Can you help me determine what I am doing wrong in this proof? Is this sufficient?
Statement: There are uncountably many surjections from $\mathbb N$ to $\mathbb N$.  Is the following proof valid:  The real numbers are uncountable.   There exists a bijection between the set of real numbers and the set of all functions from $\mathbb N$ to $\{0,1\}$.   Given a function, $f$, from $\mathbb N$ to $\{0,1\}$, construct a surjection from $\mathbb N$ to $\mathbb N$, denoted $g$, as follows:  We map the $n$th composite number to $n$, and the nth prime, to $f(n)$.   Clearly this is a surjection since the composite numbers are countable. Clearly different functions $f$ lead to different surjections $g$. Since there are uncountably many $f$s, there are uncountably many surjections $g$.
Find the integer which is closest to the value of   $\frac{1}{\sqrt[6]{5^6+1}- \sqrt[6]{5^6-1}}$  I have tried putting them between n and n+1 and tried manipulating  n but I cant seem to find a strategy. Can "whatever" be split into two words? I tend to write, "say whatever they want", but I'm always tempted to write "say what ever they want". Is it acceptable to split the word in this context? Set of branch points isn't discrete, but branch points are isolated? I refer to , which I understand says that the branch points of a nonconstant holomorphic map $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are not necessarily compact, form a discrete subset of range $Y$ of the map $F$. (Not sure if 'connected' is part of the definition of Riemann surfaces in other textbooks, but it is here.) Question 1: Do I understand correctly, and is this indeed true? What else I understand:  $A$ is a discrete subspace/subset (I just assume everything here is the usual subspace topology) of $B$ if and only if every singleton subset of $A$ is open in $A$ if and only if every point of $A$ is isolated.  This overflow question () seems to suggest this is not true but is true if $F$ is proper. However, I'm not sure the definitions of Riemann surface are the same (maybe some answers use definitions where Riemann surfaces are not necessarily connected or something).  2.1. I know $F$ is 'discrete' (meaning that its fibres are discrete; here, i refer to a ): see definition before Lemma 3.1, on p. 7), by Proposition II.3.12, and open, by open mapping theorem, which is Proposition II.3.8, but I'm not sure these help show $F$ is proper or are otherwise helpful. Ostensibly 'discrete' maps don't map discrete subsets to discrete subsets.   This stackexchange question () seems to suggest this is true.  I think this is true if $F$ is injective or at least something like locally injective. (Please don't make me type up all the stuff I did for this part.) However based on the question in (3) and based on the , which I think uses the term 'branch points' to mean the same thing as what Miranda means by 'ramification points' (and thus is different from Miranda's 'branch points'), it seems that no neighbourhood $U$ of a ramification point $p \in U \subseteq X$ of $F$ is such that the restriction $F|_U$ is injective.  Kinda related:    This page () says    Now, the ramification and branch points must form a discrete set  I don't think the page assumes anything like $F$ is proper.  The  text in (2.1) actually seems to assume proper based on the '3.7. Proper holomorphic maps' part, but said part refers to 'Lemma 3.17', so I think proper can be replaced with closed. Maybe $F$ is closed or something.  Kinda related (see Question 2 below):     says    If $F$ is proper then the image $\Delta = F(R)$ is discrete in $Y$.   This seems to have all the same assumptions as with Miranda. Ostensibly, either Miranda is wrong or there's some higher level machinery that allows us to omit $F$ proper.  Question 2: To possibly generalise this, what are the minimum requirements on $F, X$ and $Y$ to make $F$ map discrete subsets to discrete subsets? I mean, I'm not sure we really need 'holomorphic' here. I have a feeling this applies perhaps to maps that are just open/closed, continuous and discrete and spaces that are just Hausdorff and locally compact or something.  (again) Kinda related: Playing a remote movie on the remote computer How does Star Trek bridge crew spend their time when on duty? It's implied that the Enterprise bridge crew, in any of the TV series, spend long durations at their duty stations, either sitting or standing.  What exactly are they doing for hours on end while the Enterprise is en route between destinations? Is it me or does this seem like an incredibly tedious if not fatiguing job, sitting in a chair staring at the same view screen? Couldn't these tasks be automated? What causes the portraits of Wizards to behave interactively? Does J.K. Rowling address how the interactive portraits work? Would a portrait painted after a Wizard/Witch died become interactive, or does the painting of the portrait have to take place during the Wizard/Witches lifetime? Nested sequence of non-empty compact subsets - intersection differs from empty set I have the following proposition:  "If $(S_n)$ is a sequence of compact subsets of a metric space such that  $S_1 \supseteq S_2 \supseteq ... \supseteq S_n$  Then $\bigcap^\infty_{i=1} S_i$ $\neq$ $\emptyset$"  My lecture notes provide the following proof, whose last step I don't understand:    Proof: Assume by way of contradiction that $\bigcap^\infty_{i=1} S_i$ = $\emptyset$. Let $V_n := (S_n)^c$ for each $n$. The class $\mathbb{C} :=\{V_n: n \in \mathbb{N} \}$ is an open covering of $S_1$. Since $S_1$ is compact, $\exists I$ finite, $I \subset \mathbb{N}$, s.t $\{V_i: i \in I \}$ covers $S_1$. Let $n:=max(I)$. The class $\{V_1, ..., V_n \}$ also covers $S_1$ and we get the following contradiction   $\emptyset = \bigcap^n_{i=1} S_i = S_n$    So, I don't get the part from "Let $n:=max(I)$..." onwards, especially the contradiction. I have tried applying DeMorgan's Law and the only way the last equalities makes sense is if the union of $\{V_1, ..., V_n \}$ equals the whole metric space, a property I haven't been able to understand from the steps of the proof. This is a corollary to the finite intersection property, so it must have a connection but again I can't quite get it.  Thanks in advance I have created screenShot() method to take the screenshot of the web page. When I used this method in my code first time it will capture the screen and it will save the image with image1 name. When I used the same method screenShot() to capture screens for other pages with the use of this.screenShot(); this will again capture the current web page screen but it stores With the same name image1 in the same folder (which overrides to previously captured screen image1).  Now, I have to take screenshots for the number of web pages &amp; want to store it into the same folder with different names like image1, image2, image3...etc. by using the same method screenShot().  public void screenShot() throws IOException, InterruptedException {     File scr = ((TakesScreenshot)driver).getScreenshotAs(OutputType.FILE);     File dest = new File("filPath/image1.png");     FileUtils.copyFile(scr, dest); }   Any workaround to achieve this scenario? Prove that $T$ is an orthogonal projection Let $T$ be a linear operator on a finite-dimensional inner product space $V$. Suppose that $T$ is a projection such that $\|T(x)\| \le \|x\|$ for $x \in V$. Prove that $T$ is an orthogonal projection.    I can't understand well. The definition of orthogonal operator is $\|T(x)\| = \|x\|$. But why that $\|T(x)\| \le \|x\|$ for $x \in V$ means orthogonal projection? What is the best way to execute a script when entering into a directory? When I move into a new directory I would like bash to execute the projectSettings.bash script much like RVM does. How much hard drive space do I need to install Windows 7 on my Mac via Boot Camp? I have a Macbook Air and would like to install Windows 7 on it via Boot Camp. How much hard drive space should I make sure to have free before doing this? Prove that $\lim_{x \to \infty} \frac{\log(1+e^x)}{x} = 1$ Show that $$\lim_{x \to \infty} \frac{\log(1 + e^x)}{x} = 1$$ How do I prove this? Or how do we get this result? Here $\log$ is the natural logarithm. Orthogonal matrix and orthonormal columns Ex-dividend date: How long do I have to hold a stock in order to get the next dividend? I want to be a stockholder on record for a dividend. So lets assume I bought stock before the ex dividend date. Can I immediately sell the stock after the ex-dividend date to collect the dividend, or do I need to hold for a few weeks to collect the dividend?  Lets say the ex-dividend date is the 15th of the month. I buy stock on the 14th (I think this is soon enough to be a shareholder on record) and then I sell on the 15th. Do I get the dividend in my account when it is distributed?  Don't worry &ndash; I have a proprietary way of hedging the stock's post-dividend price adjustment, and it costs less than the actual dividend so all is well.  I just don't know what the shortest time I can hold a position is. I'd like to know the shortest theoretical time possible in order to get the dividend.  Insight appreciated. Is it possible to run Android apps on Ubuntu? And I don't mean inside an emulator, but natively, as you would with any other application installed using the Ubuntu Software Center. Avoid synchronized(this) in Java? Whenever a question pops up on SO about Java synchronization, some people are very eager to point out that synchronized(this) should be avoided. Instead, they claim, a lock on a private reference is to be preferred.  Some of the given reasons are:    (very popular this one, also has an "accidentally" variant)     Other people, including me, argue that synchronized(this) is an idiom that is used a lot (also in Java libraries), is safe and well understood. It should not be avoided because you have a bug and you don't have a clue of what is going on in your multithreaded program. In other words: if it is applicable, then use it.  I am interested in seeing some real-world examples (no foobar stuff) where avoiding a lock on this is preferable when synchronized(this) would also do the job.  Therefore: should you always avoid synchronized(this) and replace it with a lock on a private reference?      Some further info (updated as answers are given):   we are talking about instance synchronization both implicit (synchronized methods) and explicit form of synchronized(this) are considered if you quote Bloch or other authorities on the subject, don't leave out the parts you don't like (e.g. Effective Java, item on Thread Safety: Typically it is the lock on the instance itself, but there are exceptions.) if you need granularity in your locking other than synchronized(this) provides, then synchronized(this) is not applicable so that's not the issue .fbx export why there are no materials or textures? Legendre's formula counts the number of positive integers less than or equal to a number $x$ which are not divisible by any of the first $a$ primes: $$\begin{align} &amp;\phi(x,a)=\lfloor x \rfloor-\sum_{p_i\le a}\left\lfloor \dfrac{ x }{(p_i)}\right\rfloor+\sum_{p_i&lt;p_j\le a}\left\lfloor\dfrac{ x}{(p_ip_j)}\right\rfloor-\sum_{p_i&lt;p_j&lt;p_k\le a}\left\lfloor \dfrac{x}{(p_ip_jp_k)}\right\rfloor+\dots \end{align}$$ then $\pi(x)=\phi\bigl(x,\pi (\sqrt{x})\bigr)+\pi (\sqrt{x})-1$ as given .  My question is, why does $\bigl\lfloor\phi(x,\lfloor\sqrt{x}\rfloor)+\sqrt{x}-1\bigr\rfloor$ also equal $\pi(x)?$ Why does my transparent PNG not look good? I want to convert a JPG to transparent and using GIMP I added alpha layers and transparency same way I make a GIF transparent I converted it to PNG but it doesn't display well when I load it on my template:    The original image is  Why doesn't it look good when it's made into a transparent PNG? Could I have better luck making a transparent GIF? OR is it the blue colors that don't mix well with the black and I could have more luck using some of the other logos:    Thanks Is it true that if $A$ is a nonnegative primitive matrix, then $AA^T$ is also primitive? Obviously $A^T$ is primitive but in general product of primitive matrices is not primitive.  Any hint?