Verify that $S=\{0\}\cup\bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$ is compact subset of $\mathbb{R}$ while $\bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$ is not.  If I take the sequences $\{\frac{1}{i}:i\in\mathbb{N}\}$ there is no convergent subsequnce once $\{0\}\notin \bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$ . So $\bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$ is not compact. However when I examine $S=\{0\}\cup\bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$. I do not know how can I address all the coverings in order to prove it is compact. Questions: Is my proof right insofar? How should I prove $S$ is not compact? What is the strategy for that sort of cases? Thanks in advance!
Prove that $ S=\{0\}\cup\left(\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}\right)$ is a compact set in $\mathbb{R}$, but $\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}$ is not a compact set.   (Can we use definition of compact set for this problem? )
I can't see anything - Teams banner and cookie toaster obscure my whole screen There have been  for a while, but this is ridiculous. It reminds me of , but this time it's network-wide.   The banner must be scrolled past (or dismissed) on every page unless both first-party JavaScript and first-party cookies are enabled. If first-party JavaScript is enabled, and the banner is visible, the rest of the page jumps up and down every so often.  Under certain circumstances (not quite sure what they are, yet – a direct link to an answer, maybe? I'm seeing it ) the entire page jumps up and down a line every few seconds even when the banner is off-screen.   The cookie banner keeps coming back unless both third-party JavaScript  are enabled (and  is solved). The cookie box occludes the “Learn more about Teams” link in the banner (with default Firefox window sizes) and the Job / Teams link in the sidebar.  As it stands,  – a site for hobbyists and researchers, whose audience is not very likely to purchase Teams – is much, much harder to use, for no benefit that I can see. Can you roll back this change, please? $f:\mathbb{R}^3\to\mathbb{R}, f(rx)=r^{\alpha}f(x)$ for some $\alpha&gt;0$, $\forall x\in\mathbb{R}$, for any $r\in\mathbb{R}$, could any one tell me which of the following is true?  $1.$ If $f(x)=f(y)$ whenever $||x||=||y||=\beta&gt;0$, then $f(x)=\beta ||x||^{\alpha}$  $2.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)= ||x||^{\alpha}$  $3.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)=c||x||^{\alpha}$ for some constant $c$  $4.$ If $f(x)=f(y)$ whenever $||x||=||y||$, then $f(x)$ must be a constant function. Thank you. Expanding the equation out gives  $(3^{2n}\times3)+(2^n\times2^2) \equiv 0\pmod{7}$   Is this correct? I'm a little hazy on my index laws.  Not sure if this is what I need to do? Am I on the right track? Converting/using ArcGIS *.style file for/in QGIS (QML or SLD)? I have downloaded a dataset from a government organisation. it comes with a .style file.  I use QGIS 1.8 on windows 7.    Is there a way to use the .style file as a qml or sld? I have tried simply changing the extension but to no avail. Number Theory: Prove that $m | \phi(a^m - 1)$ Uniform convergence of $\sum_{n=1}^{\infty}\frac{1}{n(1+(x-n)^2)}$ Is $$\sum_{n=1}^{\infty}\frac{1}{n(1+(x-n)^2)}$$ uniformly convergent on $(0,+\infty)$? Sadly, Weierstrass' criterion nor Cauchy's condition don't seem to work here and I just ran out of ideas. I just installed Ubuntu 13.10, and I can't configure my wi-fi Epson TX430W printer.  Using the Settings-&gt;Printers interface, I can detect the printer. When I click add, I briefly see a "Searching for drivers" dialog. It dissapears, and then I'm left with the printer forever in "Installing" status.  If I close the window and reopen Printers, it doesn't show up on the list.  How can I solve, or at least diagnose the problem? How to populate the geometry column in a PostGIS table? After loading a polygon shapefile into PostGIS, the geometry column (geom) is blank for several features. In one case, only a single feature has the geometry calculated and all others are blank. I'm using the PostGIS SHP and DBF loader v.2.0. I am creating a spatial index, and using COPY rather than INSERT. I have loaded several other point datasets and they geom column is populated just fine... (Otherwise, things are working just fine for the polygons: when I create a new feature, the geometry is calculated / geom field populated...etc.) Have I done something wrong when importing the data?  Solution by @ThomasG77 (in comments) PgAdmin3 [doesn't] display features that are too complex but they are present.  You can try SELECT length(ST_astext(the_geom)) from your_table to confirm or you can find &quot;true&quot; empty geometry with SELECT * from your_table WHERE the_geom IS NULL;. Replacing Raster Values in Ras1 with Ras2 Constraints of relativistic point particle in Hamiltonian mechanics I try to understand constructing of Hamiltonian mechanics with constraints. I decided to start with the simple case: free relativistic particle. I've constructed hamiltonian with constraint:  $$S=-m\int d\tau \sqrt{\dot x_{\nu}\dot x^{\nu}}.$$  Here $\phi=p_{\mu}p^{\mu}-m^2=0$  $-$ .  Then $$H=H_{0}+\lambda \phi=\lambda \phi.$$   So, I want to show that I can obtain from this Hamiltonian the same equation of motion, as obtained from Lagrangian.  But the problem is that I'm not sure what to do with $\lambda=\lambda(q,p)$. I tried the following thing:  $$\dot x_{\mu}=\{x_{\mu},\lambda \phi\}=\{x_{\mu},\lambda p^2\}-m^2\{x_{\mu},\lambda\}=\lambda\{x_{\mu},p^2\}+p^2\{x_{\mu},\lambda\}-m^2\{x_{\mu},\lambda\}$$$$=2\lambda \eta_{\mu b} p^b+p^2\{x_{\mu},\lambda\}-m^2\{x_{\mu},\lambda\}=2\lambda \eta_{\mu b} p^b+p^2\frac{\partial \lambda}{\partial p^{\mu}}-m^2\frac{\partial \lambda}{\partial p^{\mu}},$$  $$\dot \lambda=\{\lambda, \lambda \phi \}=\{\lambda,\lambda p^2\}-m^2\{\lambda,\lambda\}=\lambda\{\lambda,p^2\}+p^2\{\lambda,p^2\}=2\lambda\eta_{ak}p^{a}\frac{\partial \lambda}{\partial x^{k}},$$  $$\dot p_{\mu}=\{p_{\mu},\lambda p^{2}-m^2\lambda \}=p^{2}\{p_{\mu},\lambda\}-m^2\{p_{\mu},\lambda\}=-p^{2}\frac{\partial \lambda}{\partial x^{\mu}}+m^2\frac{\partial \lambda}{\partial x^{\mu}}.$$  If we recall that $p^2-m^2=0$, then we get from the third equation: $\dot p=0$, and from the first: $$\dot x_{\mu}=2\lambda\eta_{ak}p^{a}.$$  So we have    $\dot x_{\mu}=2\lambda\eta_{\mu b}p^{b}.$ $\dot \lambda=2\lambda\eta_{ak}p^{a}\frac{\partial \lambda}{\partial x^{k}}.$ $\dot p=0.$   But I don't know what to do next. Can you help me? combining results of two select statements How can I type a Unicode character (for example, em-dash —?) The title pretty much says it all: how can I type Unicode characters - specifically the em-dash —? What lifetime can be expected of the typical hard disk? Or are there big differences between different types? And does it make a difference if it is used heavily instead of never being connected to system (for example serving as a backup medium)? Generating ordered combination of numbers I can form numbers with only 0,2,4,6,8.  The sequence is as follows  0,2,4,6,8,20,22,24,26,28,40,42,.....  How to generate an ordered sequence of numbers from combinations of 0,2,4,6,8. Create subdomains on the fly with .htaccess (PHP) I am looking to create a system which on signup will create a subdomain on my website for the users account area.   e.g. johndoe.website.com  I think it would be something to do with the .htaccess file and possibly redirecting to another location on the website? I don't actually know. But any information to start me off would be greatly appreciated.  Creating a sign up area is not the problem - I have done this many a time. I am just unsure where to start with the subdomain. Sum of reciprocals of numbers with certain terms omitted Given $\frac{a}{c} + \frac{b}{d} = 1$ Prove that $$\lim_{(x,y)\rightarrow(0,0)} \frac{|x|^{a}|y|^{b}}{|x|^{c} + |y|^{d}}$$ does not exist.  So I have done the proof for strict inequalities. And the limit only exists when the fractions add up to an integer greater than $1$. I'm not sure how to approach this other than counter examples. I was reviewing  for , and noticed that nobody voted to close it, even though the page reports a duplicate.      Why does this happen? Word for sentences that make sense when read backwards? Is there a word for sentences that make sense when read normally or backwards (not necessarily by reversing the words)? Example:     "Are you as bored as I am?" Why Objective-C,PHP do not support method overloading?