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Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous map such that $\overline{f(\mathbb{N})} = \mathbb{R}$. Prove that $f$ is surjective.  I tried to use the definition of the continuity and $\overline{f(\mathbb{N})} = \mathbb{R}$ to show that $f$ is surjective but it didn't work. Since $f$ is continuous, the preimage of every open set is open. Also it results in $f(\overline{A})\subseteq \overline{f(A)}$ for every set $A$. How this couples with $\overline{f(\mathbb{N})} = \mathbb{R}$ to show surjectivity? Adding 2 favorite tags at one time combines tag after page refresh I was adding some tags to my favorite tags today and noticed tags being joined together when the page is refreshed.  Steps to reproduce:   Click edit on Favorite Tags Create the first tag and click add [I added django in this case] Add a second tag and click add [I added python in this case] Refresh page  Favorite tags now show previous tags + djangopython   In the edit section, it appears as if the tags are added separately, but when I refresh the page they get joined  OS: MacOSX-10.8  Browser: Chrome, Firefox "There is no point in" or "There is not a point in" I was thinking about these negations. Do these mean the same thing?     There is no point in ...   There is not a point in ...   or:     I have no clue   I do not have any clue     etc. I posted a question and a couple of respondents asked for the source code. I could only post portions of the code.  If attachments were allowed I could have uploaded a .zip file with the whole application.  Will this be a feature in SO? Why sets of nonzero measure contain almost all the point from Borel-cantelli Leems? An integral domain $A$ is exactly the intersection of the localisations of $A$ at each maximal ideal This result appears to be ubiquitous as an algebra exercise. How do you prove this result?     Let $A$ be an integral domain with field of fractions $K$, and let $A_{\mathfrak{m}}$ denote the localisation of $A$ at a maximal ideal $\mathfrak{m}$ considered as a subring of $K$. Prove that $$A = \bigcap_{\mathfrak{m}} A_{\mathfrak{m}}\,,$$ where the intersection is taken over all maximal ideals $\mathfrak{m}$ of $A$. I am trying to capitalize Western Canada or western Canada properly and am wanting a definition for when the "ern" is added as a suffix to a locational distinction of a proper noun.  I believe the capitalization is related to the suffix and am grouping two questions.  After researching on the internet I have found a trend that areas of discrete area become capitalized along with their proper noun   East New York West Hollywood East London (Britain) East Los Angeles   Some locations have "ern" added to their locational modifier.    Web Examples (from government sites):   eastern Canada ( Western Canada ()   All the examples listed contain discrete areas.  The "ern" suffix is used for larger areas, but I am unable to determine when an area becomes large enough. I've got a problem while unwrapping UVs. When I'm unwrapping UVs it showing me this message.     Why is this happening? Does gravity affects on electrons in a noticeable way? Is there a way to force a free electron move in a low speed in order to observe it actually hitting 'on the floor'? Every finite group is isomorphic to some Galois group for some finite normal extension of some field. Prove that $A = B = C = D$ $A, B, C, D$ are sets, where $A \not = \emptyset$ and $B \not = \emptyset$. Equation $(A \times B) \cup (B \times A) = C \times D$ is true.  Prove that $A = B = C = D$.  I tried to find $A, B, C, D$ separately, but that didn't get me anywhere. Any ideas how to solve it? I'm self-studying Guillemin and Pollack, but I'm stuck on Problem 3 of section 2. It says that if $V$ is a vector subspace of $\mathbb{R}^N$, then $T_x(V)=V$ if $x\in V$.  If $x\in V$, then since $V$ is a manifold, there is a local parametrization $\phi: U\to V$ where $U$ is open in $\mathbb{R}^k$. Without loss of generality, we can require $\phi(0)=x$. Then $T_x(V)$ is defined to be the image of $d\phi_0$ on $\mathbb{R}^k$.   An arbitrary element of $T_x(V)$ looks like $$ d\phi_0(v)=\lim_{t\to 0}\frac{\phi(0+tv)-\phi(0)}{t}=\lim_{t\to 0}\frac{\phi(tv)-x}{t} $$  but this doesn't seem very useful to show $T_x(V)=V$. What is the right approach? You should choose your Display Name when signing up Dense sets in Hilbert space of square summable sequences? Let $l^2 = \{x = (x_n)| x_n \in \mathbb{R}, \sum_{n=1}^{\infty} x_n^2 &lt; \infty\}$  be the Hilbert space of square summable sequences and let $e_k$ denote the $k^{th}$ co-ordinate vector (with $1$ in $k^{th}$ place, $0$ elsewhere). Which of the following subspaces is NOT dense in $l^2$?   span$\{e_1-e_2, e_2-e_3, \ldots\}$ span$\{2e_1-e_2, 2e_2-e_3, \ldots\}$ span$\{e_1-2e_2, e_2-2e_3, \ldots\}$  span$\{e_2, e_3, \ldots\}$   By definition, we have to prove that for any $x \in l^2$ and for any $\epsilon &gt; 0$, there exists $y$ in the respective sets such that $\|x-y\| &lt; \epsilon$. How to proceed further? Is there a way to unlock PDF documents? I realize most PDF documents are locked for a reason, but I spend several hours a week editing datasheets provided directly from my vendors to be sent to customers. Obviously it's very annoying and somewhat of a show-stopper when I download a datasheet in PDF format, then I am not able to draw boxes, add comments, or draw arrows to the supplied product.  One of our vendors told us the way around that was to print a physical copy, then scan it in as a PDF - not the most technological or aesthetically pleasing solution I have found.  I at least would like to be able to print the PDF to another PDF so that the new copy is unlocked for editing. Is there any way around this? which of the following is NOT a possible value of $(e^{f})''(0)$?? Let $f$ be an analytic function on $\bar{𝐷} = \{z \in \mathbb{C}: |z| \le 1\}$. Assume that $|𝑓(𝑧)| ≤ 1$ for each $z\in \bar{D}$. Then, which of the following is NOT a possible value of $(e^{f})''(0)$??  $(A) 2$  $(B) 6$  $(C) \frac{7}{9}e^{\frac{1}{9}}$  $(D) √2 + 𝑖√2$  So I take a look at Cauchy Integral Formula and then $$(e^{f})''(0)=\frac{1}{\pi}\int_{|z|=1}\frac{e^{f(z)}}{z^3}$$  Taking modulus on both sides we have $$|(e^{f})''(0)| \le \frac{e}{\pi}2\pi=2e$$  Hence (b) cannot be the choice.   Is this alright!!  Thanks for the help! My linear algebra professor proved that      Diagonalizable matrices with complex values are dense in set of $n \times n$ complex matrices.   He defined a metric (I believe) that was somehow related to the usual metric on $\mathbb{R}^{n^2}$.  Then somehow proved that diagonalizable matrices were dense because for any matrix $A$ if $\det(A - \lambda I) = 0$ on an open subset, then $\det(A - \lambda I)$ was the zero polynomial.  I Googled around a bit and found some stuff talking about the Zariski topology, and I am not sure this is what I want. Most of what I have found on this topology is much more general than what he was doing.  Does anyone know any resources that would explain how to go about proving this statement?  Edit: I found out how to prove this the way my professor did.  We define the norm of a matrix by $$ |A| = \max\{|Ax| : |x| = 1 \}.$$ Now we have a distance $d(A, B) &lt; \epsilon$. You can prove that if $(A - B)_{ij} &lt; \epsilon/n$, then $|A - B| &lt; \epsilon$.  Let $p(x)$ be the characteristic polynomial of $A$, an $n \times n$ matrix.  $$p(x) = \prod_1^n (x - x_i) = x^n + (-1)\sigma_1 x^{n - 1} + \cdots + (-1)^n\sigma_n$$ where $\sigma_1, \ldots, \sigma_n$ are the elementary symmetric polynomials of $x_1, \ldots, x_n$ which are the eigenvalues.  The discriminant of the characteristic polynomial is a symmetric polynomial, therefore it can be written in terms of the elementary symmetric polynomials, which in turn can be written in terms of the entries of the matrix.  But since the discriminant is a polynomial, it only has finitely many roots. Therefore for $\epsilon &gt; 0$, by changing the entries of the matrix less than $\epsilon / n$ we can find a new matrix $B$ such that $|B - A| &lt; \epsilon$ and the discriminant is not zero.   The discriminant being not zero means $B$ has distinct eigenvalues, thus has a basis of eigenvectors. Therefore, it is diagonalizable.  Thus the set of diagonalizable matrices is dense in the set of matrices with respect to that metric. I've created a new user by using "User Accounts" in "System Settings". How do I grant rights so that this new user can have the same level of access as root? Can I delete unnecessary device simulators of Xcode? I found that Xcode on my laptop consumes much of its storage, so I am trying to delete unnecessary files in Xcode.  There are four directories under Application/Xcode.app/Contents/Developer/Platforms/. I feel that WatchSimulator.platform and AppleTVSimulator.platform are not necessary, because I use Xcode almost only for basic iOS development.  I would like to know if I can delete those platforms directories directly? How to specify a short int literal without casting?