What's the cardinality of the infinity introduced in Riemann sphere model? In the Riemann sphere model, people extend the complex plane with a point at infinity: $\mathbb {C} \cup \{\infty \}$ and claim this expression: $\frac{1}{0}=\infty$. I'm confused with what is this $\infty$. Is it a member of $\{\aleph_n\}$ numbers? When we study infinities, we usually discuss cardinality. Does this $\infty$ corresponds to something's cardinality?
Is $\lim\limits_{n\rightarrow\infty} n$ comparable to $\aleph_0$? For example, can we say: $\infty=\lim\limits_{n\rightarrow\infty} n &lt; \aleph_0$?  These are two different types of structures. The limit being like the length, extension, or just generic magnitude and the other being cardinality of a set. Can we compare magnitude to cardinality?  Intuitively, we can reach $\aleph_0$ by counting the natural numbers on the number line and in the process will be approaching $\infty$. Which leads me to believe $\infty\leq\aleph_0$. But I can't see why it should be a strict inequality. I feel like they should be of equal magnitude.  I saw on a recent comment that $2^\infty=\infty$, but are those infinities really the same? It seems not to me. Of course we (usually) have that  $2^{\aleph_0}=\aleph_1$ where ${\aleph_0}$ and $\aleph_1$ are clearly two very distinct infinities, countable vs uncountable at least. Maybe one might argue that as far as the concept of magnitude is concerned, all infinities are "equal".
How to open nautilus without a desktop? I'm using xmonad, and I would like to open a nautilus window to a certain directory, because I like the convenience of browsing inside an encrypted zip file without unzipping it, which nautilus allows.   however, opening nautilus, even with a path argument starts a desktop on the monitor I'm using (There are 8 xmonad monitors) The desktop is inoperative and renders that monitor useless until nautilus is killed.  Is there any way to open just a nautilus window without any desktop starting? Evaluate $\lim_{n\rightarrow\infty} \int_0^1 f(x^n) dx$ Let $f$ be a real, continuous function defined for all $0\leq x\leq 1$ such that $f(0)=1$, $f(1/2)=2$, and $f(1)=3$. Show that  $$\lim_{n\rightarrow\infty} \int_0^1 f(x^n) dx$$  exists and compute the limit.  Attempt:   Since $f$ is real-valued continuous on $0\leq x\leq 1$ and the boundaries $f(0)=1$ and $f(1)=3$, $f$ is bounded on the interval  and the integral $\displaystyle\int_0^1 f(x^n)dx$ exists for any positive $n$. Thus, we can interchange the limit and the integral and compose the limit, \begin{align*} \lim_{n\rightarrow\infty} \int_0^1 f(x^n)dx&amp;=\int_0^1\lim_{n\rightarrow\infty}f(x^n)dx\\ &amp;=\int_0^1 f\left(\lim_{n\rightarrow\infty} x^n \right)dx\\ &amp;=\int_0^1 f(0)dx=1. \end{align*}  Questions: So, my first question concerns whether interchanging the limit and the integral is correct. It seems it would be justified by the dominated convergence theorem, where $f(x^n)$ is dominated by a function $g(x)=\displaystyle\sup_{0\leq x\leq 1} f(x^n)$ and $f_n(x^n)$, on $0\leq x\leq 1$, converges pointwise to a function $f$ that takes value 1 for $x\neq 1$ and 3 for $x=1$.  My next question is whether the interchange between the limit and composition of the function is allowed. Since the function is continuous and the limit exists at that point, it seems the interchange would be justified. Calculate the last digit of $3^{347}$ How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$ Pulling a CSV file from website into QGIS with ability to automatically refresh? How do I set up file sharing between two Ubuntu laptops on my wireless network? I have two laptops running Ubuntu 12.04. Both connect to the same WIFI network to access internet.  How do I connect them to each other so that I can access files on one from another and vice versa?  Also, how do I manage the user permissions, etc. on them? Error to calculate date diff with DateTime class Two different expansions of $\frac{z}{1-z}$ This is exercise 21 of Chapter 1 from Stein and Shakarchi's Complex Analysis.  Show that for $|z|&lt;1$ one has $$\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots +\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots =\frac{z}{1-z}$$and   $$\frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots \frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots =\frac{z}{1-z}.$$  Justify any change in the order of summation.   [Hint: Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+\cdots +2^k$.]  I don't really know how to work this through. I know that $\frac{z}{1-z}=\sum_{n=1}^\infty z^n$ and each $n$ can be represented as a dyadic expansion, but I don't know how to progress from here. Any hints solutions or suggestions would be appreciated. How does a pharmacy discount card work? Intuitive Explanation of Tippe Top Effect? Least rank of a matrix Let $A$ be a $7\times 7$ matrix such that it has rank $3$ and $a$ be $7\times 1$ column vector.  Then least possible rank of $A+(a a^T)$ is? ($a^T$ is transpose of the vector)  Intuitively I think it's $2$ as $a a^T$ has rank $1$, and the most it could affect is one row of $A$. How can I install Steam? This is a Putnam Problem that I have been trying to solve (on and off) for two years, but I have failed. I am in Calculus BC. This problem comes from the book "Calculus Eighth Edition by Larson, Hostetler, and Edwards". This problem is at the end of the first section of the chapter 8 exercises. Here's the problem:  Evaluate $$\int_2^4 \frac{\sqrt{\ln(9-x)}\,dx}{\sqrt{\ln(9-x)} + \sqrt{\ln(x+3)}}.$$  Please. Any help is very much appreciated. So are solutions. Thank you!  Edit: I like the solution given, but I was interested to see if there is any other way of doing the problem? I'm excited to see the results. What is the best way to convert seconds into (Hour:Minutes:Seconds:Milliseconds) time? When exactly does tmpwatch clear out files I place in /tmp? Launch an app from within another (iPhone) Is it possible to launch any arbitrary iPhone application from within another app?, For example in my application if I want the user to push a button and launch right into the Phone app (close the current app, open the Phone app).  would this be possible? I know this can be done for making phone calls with the tel URL link, but I want to instead just have the Phone app launch without dialing any specific number. Possible to create new Application Handler? Blender's application handlers are listed , but is it possible to create a new one?  For example, I'd like an application handler which is a callback for every time I add a strip to the Video Sequence Editor. Let's assume a very simplified situation. There is only one seller asking for 10$ and there is only one buyer bidding 20$. At what price would the transaction go?   Some answers say it won't, because the ask price has to match the bid price. () Other answers say, that the time ordering is important. () Which answer is correct? What is the final price in the presented situation? How to parse a query string into a NameValueCollection in .NET I would like to parse a string such as p1=6&amp;p2=7&amp;p3=8 into a NameValueCollection.  What is the most elegant way of doing this when you don't have access to the Page.Request object? I want to do:  cat update_via_sed.sh | sed 's/old_name/new_name/' &gt; new_update_via_sed.sh   in my program.  But I want to use variables, e.g.  old_run='old_name_952' new_run='old_name_953'   I have tried using them but the substitution doesn't happen (no error). I have tried:  cat update_via_sed.sh | sed 's/old_run/new_run/' cat update_via_sed.sh | sed 's/$old_run/$new_run/' cat update_via_sed.sh | sed 's/${old_run}/${new_run}/'