Let $(X, \rho)$ be a compact metric space.   Suppose that $T: X \rightarrow X$ and for all $u \neq v \in X$,  $$ \rho(T(u),T(v)) &lt; \rho(u,v) $$  Then show that $T$ has a unique fixed point.    I am thinking to first show existence, and then show that it has to be unique.  To show existence, there is a hint to use the fact that $x$ is fixed if and only if $\rho(T(x),x) = 0$.  It is similar to the contraction mapping principle, but not enough that you can use it to prove this.  Also, if the above strict inequality is replaced by a weak inequality ($\leq$) then must $T$ have a fixed point?   I am thinking no, given that the requirement for a fixed point in the contraction mapping principle is that there be a $c \in (0,1)$ such that $ \rho(T(u),T(v)) \leq  c\rho(u,v)$.  I am thinking that a counterexample would be enough to show that this is not true.
Let $A$ be a closed and bounded subset of $\mathbb{R}^{n}$. Given that $f: A \to A$ satisfies $d(f(x),f(y)) &lt; d(x,y)$ for all $x,y \in A$, $x\neq y$ where $d$ is the usual Euclidean metric. Show that $f$ has a unique fixed point.    My observations:    Since $\mathbb{R^{n}}$ is complete, then $(A,d)$ is complete since $A$ is closed; Since $A$ is bounded, then $A \subset B_{R}$ for some $R &gt; 0$; Banach Fixed Point Theorem guarantees a unique fixed point if $f$ is contraction on $A$.   Please correct me if I am wrong. My quandry here is showing that $f$ is a contraction, i.e.  $$ d(f(x),f(y)) \leqslant k\cdot d(x,y), \hspace{3mm} k \in (0,1) $$  I'm looking for some hints or tips. Thank you.
What if the universe is rotating as a whole? Suppose in the milliseconds after the big bang the cosmic egg had aquired some large angular momentum. As it expanded, keeping the momentum constant (not external forces) the rate of rotation would have slowed down, but it would never reach zero.  What implications would that have to measurments to distant supernova and CMB radiation? Do we have any experimental data that definitely rules out such as scenario? And to what confidence level?  Edit A  suggests that the universe might indeed be spinning as a whole. Anyone care to poke holes at it?  Edit 2 Even more  places limits on possible rotation. Is there a way to run Python on Android? We are working on an  version and this platform has a nice Python API..  However, there is nothing official about Python on Android, but since  exists, is there a way to let the snake and the robot work together?? Union of ascending chain of Topologies Fastest Gun in the West Problem I feel like there's a problem with Stack Overflow, as the number of people prowling it increases.  Each question's answers are sorted by descending score and then by descending time of posting. This means that if a person sits down and answers a question in a long, thorough way, going through every nook and cranny, once they post their answer, it will already be one of about seven different ones, some of which have already been upvoted. This wouldn't be a problem if those answers were as thorough as the one this guy's posting, but they usually aren't. Some of them are downright wrong, some aren't even answers to the question asked because their poster didn't bother to read the question all the way through.  This causes a problem I like to call Stack Overflow's Fastest Gun in the West Problem. I've come to a point where I'd rather just send a short, simple, correct explanation, than to go and do some proper research, write a whole blog post about it or even make sure the code I post even compiles, just so it will be noticed, as opposed to the incorrect ones.  I'm sure I'm not the only one doing this and that it despairs many people from even trying to answer questions.  I've long ago learned to try to always raise a problem with a solution in hand, rather than just say "This is a problem, handle it," so my question, after this long tirade, is:     How do you think this can be changed? What would you change in Stack Overflow to make this problem go away without hurting the site?   I promise to vote on answers I like, even if it takes you a long while to post them. :) $g^k$ is a primitive element modulo $m$ iff $\gcd (k,\varphi(m))=1$ How to pass multiple operator properties via UI layout? Basically, I want to have a single operator which can do multiple things, based on some attributes values.  class MyPanel(bpy.types.Panel):     # (...)     layout = self.layout     row = layout.row()     row.operator("my.button", text="Button text")  class MY_BUTTON_OT_Button(bpy.types.Operator):     bl_idname = "my.button"     bl_description = "Button description"     bl_label = "Button"     foo = bpy.props.IntProperty()     bar = bpy.props.BoolProperty()      if bar:         # Do something with foo     else:         # Do something else :P   I know how to set one attribute, like so:      row.operator("my.button", text="Button text").foo=5   But what about multiple arguments? I tried several things that didn't worked, like (e.g.):   With setattr(): row.operator("my.button", text="Button text").setattr(foo=5,bar=True)   With a custom method setVal() in the operator class: row.operator("my.button", text="Button text").setVal(foo=5,bar=True)   Maybe I must use a single EnumProperty? I hope to avoid this, so my question is: Is it possible to pass multiple custom arguments to set attributes in an operator class? And if yes, how ?  Thanks for your help. Sum of two co-prime integers I need some help in a proof: Prove that for any integer $n&gt;6$ can be written as a sum of two co-prime integers $a,b$ s.t. $\gcd(a,b)=1$.  I tried to go around with "Dirichlet's theorem on arithmetic progressions" but didn't had any luck to come to an actual proof. I mainly used arithmetic progression of $4$, $(4n,4n+1,4n+2,4n+3)$, but got not much, only to the extent of specific examples and even than sometimes $a,b$ weren't always co-prime (and $n$ was also playing a role so it wasn't $a+b$ it was $an+b$).  I would appriciate it a lot if someone could give a hand here. The kernel and range of the powers of a self-adjoint operator Suppose $L: V \rightarrow V$ self-adjoint (on a finite-dimensional space $V$); $b$ is a real number. Show   The minimal polynomial has distinct roots; $\ker(L) = \ker(L^k)$ for $k\geq1$; $\operatorname{im}(L) = \operatorname{im}(L^k)$ for $k\ge 1$; $\ker(L-bI) = \ker((L-bI)^k)$ for $k\geq1$;     My thoughts:  1) I think I'm supposed to assume root is repeated and reach a contradiction but I'm not sure how to proceed mathematically with that.  2-4) pretty sure there's some huge theorem that I'm missing. When reading a stack trace like:  [FormatException: Input string was not in a correct format.]    System.Number.StringToNumber(String str, NumberStyles options, NumberBuffer&amp; number, NumberFormatInfo info, Boolean parseDecimal) +2755599    System.Number.ParseInt32(String s, NumberStyles style, NumberFormatInfo info) +112    System.Convert.ToInt32(String value) +68   What do the +68, +112, etc. numbers mean. I'm guessing that they're offsets to something. If so, what? Is there a page listing all fields? What's difference between shadowing and overriding a method in C#? How to use fakeslant as slanted but not italic font with fontspec? Using fontspec with Luatex it is possible to create a fake slant and use that as italic should no italics be around:  \setmainfont[ AutoFakeSlant=0.15 ]             {FontName}   However, also if an italic version exist this faked slant will be used both for \itshape and \slshape. How can I keep using the shipped italic font defined by the ItalicFont= parameter to \setmainfont but use fake slant for \slshape? In Thor: The Dark World, Odin and Loki are discussing Loki's actions during The Avengers.  Loki argues that he would have ruled Earth as &quot;a benevolent god&quot;, and Odin replies:  Odin: We are not gods! We are born, we live, we die, just as humans do.  Now that's pretty definitive.  And Loki's snarky reply (&quot;Give or take five thousand years...&quot;) would seem to confirm that Asgardians have a limited lifespan, and will eventually die of old age.1  And we know that Asgardians can die in battle, because we've seen it happen over and over with supporting characters and Asgardian soldiers. Now, I've always been more of a fan of Thor stories (like Jason Aaron's Thor: God of Thunder arc) in which Thor and the Asgardians are gods.  Straight-up, genuine, higher-level-of-existence gods, not von-Däniken-esque wannabes playing on the tired old notion of magic just being super-advanced science.  BUT, despite my preferences, this exchange would seem to confirm once and for all that in the MCU, the Asgardians are not gods, but rather were mistaken for gods because of their ram-horns full of awesome-sauce and their magnificent bulging pecks. But wait!  Watch that movie for about 30 seconds longer and we come across this exchange:  Thor: She is ill. Odin: She is mortal. Illness is their defining trait.  Wait, what?  I thought Asgardians weren't immortal?  They can die in battle, and they naturally die of old age.  In my book, that makes them long-lived, but entirely &quot;mortal&quot;.  Surely Odin isn't splitting a hair so fine as to say that resistance to disease alone makes one &quot;immortal&quot;? So, if the Asgardians of the MCU aren't actually gods (grumblegrumblegrumble...), and they can't live forever (or even particularly long in a cosmic sense), and they can be killed in battle, in what sense are they &quot;immortal&quot; at all?  1 Fun fact: adjusted for a roughly 5,000 year lifespan, 1,048-year-old Loki is currently in the tail-end of his rebellious teenage phase. I think it would be cool if a question would eventually open up to moderators to accept answers on the user's behalf, once a time period has passed. The asking user would get the rep score for the accept.  I know I've seen questions successfully answered with lots of upvotes and no accepted answer that sit there for months unaccepted. This would give superusers a little more functionality and would tie up a lot of loose ends.  What do you think? I am creating a chat using Ajax requests and I'm trying to get messages div to scroll to the bottom without much luck. I am wrapping everything in this div: #scroll {     height:400px;     overflow:scroll; }  Is there a way to keep it scrolled to the bottom by default using JS? Is there a way to keep it scrolled to the bottom after an ajax request? The Earth is spinning, so why don't we jump and land on a different location? I know there are similar questions on StackExchange but I think it is different and detailed.  The earth is spinning 465 meters/second so why don't we jump and land on a different location?  I googled about this question and I got some answers:   When we jump, we (and the atmosphere itself) also spin along with the earth so we don't land at a different location. (But why we are also spinning along with earth ?) The earth is so big and we are very small relative to earth so the tiny jump won't make any difference.   But according to answer 1, We are also spinning with the earth. But the question is, "why are we also spinning along with earth" ??  Regarding the answer 2, it won't make any difference but logically it doesn't make sense or I just don't understand it.  Here is the followup question,   "When we jump why don't we thrown out of earth due to centrifugal force of the spinning earth ??"  Please provide some detailed answer to these questions. With logic if possible.   This phenomenon is not possible I know, but even if it exists, then traveling will be so easy, we just have to hang on space (with help of helicopter etc.. ) and we can travel the whole world in 24 hours :D Math.cos() gives wrong result How to remove the top horizontal bar in fancyhdr? When using fancyhdr, how do you remove the top horizontal rule?  I do not want a header at all, just a footer.  This is what I do now:  \documentclass{article} \usepackage{fancyhdr} \pagestyle{fancy}  \lfoot{text} \cfoot{text 2} \rfoot{text 3}  \begin{document} Foo Bar \end{document} Will hardware changes affect system? explicit formula for recurrence relation $a_{n+1}=2a_n+\frac{1}{a_n}$