I'm using Stack&nbsp;Exchange to learn new things and a little bit like a wiki... And from time to time I come across new topics with many questions and answers. I upvote them because they are really good and/or I thought about a similar question.  It would be nice to easily find them again after some time, when facing a problem which is related and not have to go through all (in my personal point of view low-quality or not appropriate) questions/answers again. I know I can mark questions as favorites, but marking them all will result in too many favorites which going through all the time is also tedious.  I'm thinking about a (live-)search/filter at the votes-section beside the other filters. If not at all, it would also be an improvement for the answers-, questions- and favorites-sections. Additional, I think this would be easier to handle for new (non-tech) users than a advanced search-syntax.    The same problem occurs for example in the flagging history: For flagging recurring questions as duplicates I use the same high-voted question from time to time. - I can't mark them all as favorite to easily find them and using the browser-integrated search for >1000 flags and therefore >25 sites isn't really applicable.    Based on  as an alternative an  like voted:byme, maybe also upvoted:byme and downvoted:byme would be a little bit more inconvenient option (whereby :byme may be unnecessary because of the "anonymous votes"-policy).
When I vote up questions or answers, I'm not only increase poster's reputation but also saving a link to the post in my profile — the "Votes" tab shows all posts I have voted on, so it's possible to view the posts again without searching.  But finding the needed answer is uncomfortable when the number of posts voted on is too large.  Please add an option to search in my voted posts.  Adding a new  would be easier. But I think that a filter in the "Votes" tab will be more comfortable (at least for me). Also this filter could be used in other tabs in the profile, e.g. the "Answers" and "Questions" tabs.
Suppose that C is a closed convex subset of $\mathbb R^n$ and $x \in \mathbb R^n$. The projection of $\mathbf x$ onto C is the closest point $\mathbf y \in C : \mathbf z = \mathbf y$ minimizes ||$\mathbf z - \mathbf x$||$_2$ over all $\mathbf z \in C$.  (a) Show that the minimum exists.  (b) Show that there is only one minimizer: that is, show that if $\mathbf y_1$ and $\mathbf y_2$ both minimize ||$\mathbf x - \mathbf z$|| then $\mathbf y_1 = \mathbf y_2$   My ideas:  For (a), to show that the minimum exists I will need to show that it is closed and bounded. We already know it's closed because C is a closed convex subset so I just need to show it's bounded. Am I wanting to minimize ||$\mathbf x - \mathbf z$||$_2$? This is just the Euclidean norm which would be $\sqrt{(\mathbf z - \mathbf x)^2}$=$\sqrt{\mathbf z^2 -\mathbf z \mathbf x -\mathbf x \mathbf z + \mathbf x^2}$. I'm not sure where to go from here...  For (b) I want to use proof by contradiction- so I will assume that $\mathbf y_1$ and  $\mathbf y_2$ both minimize the function and then I'll somehow see that they are in fact equal. How do I set up this argument? What is the difference between String and string in C#? Example (note the case):  string s = "Hello world!"; String s = "Hello world!";   What are the guidelines for the use of each? And what are the differences? While proving a universal quantification, we begin by "Let ... in ..." or "Arbitrary choose ..."? Asume the following search queries: osx, cocoa, iphone, xcode, applescript, mac, carbon    Searching for osx, cocoa, iphone or xcode like:   http://stackoverflow.com/search?q=osx   Results into searching by tag:   http://stackoverflow.com/questions/tagged/osx   While searching for applescript, mac and carbon like:   http://stackoverflow.com/search?q=applescript   Results into searching by string.  Yes, I know how to search by tags using brackets: [applescript] But why are the queries above interpreted differently? All queries are valid Stack Overflow tags! Why not convert all queries into search by string or search by tag? What is the logic behind? Tag popularity? It seems rather inconsistent to me. Prove existence of disjoint open sets containing disjoint closed sets in a topology induced by a metric. Finite Cartesian Product of Countable sets is countable? Is the finite Cartesian Product of Countable sets countable?  If possible, could you give a bijection that would work for any example?  Any help would be appreciated. As figured out in  localized SE sites must not be displayed in the HNQ's. But  actually is.        Has to be fixed I guess? What is the difference between POST and PUT in HTTP? According to , POST is used to create a resource:  The POST method is used to request that the origin server accept the entity enclosed in the request as a new subordinate of the resource identified by the Request-URI in the Request-Line.  According to , PUT is used to create or replace a resource:  The PUT method requests that the enclosed entity be stored under the supplied Request-URI. If the Request-URI refers to an already existing resource, the enclosed entity SHOULD be considered as a modified version of the one residing on the origin server. If the Request-URI does not point to an existing resource, and that URI is capable of being defined as a new resource by the requesting user agent, the origin server can create the resource with that URI.  So which HTTP method should be used to create a resource? Or should both be supported? I'm writing a script which shows the git log for a directory when I cd into it. Such a log can be overwhelming, containing hundreds of lines. So far I have been limiting that to a hard-coded 20 lines (... | head -n 20), which is fine on the screen at work, but too much on the smaller MacBook screen at home.  I would prefer the log to take up about half the (vertical) screen on either terminal. And "terminal" also changes: it's Gnome terminal at work, but iTerm2 at home. And I do not use screen or tmux.  How do I find the number of vertical lines available in a terminal from command line? If $E[X|Y]=Y$ almost surely and $E[Y|X]=X$ almost surely then $X=Y$ almost surely Assume that $X$ and $Y$ are two random variables such that $Y=E[X|Y]$ almost surely and $X= E[Y|X]$ almost surely. Prove that $X=Y$ almost surely.   The hint I was given is to evaluate: $$E[X-Y;X&gt;a,Y\leq a] + E[X-Y;X\leq a,Y\leq a]$$  which I can write as: $$\int_A(X-Y)dP +\int_B(X-Y)dP$$ where $A=\{X&gt;a, Y\leq a\}$ and $B=\{X\leq a,Y\leq a\}$.  But I need some more hints. I have a correlation matrix of security returns whose determinant is zero. (This is a bit surprising since the sample correlation matrix and the corresponding covariance matrix should theoretically be positive definite.)  My hypothesis is that at least one security is linearly dependent on other securities. Is there a function in R that sequentially tests each column a matrix for linear dependence?  For example, one approach would be to build up a correlation matrix one security at a time and calculate the determinant at each step. When the determinant = 0 then stop as you have identified the security who is a linear combination of other securities.  Any other techniques to identify linear dependence in such a matrix are appreciated. Proving a distance between molecules defines a metric space. A DNA molecule can be represented as a string of symbols $A$, $C$, $G$ and $T$, such as  $$GGATAATTCTAG. . .GACCGTACCC$$ For the purposes of this question, we will assume that all DNA molecules contain the same (large!) number $N$ of symbols. Thus, a DNA molecule is an $N$-tuple $x = (x_1, . . . , x_N)$ where $x_i  ∈ \{A,C,G,T\}$ for each $i$.  Define the distance between two DNA molecules $x$, $y$ as the number of elements $i ∈ {1, . . . ,N}$ such that $x_i \neq y_i$. Prove that this defines a metric on the set of DNA molecules. Hide/Show field on list form based on value from another field I need to hide field 'B' when the value in field 'A' is NOT equal to 'Other'.  This is a custom list form.  Is jquery the way to go? Cam I have two GFCIs connected to one double gang box in my laundry room and if so can I do a single line connection so that they both can work independently?  I have a 12/2 going from the panel through conduit (copper wire) to the double gang box. I want to have each GFCI to run independently without using the load. Why, in case one shuts off the other isn't affected. the double gang box is standard. the device is a Leviton 20 amps. I was told that you can run 12/2 from the panel to the first GFCI black and white. Then connect the second GFCI using a jumper and attaching it to he second GFCI. Is this acceptable by code. we use a washer,dryer in the laundry and the garage heavy machinery. so why should this be a problem running two independent? When I upload an image to Stack Overflow, I cannot see my JPEG file in my post. It's all blank. I've tried it with different files, but gotten the same result. Induction: Prove that $4^{n+1}+5^{2 n - 1}$ is divisible by 21 for all $n \geq 1$. Induction: Prove that $4^{n+1}+5^{2 n - 1}$ is divisible by 21 for all $n \geq 1$. How can I recover a game center account where I forgot ID and password? A long time ago, I made a game center account to which I attached my CoC village. This account hasn't been used anymore, and I have forgotten the ID and the password.  Now I want to transfer that game to my new phone which has a new game center ID, but need a transfer code from the old account to do so. Is there a way to recover my old game center account details? So this homework question is in the context of $\mathbb{R}$ only, and we are using Lebesgue measure.  The sum $A+B$ is defined to be $A+B=\{a+b|a\in A,b\in B\}$. The question is: If $m(A),m(B)&gt;0$, is it necessarily the case that $A+B$ contains an open interval?  This question is quite annoying, since it is quite difficult to calculate the sum even after you construct an example. However, it seems like no matter how I constructed a set of positive measure, there is a region that have a lot of point in it (yet this region is not necessarily dense). This suggest that when summed up, the holes should be covered up somehow and thus there would be an open interval, and so the answer is yes. Sorry for a rather vague idea, but I'm just not sure what to do with this problem. I have a feeling that somehow topology would have a way to capture precisely the idea above about a region with a lot of point.  (also, I think my middle school teacher mentioned this once, and I remembered the answer to be yes; but that is quite a while ago though)  Thank you. Any help will be appreciated.  EDIT2: So Camilo's answer is gone. I am looking at the convolution method right now. Because the convolution is continuous, if it is nonzero I am done. But then there is no guarantee that the convolution is nonzero even if $x\in A+B$. I still need help with this direction. Can someone give me some hints?  EDIT: Camilo's answer below essentially suggest the method of getting 2 open interval from regularity theorem, one cover a subset of A and one cover a subset of B such that the sum of the measure of these two subset is bigger than the measure of the larger interval. user99680's answer point to a wiki which show a method that essentially simply acquire an open set. After some thinking, I believe that these method simply can't be made to work. I constructed this example in the hope that it would convince you that it is the case:  Let $I_{0},I_{1},\ldots$ be disjoint open interval such that the distance between any 2 of them is nonzero and $m(I_{n})=9^{-n}$ and all of them are inside $[0,2]$. Let $J_{0},J_{1},\ldots$ be disjoint open interval such that the distance between any 2 of them is at least $3$ and $m(J_{3n})=m(J_{3n+1})=m(J_{3n+2})=\frac{m(I_{n})}{3}$. Let $O=\bigcup I_{n}$ and $H=\bigcup J_{n}$. Construct $A\subset O$ such that $m(A\bigcap I_{n})=\frac{3m(I_{n})}{4}$. Construct $B\subset H$ such that $m(B\bigcap J_{n})=\frac{3m(J_{n})}{4}$. Now if you apply Camilo's method, it is possible that your $I$ would be one of the $I_{n}$ above, and your $J$ is one of the $J_{m}$ above. No matter which $I_{n},J_{m}$ you have, the argument does not work. Rewrite $m=3q+r$ where $0\leq r\leq 2$. If $q&lt;n$ then $m(J_{m})&gt;m(I_{n})$ so the longer interval have length $\frac{m(I_{q})}{3}$; $m(A\bigcap I_{n})$ is at most $\frac{3m(I_{q+1})}{4}=\frac{m(I_{q})}{12}$ and $m(B\bigcap J_{m})=\frac{3m(J_{m})}{4}=\frac{m(I_{q})}{4}$. Thus $m(A\bigcap I_{n})+m(B\bigcap J_{m})$ is no larger than the larger interval. If $n\geq q$ then $m(I_{n})&gt;m(J_{m})$ so the larger interval is of length $m(I_{n})$; $m(A\bigcap I_{n})=\frac{3m(I_{n})}{4}$ and $m(B\bigcap J_{m})$ is at most $\frac{3m(J_{3n})}{4}=\frac{m(I_{n})}{4}$. Once again $m(A\bigcap I_{n})+m(B\bigcap J_{m})$ is no larger than the larger interval. Hence the Camilo's argument does not work. user99680's method does not work either, since you get $O$ and $H$ which is even bigger than any interval. Non-finite clause complementation of complex transitive verbs This question has been bothering me for a while. It came up when I was reading Chapter 16 of "A Comprehensive Grammar of the English Language."   How to explain the grammatical structure of the following four sentences?    "We knew him to be a spy." "I saw her leave the room." "I heard someone shouting." "I got the watch repaired."    What category do the four non-finite clauses in boldface fall into? They do not seem to fall into any of relative, nominal, comparative and adverbial clauses. Let me use the following example to explain my question: public string ExampleFunction(string Variable) {     return something; }  string WhatIsMyName = &quot;Hello World&quot;; string Hello = ExampleFunction(WhatIsMyName);  When I pass the variable WhatIsMyName to the ExampleFunction, I want to be able to get a string of the original variable's name. Perhaps something like: Variable.OriginalName.ToString()  Is there any way to do this?