^x.$$  But suppose that I didn't make this observation and instead tried to apply the following differentiation rules:  $$\frac{d}{dx}x^c=cx^{c-1} \qquad (1)\\ \frac{d}{dx}c^x = \log{c}\ \cdot c^x \quad (2)$$  which are valid for any constant $c$. Obviously neither rules are applicable to the form $x^x$ because in this case neither the base nor the exponent are constant. But if I pretend that the exponent is constant and apply rule $(1)$, I would get $f'(x)\stackrel{?}{=}x\cdot x^{x-1}=x^x.$ Likewise, if I pretend that the base is constant and apply rule $(2)$ I obtain $f'(x)\stackrel{?}{=}\log{x}\cdot x^x$.  It isn't hard to see that neither of the derivatives are correct. But here's where the magic happens: if we sum the two “derivatives” we end up with $$x^x+ \log{x}\cdot x^x=(1+\log{x})\ x^x$$ which is the correct expression for $f'(x)$.  This same trick yields correct results in other contexts as well. In fact, in some cases it turns out to be a more efficient way of taking derivatives. For example, consider $$g(x)=x^2 = \color{blue} x\cdot \color{red} x.$$ If we pretend the blue $\color{blue} x$ is a constant we would get $g'(x)\stackrel{?}{=}\color{blue}x\cdot 1=x$. Now if we pretend the red $\color{red}x$ is constant we get $g'(x)\stackrel{?}{=}1\cdot \color{red} x=x$. Summing both expressions we end up with $2x$ which is of course a correct expression for the derivative.  These observations have led me to the following conjecture:     Let $f(x,y)$ be a differentiable function mapping $\mathbb{R}^2$ to $\mathbb{R}.$ Let $f'_1 (x,y)=\frac{\partial}{\partial x} f(x,y)$ and $f'_2 (x,y)=\frac{\partial}{\partial y} f(x,y)$. Then for any $t$ we have: $$\frac{d}{dt}f(t,t)=f'_1 (t,t) + f'_2 (t,t).$$   (I apologise for the somewhat awkward notation which I could not seem to get around without causing undue ambiguity.)  This formulation also seems to lend itself to following generalisation:     Let $f:\mathbb{R}^N \to \mathbb{R}$ be a function differentiable in each of its variables $x_1,x_2,\ldots,x_N$. For $n=1,2,\ldots,N$ define $f'_n(x_1,x_2,\ldots,x_N)=\frac{\partial}{\partial x_n}f(x_1,x_2,\ldots,x_N)$. Let $t$ be any real number and define the $N$-tuple $T=(t,t,\ldots,t)$. Then one has: $$\frac{d}{dt} f(T)=\sum_n f'_n(T).$$   Thus my question is:   Is this true? How can it be proven? (Specifically in the case $N=2$ but also in the general case.) Is hiding the ball allowed in football? Imagine all our football team except one player form a tight circle, with our hands on the shoulders of the people next to us.  Inside the circle there is the eleventh person with the ball. The whole "construction" is moving slowly towards the opponent's goal. It is inevitable that we will score.  Is this illegal? If yes, according to what rule? Displaying layout marks in a document set with different layouts Is there any package that allows for the display of layout marks within a document where the layout changes several times throughout? I'm using the geometry facilities to issuing \newgeometry commands throughout my document, but it doesn't update the displayed layout and neither does the showframe package or the facilities of the memoir class.  EDIT: As Werner has pointed out, the showframe option of the geometry package displays correctly any updated page layout set by issuing \newgeometry commands.   My trouble came out of the fact that I'm issuing also \pagecolor commands and the page color seems to be hiding the layout lines typeset according to the geometry package. My confusion only worsened because the showframe package layout lines were not hidden by the page color (the package got the layout wrong on some pages, though). So, it all boils down to a conflict with the page color.   Even though I'm using geometry's showframe option for "debugging" ends, and I don't mind to have no page color at that stage of development, is it possible to override this problem and have at the same time both the page layout displayed by geometry as well as the page colored?   Btw, should I open a new question given that the problem is actually different from the originally posted one? Correlation between a nominal (IV) and a continuous (DV) variable I have a nominal variable (different topics of conversation, coded as topic0=0 etc) and a number of scale variables (DV) such as the length of a conversation.   How can I derive correlations between the nominal and scale variables? Show that if $n>2$, then $(n!)^2>n^n$. Show that if $n&gt;2$, then $(n!)^2&gt;n^n$.     My work: I tried to apply induction. So, at the induction step, I need to prove, $n^n&gt;(n+1)^{n-1}$ Here, I tried to use induction again without any luck. I also took log of both sides, but I did not get anything. Please help! How are shorter power-plays counted in power play percentage? # ~/.profile: executed by the command interpreter for login shells. # This file is not read by bash(1), if ~/.bash_profile or ~/.bash_login # exists. # see /usr/share/doc/bash/examples/startup-files for examples. # the files are located in the bash-doc package.  # the default umask is set in /etc/profile; for setting the umask # for ssh logins, install and configure the libpam-umask package. #umask 022  # if running bash if [ -n "$BASH_VERSION" ]; then     # include .bashrc if it exists     if [ -f "$HOME/.bashrc" ]; then     . "$HOME/.bashrc"     fi fi  # set PATH so it includes user's private bin if it exists if [ -d "$HOME/bin" ] ; then     PATH="$HOME/bin:$PATH" fi   Can anyone explain the contents of the ~/.profile file? So when you enter into the ~/.profile file what does all the writing mean? How do I get Perl to read the contents of a given directory into an array?   can do it, but is there some method using 'scandir' or a similar term? I have a table with a varchar(max) column that stores text ~4-10KB in size. Data is just being added - never modified, it is a log. My problem is that a table grows too large in size - I would like to keep about 1000000 rows in this table.   I am limited to SQL Server Express and hit the limit on database size. Compression is not available in SQL Server Express and I afraid that Filestream may not be appropriate and the number of files would be too big.   What is the best solution in this case? I would like to protect myself from unreasonable demands during my work day.  One of those is telling people that I will not respond to any emails before 3pm.  I'd like to set up a rule in outlook that replies to all emails before 3pm with a specific message. I'd like this to be done every day, no matter the date.  How do I set this up?  I've scoured Google best I can so far, and I think, based on a Exchange forum, that a script might be needed for this.  But what I found so far was only server side scripts. I need something on my own machine.  references: