Is there any way to dismiss JavaScript alert() programatically OR by pressing any key and not necessarily pressing the ENTER key?
I want to be able to close an alert box automatically using Javascript after a certain amount of time or on a specific event (i.e. onkeypress).  From my research, it doesn't look like that's possible with the built-in alert() function.  Is there a way to override it and have control over the dialog box that it opens?  Also, I don't want an override that shows a hidden div as the alert.  I need an actual dialog box.
Set of limit points of a subset of a Hausdorff space is closed. Let $X$ be a Hausdorff space and $A\subset X$.  Define $A'=\{x\in X\mid x\text{ is a limit point of }A\}$.  Prove that $A'$ is closed in $X$.  Relevant information:  (1.) Every neighborhood of a point $x\in A'$ contains a point $y\in A'$ distinct from $x$ (in fact, in a Hausdorff space, every neighborhood of $x$ contains $\infty$-many points of $A$ distinct from $x$)  (2.) In a Hausdorff space, every sequence has a unique limit.  On first glance, it should be an easy proof, but I've made little progress.  I was planning on showing $\overline{A'}=A'$.  Firstly, $A'\subset \overline{A'}$ trivially.  To show $\overline{A'}\subset A'$, proceed by contradiction.  Assume there exists $x\in\overline{A'}$ such that $x\not\in A'$.  This should yield an easy contradiction but I don't see it.  In particular, I'm unsure if the fact that $x\in\overline{A'}$ implies that there actually exists a sequence in $A'$ converging to $x$.  By (2) we know all sequences have unique limits, but do we know that elements in the closure are limits of sequences?  If this is true, it should yield an easy contradiction.  Any help? How permanent magnets work?Are they working with motion of electrons? How do I deal with a bored or distracted player bringing down the group? My group have been playing various D&amp;D style games for a few years,  I have one player however that often will drift off and start playing on his phone  and several times has actually fallen asleep (with his character having to be "nudged" along by his fellow players).  He insists he's "just tired" and "isn't bored" — and besides, he keeps insisting we all get together for D&amp;D and wants to play.  Unfortunately we're all pretty busy, so if I'm to get the group together at all it's usually late on a Sunday. After talking to the player he insists it's not the game, or the story, he just happens to be tired. But, it's the same excuse every week, and the odd times we've played at different times or situations, the falling asleep has happened.  How do I cope with:   The morale destroying (for me) idea of your players falling asleep through a campaign and story you've put lots of work into. The general tone of the room when this player loses interest. (Occasionally a couple of others zone out into Facebook when they're not "active" because the players have split up / gone unconscious).   Given that it's an "all or nothing" situation with the group (there are only 3 players and they won't play unless all 3 are around), how do I best deal with it? What is the proof that a force applied on a rigid body will cause it to rotate around its center of mass? Why this polynomial is irreducible? Let $K=\mathbb{Z}_p(t)$. How to prove $f(x)=x^p-t$ is irreducible in $K[x]$? What steps can I take to try to recover lost or inaccessible data from any storage device?  Answers:   This applies to any computer storage device, e.g. internal/external hard drives, USB sticks, flash memory. The most important thing is to STOP using it, any type of I/O can ruin your chances of a recovery.     We have separate questions covering common problems with USB flash drives in greater detail: $ d = \gcd(a,b)\Rightarrow\ \gcd(a/d,b/d) = 1$ Could someone please help me with this proof?     Suppose that $a, b \in N$, and $d = \gcd(a, b)$. Since $d$ divides $a$, we have $a = de$ for some integer $e,$ and similarly $b = df$ for some integer $f$. Prove that $\gcd(e, f) = 1$.   I understand why it works. Since d is all the common factors of $a$ and $b, e$ and $f$ had no common factors, therefore the $\gcd(e,f) = 1$. But how do I prove this?  Thanks in advance. I am currently working on an update of tkz-euclide and new documentation but I just met the following problem. If I compile this code with pdflatex or lualatex and pgf 2.1 (The library babel does not exist with pgf 2.1) then the code compiles fine but with 3.0 pgf I must comment on the call \usetikzlibrary{babel} (pgf doc gives so I recommend to always load this library by saying \usetikzlibrary{babel}).  I'd like to know where the error occurred. Especially I would like to work around this problem.   With \usetikzlibrary{babel} I get the error Latex Error: ./untitled 2.tex:46 Undefined control sequence. l.1 \tkz@line@start  Bug in babel, bug in tikzlibrary{babel} bug in my code. I would like to keep  \usetikzlibrary{babel} but to avoid the error.  %!TEX TS-program = pdflatex \documentclass[a4paper]{article} \usepackage[french]{babel} \usepackage{tikz}  \usetikzlibrary{calc} %\usetikzlibrary{babel}  \makeatletter \pgfkeys{%     tkzdrawline/.cd,     start/.code         =  {\gdef\tkz@line@start{#1}},%     end/.code           =  {\gdef\tkz@line@end{#1}},%     start style/.code   =  {\tikzset{tkzstartstyle/.style={#1}}},%     end style/.code     =  {\tikzset{tkzendstyle/.style={#1}}},%    /tkzdrawline/.unknown/.code ={\let\searchname=\pgfkeyscurrentname\pgfkeysalso{%                          \searchname/.try=#1,/tikz/\searchname/.retry=#1}}}  %&lt;--------------------------------------------------------------–&gt; %                              Draw line %&lt;--------------------------------------------------------------–&gt; \pgfkeys{tkzdrawline/.cd, start/.code       =  {\def\tkz@line@start{#1}},% end/.code         =  {\def\tkz@line@end{#1}}, start style/.code =  {\tikzset{tkzstartstyle/.style={#1}}}, end style/.code   =  {\tikzset{tkzendstyle/.style={#1}}},   /tkzdrawline/.unknown/.code   = {\let\searchname=\pgfkeyscurrentname                                  \pgfkeysalso{\searchname/.try=#1,                                 /tikz/\searchname/.retry=#1}}}%         %&lt;--------------------------------------------------------------–&gt;  \def\tkzDrawLine{\pgfutil@ifnextchar[{\tkz@DrawLine}{%                                   \tkz@DrawLine[]}}%   \def\tkz@DrawLine[#1](#2,#3){%  \begingroup   \pgfkeys{tkzdrawline/.cd,start = {}, end = {}, start style={}, end style={},}   \pgfqkeys{/tkzdrawline}{#1}   \draw[ /tkzdrawline/.cd,#1] (#2) to%     node [at start,left,/tkzdrawline/.cd,tkzstartstyle] {\tkz@line@start}%     node [at end,right,/tkzdrawline/.cd,tkzendstyle] {\tkz@line@end} (#3);  \endgroup  }%   \makeatother   \begin{document}    \begin{tikzpicture}      \node (A) at (0,1) {};       \node (B) at (5,2) {};       \tkzDrawLine[end = {$\alpha$}](A,B)       \end{tikzpicture}  \end{document} I cleared all my browser data and visited  and it logged me back in. How?  This happens on Chrome on OS X and Windows 7.  Surely the security concerns outweigh the benefits to the user. What if a family member or friend wants to use my machine? I wouldn't want them logged in as me.  Update  My mistake, I should have said data &#8212; all those checkboxes clicked/checked. I completely cleared everything, yet it still logs me back in.  Update 2  Here's a video to prove it: Nicely discouraging serial upvoting Probably everybody who is very active on SO has experienced cases of random, serial upvoting as the expression of thanks from a (usually newbie) user. The pattern is usually 5 or more upvotes on your latest answers. The upvotes will be detected within 24 hours, and removed. (Or so I assume.)  No harm done except maybe a minor irritation for people after one of the &quot;rep cap&quot; badges because they are shown a higher reputation than they actually have. Still, I was thinking whether, in an attempt to channel the positive attitude into more productive areas, it would be worth showing a tooltip on the third upvote for the same person within time frame x:  Please use one upvote, a comment, or some quality content of your own, to say thank you Large numbers of upvotes for the same user may become invalid as part of our spam detection mechanisms.  Does this make sense? Maybe it is too localized and rare, but it has happened to me three or four times in the last two weeks alone. You decide.  Clarification: I am not advocating putting up any kind of barrier on upvotes. . I'm suggesting informing people of the barrier and the futility of their ten or so upvotes for a single person because those votes are likely to be revoked.  Update: @devinb makes the good point that &quot;saying thanks&quot; is not the only possible reason for serial upvoting, and the message wouldn't make sense then. The point still stands: Your votes aren't going to count, and when it comes to up voting (which is always some sort of positive behaviour, however misguided if done indiscriminately) I think it's fair to inform people of this. Possible Duplicates:             I heard a professor of mathematics explaining that the following conslusions are true.   $\textrm{False}\Rightarrow \textrm{True}$ $\textrm{False}\Rightarrow \textrm{False}$   Could someone please explain to me why both conclusions are true? I am especially interested in the first case. I'd like to hear two explanations if possible: One for mathematicians and one for non-mathematicians. :-) Proving that $(3n)!$ is divisible by $n! \times (n + 1)! \times (n + 2)!$ if $n$ is greater than 2 Prove that:  If $n$ is greater than 2, then $(3n)!$ is divisible by $n! \times  (n + 1)! \times  (n + 2)!$  From Barnard &amp; Child's "Higher Algebra".  I know that the highest power of a prime $p$ contained in $N$! is  $$ \left\lfloor{\frac{N}{p}}\right\rfloor + \left\lfloor{\frac{N}{p^2}}\right\rfloor + \left\lfloor{\frac{N}{p^3}}\right\rfloor ... $$  I'm unable to show that the formula above gives a higher value for $N = 3n$ than for the sum of its values when $N$ = $n$ , $n + 1$ and $n + 2$, considering the condition that $n &gt; 2$. Great Theorem - Help name the Sci-fi book read long ago I'm trying to find a book I read long back. This book may have won a Hugo/Nebula (Trust me I've googled to death this one!).  Don't remember much of the story (that's why I want to re-read it). What I remember is that some of the powers in the book are called "Gods". And they have brains the size of the planets (or planets being their brains).  The protagonist in the climax, solves a theorem (The Great Theorem, perhaps) by remembering one of the obscure theorems (Jacuzi (?)) and incorporating it. This [Great] theorem helps spaceships helps find the shortest route from point A to B. He surprises the antagonist by reaching the vantage point quickly with the help of his new knowledge, and blasts his enemy's spaceship.  The ending of the book - as I remember it is - that the protagonist wants to become a God himself.  Please let me know the title &amp; author of this book if you've read it.  PS #1:  BTW, It's not Asimov's The Gods Themselves :) PS #2:  It's published at the least before 1998 How can I make an overset symbol at the same size and height as \bar except with a different symbol? Apologies for the cryptic title, but I was not sure how best to phrase the question in one sentence. I am essentially trying the create a vector symbol with \rightharpoonup as the top instead of the arrow.  The output to the following code will feature an overset harpoon over x, next to a simple \bar{x}. Observe that on the left, the top symbol is significantly higher and bigger than on the right one.  \documentclass{article}  \usepackage{amsmath,amsfonts,amsthm,amssymb}  \newcommand{\vect}[1]{\overset{\rightharpoonup}{#1}}  \begin{document} $$ \vect{x} \quad \bar{x} $$ \end{document}   I would like to create something that looks in every way like the right hand symbol, except using something that looks like \rightharpoonup at the top.  If this command sequence already exists in LaTeX, please use that as an answer, because I have searched for and failed to find it so far. Does $\pi$ contain all possible number combinations? Isomorphism between two models of Hilbert's Axioms with different notions of congruence I've been reading about Hilbert's Axioms of geometry, and I have two models on $\mathbb{R}^2$ which satisfy the usual axioms of incidence, betweenness, and congruence.   The first model has a distance function given by  $$ d(A,B)=|a_1-b_1|+|a_2-b_2| $$ where $A=(a_1,a_2)$, $B=(b_1,b_2)$, and two segments $AB\cong CD$ if $d(A,B)=d(C,D)$.  The second model has a distance function given by $$ d(A,B)=\sup\{|a_1-b_1|,|a_2-b_2|\}, $$ and again $AB\cong CD$ if $d(A,B)=d(C,D)$.  The author claims that these two models are isomorphic in that there is a bijection between the sets of points in $\mathbb{R}^2$ that preserves lines, betweenness and congruence. I've been toying with it for a while, but it seems to me that the notions of congruence in particular are too different.   For example, in the first model, if $A=(0,0)$ and $B=(2,1)$, and $C=(2,0)$ and $D=(5/2,5/2)$, then $AB\cong CD$ since both segments have length $3$. To preserve congruence, I suppose two segments are congruent if the maximum of the displacement along the $x$-axis or $y$-axis by traveling along one segment is equal to the maximum of the displacement along either axis when traveling along the other segment. In this sense, how could you shape the segment $CD$ so that the maximum of the displacements along either axis is the same as the maximum of the displacement for whatever the image of $AB$ is?  In the first model, $AB$ is congruent to any segment which is the hypotenuse of triangle with legs that sum to $3$, and this sum could split into any two terms. I just don't see a function that could handle all these possibilities at once. Does anyone know of a particular isomorphism between these two? Thanks. What would happen to a star if a Dyson sphere lined with mirrors reflected a significant portion of the stars light back to the star I have looked for similar questions here on stack exchange. The closest example to this that I found is . That question assumed less than perfect absorption of material lining the inside of the sphere, and the effect of thermally re-radiating it inside the sphere.  A lot of the answers focused on that much of the thermal radiation would not be reflected back to the star, and is likely to be reabsorbed by the sphere at another location. On the other hand, this question asks what would happen if the Dyson sphere, using mirrors, deliberately refocus a significant portion of the rays directly back at the star. I want to buy a new notebook in the near future to play games on it. I'm looking forward for the new Steam for Linux client. My question is that where does the Steam install the games, on the home partition, or in the usr folder, or somewhere else? I do not use pre-release (i.e. beta) versions of either Steam or Ubuntu. Why did the Ministry of Magic choose an ax for carrying out a death sentence? My prof gave us this definition of an Echelon system:     A system of m linear equations in n variables is called an echelon system if         m ≤ n.   Every variable is the leading variable of at most one equation.   Every leading variable is to the left of the leading variables of all lower equations.   Every equation has a leading variable.      I don't understand this definition at all.  Q1)If you have this system of equations:  $ax_1 + bx_2 + cx_3 = z$  $dx_1 + ex_2 + fx_3 = z$  its not an echelon system because of multiple equations having the same variable as the leading variable and therefore violating rule #2.  But if you switch around variables and rewrite the equation as:  $ax_1 + bx_2 + cx_3 = z$  $ex_2 + dx_1 + fx_3 = z$  It becomes an echelon system! Why would the way you write a system of equations affect its definition?  Q2)Why would point number three matter? Can't you just move around the equations so that they are in order of variables?  Q3)Why would point number four matter? If you have a equation in the system that doesn't have a leading variable can't you just disregard it? For example if you have the equation $0x = 0$ in the equation can't you just ignore it to have a echelon system since it doesn't affect the solution of the system?