Let $\emptyset, A$ be an empty set and a set respectively.  Now consider Catesian product of the above.  $\emptyset  \times A$  I think it should be $\emptyset $ but I can't prove it.
Let $A \times \emptyset = \{(x,y)| x\in A, y \in \emptyset \}$. We know there is no element in $\emptyset$. But how does it follow that $A \times \emptyset = \emptyset $?
I'm trying to load a field value (machine name: field_theme_image) from a content of node value 3, in "page--front.tpl.php".  &lt;?php   $node=node_load(3);   print $node-&gt;field_theme_image[$node-&gt;language][0]['uri']; ?&gt;   Using the above code I'm not getting full URL value. Please see the image below. I can see the path value in Devel themer. I know how much space is required to install Ubuntu.  How much space does a Ubuntu installation take up on the disk? When I go to any github.com page in Chrome, I get a big ugly error:  You attempted to reach github.com, but the server presented a certificate issued by an entity that is not trusted by your computer's operating system. This may mean that the server has generated its own security credentials, which Chrome cannot rely on for identity information, or an attacker may be trying to intercept your communications. You cannot proceed because the website operator has requested heightened security for this domain.  The same thing happens (in Chrome and with curl) when I go to  too. This strange problem started about a week and a half ago. Here's what I see when I click the broken lock icon in the address bar:   But gist.github.com works just fine:   It doesn't work with curl either:  Everything works fine in Firefox. How can I fix my root CA problem? Here's what it looks like in Firefox:   Update: I noticed that the first certificate in the chain is different in my broken Chrome/Safari as compared to Chrome on my other computer.   (There's no nasty red X anymore because I trusted it in Safari.) See how the issuers are different? What can I make of that? How to execute .sh file on Windows? When I am trying to execute a file(name.sh) in the command line by the command  ./name.sh  , I am getting the error that:     "." is not recognized as an internal   or external command, operable or batch   file   please help me execute the .sh file Some sites are . What additional sites should the chat support?  One suggestion per answer. For all $1 \leq i < j \leq k$, the subtrees $T_i$ and $T_j$ have a vertex in common. Show that $T$ has a vertex which is in all of the $T_i$. Can someone please verify my proof or offer suggestions for improvement? I am aware that there is a similar question elsewhere, but I want help with my proof in particular.     Let $T_1, \ldots, T_k$ be subtrees of a tree $T$ such that for all $1 \leq i &lt; j \leq k$, the trees $T_i$ and $T_j$ have a vertex in common. Show that $T$ has a vertex which is in all of the $T_i$.   Suppose, for the sake of contradiction, that for every vertex $v \in V(T)$, there is a subtree $T'$ in the list $T_1, T_2, \ldots, T_k$ such that $v \notin V(T')$.  Pick $v_1 \in T_1$. There is a subtree $T^{(1)}$ such that $v_1 \notin T^{(1)}$. $T_1$ and $T^{(1)}$ have a vertex $y_1$ in common. Now, there exists a tree $T^{(2)}$ such that $y_1 \notin T^{(2)}$. But now, $T^{(2)}$ and $T^{(1)}$ have a vertex $y_2$ in common. Also, $T_1$ and $T^{(2)}$ have a vertex $y_3$ in common.  There exists a path $p_1$ in $T_1$ from $y_3$ to $y_1$. Also, there is a path $p_2$ in $T^{(1)}$ from $y_1$ to $y_2$ and a path $p_3$ in $T^{(2)}$ from $y_2$ to $y_3$. Note that the path formed by the union of $p_1$ and $p_2$ passes through $y_1$, while $p_3$ does not, since $y_1 \notin T^{(2)}$. Therefore, the union of $p_1, p_2$, and $p_3$ forms a cycle, contradicting the fact that $T$ is a tree. Therefore, it must be the case that there is a vertex $x$ in all of the $T_i$. I am trying to find all 1- or 2- dimensional Lie Algebras "a" up to isomorphism. This is what I have so far:  If a is 1-dimensional, then every vector (and therefore every tangent vector field) is of the form $cX$. Then , by anti-symmetry, and bilinearity:  $$[X,cX]=c[X,X]= -c[X,X]==0$$  I think this forces a unique Lie algebra because Lie algebra isomorphisms preserve the bracket.  I also know Reals $\mathbb{R}$ are the only 1-dimensional Lie group, so its Lie algebra ($\mathbb{R}$ also) is also 1-dimensional. How can I show that every other 1-dimensional algebra is isomorphic to this one? Do I use preservation of bracket?  For 2 dimensions, I am trying to use the fact that the dimension of the Lie algebra g of a group $G$ is the same as the dimension of the ambient group/manifold $G$. I know that all surfaces (i.e., groups of dimension 2) can be classified as products of spheres and Tori, and I think the only 2-dimensional Lie group is $S^1\times S^1$, but I am not sure every Lie algebra can be realized as the Lie algebra of a Lie group ( I think this is true in the finite-dimensional case, but I am not sure).  I know there is a result out there that I cannot yet prove that all 1- and  2-dimensional Lie algebras are isomorphic to Lie subalgebras of $GL(2,\mathbb{R})$ (using matrix multiplication, of course); would someone suggest how to show this last? Thanks. What exactly are the Python scoping rules?  If I have some code:  code1 class Foo:    code2    def spam.....       code3       for code4..:        code5        x()   Where is x found?  Some possible choices include the list below:   In the enclosing source file In the class namespace In the function definition In the for loop index variable Inside the for loop   Also there is the context during execution, when the function spam is passed somewhere else. And maybe  pass a bit differently?  There must be a simple reference or algorithm somewhere.  It's a confusing world for intermediate Python programmers. Why is rabies incurable? I'm still not sure about the mechanics that lead to rabies being incurable. I know that it can be treated before any symptoms show up, but why is it that once symptoms show the person is a dead man walking? How to draw circle, square, and triangle marks in TikZ picture? Is $0 \, \mathrm m = 0 \, \mathrm s = 0 \,\mathrm {kg} = 0$? How do we define $[0 \, \mathrm m]$?  I once was given an assignment where I was asked to deduce and write down some physical quantity. It turned out that this quantity was $0$ in one unit or another, so I decided to drop the physical units because I figured that it didn't matter anyhow.  But I got to thinking:  If $0 \, \mathrm m = 0$, then we could add units of different dimensions (an operation not normally defined) like so: $5 + 0 \,\mathrm m = 5$.   Does $[ \cdot ]$ map $0 \, \mathrm m$ to both $0$ and $L$? Is there an established regime of thinking about physical quantities as an abstract algebra? Perhaps as a vector space where formal sums such as $5 \, \mathrm m + 10 \, \mathrm s$ are allowed? Would it be improper to drop the units of measurement of $0 \,\mathrm m$ like this in an academic paper? CString to char* We are using the CString class throughout most of our code. However sometimes we need to convert to a char *. at the moment we have been doing this using variable.GetBuffer(0) and this seems to work ( this mainly happens when passing the Csting into a function where the function requires a char *). The function accepts this and we keep going.  However we have lately become worried about how this works, and whether there is a better way to do it.  The way i understand it to work is it passes a char pointer into the function that points at the first character in the CString and all works well.  I Guess we are just worried about memory leaks or any unforseen circumstances where this might not be a good idea. My father in law is travelling to France, for 4 days, (01/07-04/07). Is there a minimum duration for which the travel insurance needed to apply for a Schengen visa must be valid? In other words, can his travel insurance cover just the duration of the trip, or should it last longer? A set with n elements has 2^n subsets I don't understand why a set with n elements has 2^n subsets. How is this calculated? I realize that {123} has empty set - 1-2-3-1,2-1,3-2,3-1,2,3 but how is the formula derived? How to make a circular array using offset object as center? We know that the there is an Object Offset option for Array modifier, but when trying to make a circular array with that offset object as the center of the array circle, it always doesn't behave quite like what's expected. So how to make a nice and controllable circular array?  P.S.: There is an old  about moving the object's origin. I wonder if there is any way to do this without changing the location of the origin. :) How do I space quads evenly? Is there a way in blender or an addon that allows me to select a bunch of faces (or verts) and evenly distributes (or relaxes) them into quads clean quads.  Without screwing up the vert ID/order.  Edit: I would like to have the shapes of my quads to turn into perfect squares while maintaining the volume of the object and the vertex order. RC4 , Is it possible to find the key if we know the plaintext and ciphertext? Is it possible to find the key if we know the plaintext and ciphertext with RC4. How should I write the algorithm? How can I pause a render in progress and then resume it at a later time? Is there any pair of 128-bit strings M and K such that AES$_K$(M) = M?  If yes, how do I go about (efficiently) finding such a pair? I have a Google Apps email (me@business.com) set to auto-forward all mail to my personal Gmail (me@gmail.com).  I have added the me@business.com address to my personal Gmail so I can send mail as me@business.com from within my personal email account.  It's a fairly common way to consolidate email accounts.  Here's my problem.  I want me@gmail.com to filter all emails being forwarded from me@business.com and apply the Business label.  So I tried it two different ways:  A) Use an Alias:   me@business.com forward to me+business@gmail.com Create filter "To:me+business@gmail.com" in me@gmail account Have filter apply Business label   B) Use Googlemail.com address:   me@business.com forward to me@googlemail.com Create filter "To:me@googlemail.com" in me@gmail account Have filter apply Business label   However, neither of these filters work.  I've tested it using a third email address, me@yahoo.com.  Whenever I send an email from me@yahoo.com to me@business.com, the email is properly forwarded to me@gmail.com, and I can reply to me@yahoo.com using the me@business.com address.  But Gmail does NOT apply the Business label.  This is case using either filtering method A or B.  But if I send an email from me@yahoo.com DIRECTLY to me+business@gmail.com or me@googlemail.com, the filter works properly and applies the label.  So there must be something going on with the forwarding of me@business.com to me@gmail.com.  Does anyone know how to get this working?