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Let $T$ and $T^{'}$ be topologies on set $X$ such that $T \leq T^{'}$. If $(X,T^{'})$ is path-connected, is $(X,T)$ path connected? Let $T$ and $T^{'}$ be topologies on set $X$ such that $T \leq T^{'}$. If $(X,T^{'})$ is path-connected, is $(X,T)$ path-connected?  I think it is, and I tried to prove it by contradiction, but I got stuck. Any hint helps! How to find $\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}$? Problem was to find $$\sum_{r\ge 0} \binom{n}{r}\binom{n-r}{r} 2^{n-2r}.$$ My partial progress i tried to motivate such that upper term in binomial terms gets constant rather than variable , so i thought of changing it to form $\binom{n}{n-r}\binom{n-r}{n-2r}$. Now it's of the form $\binom{n}{m}\binom{m}{k}$. We know it is equivalent to $\binom{n}{k}\binom{n-k}{m-k}$. Doing this we get $\binom{n}{2r}\binom{2r}{r}2^{n-2r}$. Now what I need to do as next step as such still nothing the upper or lower terms r any constant values(sum to ) , all are variables still. Now by one answer by Robpratt i got by snake oil , is there any other way to solve this problem ? css background color with floating elements I am looking for help to point me out at a very old novel (likely from the '60s) about two empires at interstellar war. When it comes to starships, one is technologically very advanced, and the other has very low and inefficient technology.  The highly advanced empire has apparently strong chances to win the war, in particular considering a new technology they were developing. I remember the technology to work like this: consider the space as a flat surface, and on this surface, a disk around the spaceship. When the technology was turned on, this disk was stretched out, as made of rubber, so that it produced a cone, with the tip being the spaceship. The spaceship was therefore teleported 10 parsec away (if I remember correctly), but retained all the connections to the original position. If the spaceship moved, the cone moved as well. This allowed to move in position among enemy ships completely unnoticed, by stretching while afar, then moving in the proper position 10 parsecs away, then destretching.  Despite this technology, the empire lost the war because of delays. It's a perfect example of how having a disruptive technology does not guarantee to win against the competitors.  I am trying to find out the author and name of this novel, hopefully with the date of publication and in which journal/magazine. Positive semidefinite cone is generated by all rank-$1$ matrices. "All sites" overview should not require scrolling when selecting a site Is there a way to programmatically deactivate Proces Builder & flows? Is there a way to soql procesbuilder to tun activation off?  For example  List&lt;Processbuilder&gt; pbColl = new List&lt;Processbuilder&gt;();   Processbuilder pb=[SELECT status FROM Processbuilder WHERE status Active];  for(Processbuilder turnoff : pb){             turnoff.status = false;             pbColl.add(pb)} update pbColl;   If not is there a simple way to turn all activated Processes off/on? Where do I find the power cells to restore the bunker door in “Ancient armory”? I am trying to find the power cells to access the amor. Yet the placement is unmarked and I thought I would eventually stumble over them clearing the cauldrons.  Yet after clearing three cauldrons (RHO, XI, and Sigma) I still only have the one from the very beginning.  Where do I find all the power cells? How are local variables referenced in closures? Why is the ISO size for 12.04.3 larger than a CD? Why is the Ubuntu 12LTS 4MB bigger than all the blank CDs I have?  I had to burn it on to a DVD.  This never used to be the case. I'm looking to write some Groovy code to perform tasks inside of install anywhere but because of the platforms we support I'm restricted to java 1.5.  Any idea if this will be sufficient to run the latest groovy? I'm trying to merge two directory trees that have many common elements but also each have elements that are only present in one of the two trees. The main issue I am having is that when mv encounters two sub-directories with the same relative paths it either keeps the source (with -f) or the destination (with -n) but I can't make it take the union of both sub-directories. I could of course use rsync with --remove-source-files but this will actually copy the data and then delete the old files as opposed to a true move. The two directory trees contain several hundred GB of data and are both on same partition so I would love to do a true move if possible for the sake of time.   Right now I have find * -type file -exec mv -n {} /destination/{} but this only moves  files from the source to the destination when the target directory already exists. I can preface this command with something like mv -n * /destination/ but this only brings over the top level directories. Is there any way to do this in a single line? I could  always write a script to check if the directory exists before copying the file but this seems like such a basic task, it seems like there should be an easier way. Is it possible to make my laptop's screen act as a second monitor for my desktop computer? Verification that $\lim_{x \to a}\frac{f(x)-f(a)}{x-a}$ and $\lim_{h\to 0}\frac{f(a+h) -f(a)}{h}$ are equivalent definitions of the derivative. I wanted to verify that for definition of the derivative it is true that:  $$ \lim_{x \to a}\frac{f(x)-f(a)}{x-a}= \lim_{h\to 0}\frac{f(a+h) -f(a)}{h}$$  If I denote $h=x-a$,  we can  let $x\to a$, this means that $h \to 0$. Also notice that $f(x)=f(a+h)$ $$ \lim_{x \to a}\frac{f(x)-f(a)}{x-a}=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$$ As desired. Is the main idea here really just to use a variable substitution? I suppose there is some formal theorem I also use here that when $f(x) \to y$  we can let $g(f(x)) \to g(y)$ if the function $g$ is continuous. "When I grow up, I am going to be a doctor"- Is it correct? When I grow up, I am going to be a doctor.   Is this sentence correct? Personally I think my growing up will happen in future and "I am going to be a doctor" sounds like it is going to happen in next month! But the sentence clearly indicates the speaker is a child. So according to me, "I am going to be a doctor" should be replaced by "I will be a doctor". So I think a more clear sentence would be     When I will grow up, I will be a doctor.   Is my concept correct here? Restrict the access to suggested edit reviews on deleted posts on an answer deleted as rude/abusive is visible, even without logging in (and for users with less than 10k reputation). These users can't see the answer , via the , or a direct link to the .    This is a potential avenue for spammers, just like the suggested edit spam used to be, but is ; according to the :     In addition, in order to reduce the incentive for spammers to make spam edits that get rejected, suggested edit spam is hidden from logged-out users, so they can't show their clients public-facing pages with spam (so they won't get paid).   For this to work, the spammer would need two accounts, one to post the spam and a second one to suggest an edit. Even if the spam post is quickly nuked, the review link will be active forever. One could argue that logged-in users without 10k reputation shouldn't be able to see the suggested edits either; I'm partial to that. Winter 18  does not get values from a parent record Possible Duplicate:          As answered in  I want to use git diff --color=always filename | less -r to get a colored output of my git diffs. Since I use this a lot it would be nice to set up an alias for this. But how to include the filename in this command?  I tried the following without success:  alias gdiff='git diff --color=always $1 | less -r'   Typing gdiff example.php displays the content of example.php with less but ignores the diff part.  Any ideas? In an exam paper, there was a picture of an air stewardess in the aeroplane serving passengers. One of my pupils wrote the following:      The air stewardess works on an aeroplane.    Shouldn't it be the air stewardess works in an aeroplane? Given a Möbius transformation that maps one pair of concentric circles to another pair of concentric circles, why is the ratio of the radii preserved through the map?   I thought about how Möbius transformations are compositions of rotations, scaling, inversion, and translation, and that intuitively, these types of maps shouldn't change the ratio of radii between two circles.   Would it be correct to just say that if $\frac{r_1}{r_2}$ is the ratio of radii between the two circles, then   1) The radii are invariant under translation, $z \mapsto z+a$, so $\frac{r_1}{r_2}$ stays the same  2) Under scaling by a factor $z \mapsto az$, $\frac{ar_1}{ar_2} = \frac{r_1}{r_2}$   3) Under inversion, $z \mapsto \frac{1}{z}$, $\frac{1/r_1}{1/r_2} = \frac{r_2}{r_1}$  Or is there a different/better way to think about this problem?