I want to visit Italy at the end of September as a tourist. I have a refugee travel document, issued by the UK. In older refugee passports there was an additional page mentioning countries where we could go without having a separate visa in consequence of the European Agreement on the Abolition of Visas for Refugees (1959) but on my passport it's not there. Someone told me people are facing problems with it so I would like confirmation:  Do I need to get a Schengen visa? If yes, where can I go in Europe without getting a visa?
I have a UK Refugee passport. Do I need a visa to go to France?   I want to go for two days with my husband, but I'm not quite sure if I can go without a visa.
What limits a motherboard from supporting more memory? My specific question is with the ASUS M3N78-VM motherboard that my home server uses. I currently have 4x2GB installed for a total of 8GB but I was hoping to upgrade to say 16GB (4x4GB). The official specifications of that motherboard says it only supports up to 8GB but it was released back when only 2GB sticks were available. What, if any, would prevent it from working with 4GB sticks now that 4GB sticks are available?  My general question is with motherboards in general... what would prevent a motherboard from working with bigger sticks if its all just the same type (DDR2 for example and same speeds) of memory? What's the C++ way of parsing a string (given as char *) into an int? Robust and clear error handling is a plus (instead of ). Berlin Public Transport Ticket Validity I have purchased an AB zone Berlin 7 day transport ticket and validated it on Sunday 17th April so hopefully it will be valid until Sunday 24th.  I validated the ticket at about 10:00am so I guess 1000 is the time I validated? But what are the numbers 15 and 6 and letters 'So' correspond to? How (if checked by a guard) can they tell if the ticket is valid? And how can I as a customer tell when it is validated to? Insert list of figures in the table of contents Is it possible to create a list of all the figures at the end of a document (this I can do) and make it appear in the table of contents?  So in the the table of content there will be a line saying List of figures ... 19 for example? How do I show that $X = \left\{ (x,y) \in \mathbb{R}^2 \mid x \in \mathbb{Q}\text{ or }y \in \mathbb{Q}\right\}$ is path connected?    Note that $X$ is a topological space with subspace topology $\tau = \left\{ U \cap X \mid U\text{ is open in }\mathbb{R}^2\right\}$.  I only know two things (with respect to path connectedness):   A topological space $X$ is path connected iff for every $p,q \in X$, there is a continuous function $f: [a,b] \to X$ with $f(a)=p$ and $f(b) = q$.    Continuity $f$ is defined as for every open set in $X$, its preimage is open in $[a,b]$ (with subspace topology induced by $\mathbb{R}$ (with standard topology) )   Suppose $X$ and $Y$ are two topological spaces and $f: X \to Y$ is a continuous surjection. If $X$ is path connected, then $Y$ is path connected. divides, not divides and cardinalities In group theory it happens very often that you want to say: "The order of $H$ divides / does not divide the order of $G$." But the length of | is then always the same:  $ |H| | |G| $ produces |H|||G|  which is not ideal at all. (The negation \nmid has just this length too.) The length of "divides" should be longer than the "cardinality length". I could use   $ |H| \big| |G|$  but then there is no good way to say "does not divide" with the same length. Also, \big| is no solution when the expression has to be used in subscripts.   I thought it was just me, but I asked other group theorists, and nobody had a solution. I hope there is one.  I am seeking a division sign which is a little longer than the usual | , allows negation (with the same length) and proportionally correct use in subscripts (also under \sum etc.). It does not look professional if the "order sign" has the same length as the "divides" sign. (And what about combinations with absolute values, norms - although personally I don't use them as often as group orders.) A Terms of Service update restricting companies that scrape your profile information without your permission Symfony plugin sfDoctrineActAsTaggablePlugin not working Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$ Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$.  Writing out several terms, I think the answer is $1/(n+1)$, but I'm struggling to prove this. I would greatly appreciate it if someone could show how to prove this. Show a connected metric space (X,d) is $\epsilon$-chainable for $\epsilon &gt;0$       $\epsilon$-chainable Definition      (X,d) is $\epsilon$-chainable if given any two points $a,b \in X$, $\exists$ a finite set $\{ x_0,x_1,...,x_n\}$ of points in X with $x_0 = a$, $x_n=b$ and $d(x_{n-1},x_n)&lt;\epsilon$, $\forall$ n      Clopen Definition      A set is clopen $iff$ its boundary is empty      Connected Metric Definition      Let $(X,d)$ be a metric space. It is connected $iff$ it has no proper subsets which are clopen      The following are equivalent:         $X$ is connected   The only subsets of $X$ that are clopen are $\emptyset$ and $X$   $\nexists$ non-empty, disjoint, open sets $A$ and $B$ such that $X=A\bigcup B$   $\nexists$ non-empty, disjoint, closed sets $A$ and $B$ such that $X=A\bigcup B$         Theorem      Let $(X,d)$ be a metric space, and let $S$ be any subset of $X$. Then $S$ is closed $iff$ $S^c$ is open.     So how would someone show this exactly?  Let the set $A$ be the set of all elements in $(X,d)$ that can be reached from $a\in X$ in a finite number of steps with length less than $\epsilon$.  Does this mean that there should be no isolated points? (an isolated point would be at distance infinity from any other point, or distance zero from itself, since $\epsilon &gt;0$ we don't want this?)  Knowing that $X$ is connected $\implies$the only subsets of X that are clopen are $\emptyset$ and $X$  Also, by construction of A;   A is the set of all $\epsilon$-chains between any element and $a$  And, so $A^c = \emptyset$ if $A$ was $\epsilon$-chainable (because A has to have a path from all elements to a, for this to be true, the complement wouldn't have anything in it)  Leaving only $A$ as the remaining clopen set  Does this make any sense? Or should an alternative be used, and how to formalize it.  Thanks. Compact space and Hausdorff space A continuous map from a compact space to a Hausdorff space is closed. Why this is true?  Help me please I want to learn why this is correct. Link button in post editor escapes square brackets in code Please help me delete a file? I have a movie named Con Man. Apparently, in Windows, you cannot have a file name with the name CON in it.  Can anyone please help me delete it?  It is stored in F:\Movies\Con Man The name of the file that I want to delete is Con.Man.2018.720p.WEBRip.x264-[YTS.AM]  I know there are a lot of similar issues but I am not able to apply them that is why posting here.  Thankyou  PS :: I am a noob!  Edit: Found this on the internet but cant work it out How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$? I have a dataset with a large number of Yes/No responses. Can I use principal components (PCA) or any other data reduction analyses (such as factor analysis) for this type of data? Please advise how I go about doing this using SPSS. I am foreign graduate student in USA. I recently completed my undergrad (from USA) and have a bit of money saved up. I know that F1 students cannot work off campus and other business activities such stock trading etc are prohibited as well. I wanted to know if opening an account in an investment management firm like Charles Schwabb or Betterment is allowed. I would not be working but I will still be earning a small interest on my deposits. If that is not allowed, are there any good and legal ways for foreign students to earn a little interest on savings? Use of capitals and full stops in bullet points Possible Duplicate:          Let's say I have a bullet point list e.g:   Point 1. Point 2.  Point 3.    Notice I have put a capital letter for the starting letter of every bullet point and a full stop at the end. Is this the correct way to do it?  If not, what is the correct way to do it?  Do I put a capital letter and no full stop, or, the other way around, a full stop and no capital letter? Alternatively, do I not put a capital letter or a full stop? Is it only necessary to put a capital letter on the first bullet point and full stop on the last bullet point?  For reference, I'm from the UK so British-English grammar would apply. Aligning three figures next to each other based on text width Problem:  I am trying to align three figures next to each other with text centered to each figure. This is going well until the box for the third box does not adjust after the text width.  Minimal Working Example (MWE):  \documentclass{article} \usepackage{graphicx} \usepackage{caption}  \begin{document}  \begin{figure}[!tbp]   \centering   \begin{minipage}[b]{0.2\textwidth}     \includegraphics[width=\textwidth]{icon-password.eps}     \caption*{Login system}   \end{minipage}   \hfill   \begin{minipage}[b]{0.2\textwidth}     \includegraphics[width=\textwidth]{icon-shopping.eps}     \caption*{Shopping cart}   \end{minipage}   \hfill   \begin{minipage}[b]{0.2\textwidth}     \includegraphics[width=\textwidth]{icon-clock.eps}     \caption*{Temporary information}   \end{minipage} \end{figure}  \end{document}   Outputs:   Desired outcome:  To get the width to adjust after the text so that "Temporary information" can be written out in one line instead of two. Showing that $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$, when $f$ is even Does walking fast prevent eggs hatching? I have a couple of eggs that I've picked up from some Pokestops (I'm lucky enough to live above one).   I walk a lot. I get to work through a 2 mile walk / train journey every day, so I should be picking up loads of steps for the eggs I have. However, because I'm normally always running late... for everything, I have to rush everywhere, so I walk at an above average pace.   Today I've walked around 5 miles, but the game has registered 100 metres. Is this because I walk too fast, or because of something else?  It's worth pointing out that I have the app open when walking... mostly because I have to catch them all. (This made me miss a train this morning... luckily I was early :D)