Coffeescript existential operator sometimes checks against undefined, but not always
How does CoffeeScript's existential operator work?
Is every sub-lattice of $\mathcal P(X)$ isomorphic to a sub-lattice of $\mathcal P(X')$ containing singleton sets? Can we let users with privileges on one site have them on another? I have almost 4k reputation on Arqade. On Stack Overflow, I have to suggest edits and can't hit the close button. But I want to help with the close queue, which is at 62k at the time of this writing.  While I realize that people may be inexperienced with the system, I think I am experienced enough. If I want to edit something, I have to let it sit in the queue and get robo-reviewed. Yes, I know more about gaming than about coding. But I think I would know if this question is a duplicate of that, and if this question is off topic.  So, my request is that we should have the same privledges on every site, based on the site that we have the most reputation in. I am not asking that all the reputation be merged. How does Cantor's diagonal argument work? I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the  article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):  \begin{align} s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.  My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:  \begin{align} s_0 &amp;= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\ s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ s_4 &amp;= (0,1,1,\mathbf{1},\dots)\\ s_4 &amp;= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\ s_5 &amp;= (1,0,0,0,\dots)\\ s_6 &amp;= (1,0,0,1,\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:     For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,      $$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$      ...snip...      Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.      Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.   I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:  $$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 &amp; \mathrm{if\ } s_{m,m} = 1\\1 &amp; \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$  This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because...   Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.  Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".  What am I missing? Guest (not logged in) users can only sort questions with specific tag by newest I think this is platform wide (all SE sites).   Logout Choose a tag, eg.  on SO Try to sort questions by something other than newest   No matter what sorting is chosen, questions are sorted by newest Why does this not work with an imaginary exponent? Could someone please tell me why this exponentiation rule does not work when the exponent is imaginary?  The following seems to be true but is only true if $x$ is an integer.  $$e^{2\pi ix} = \left( e^{2\pi i} \right)^x = 1^x = 1$$ I reviewed  in the First Posts queue.  Which said      i changed it to 'localsystem' and done. thanks much   So it obviously was a potential low quality post. But sometimes such answers can be spot on, so I checked it out, better safe than sorry. I turned out the post had been deleted. Great! So I clicked "No action needed". Then I got this:    You can imagine I felt "wrongfully accused".  I think that deleted posts should not be used for review audits, because the meaning of "No action needed" is not clear in this context. Is there a functional difference between the modern police station and classic police station? What is the point of throwing candy on the ground using the "Throw 10 candies on the ground"? All that seems to happen is that the text under it becomes more and more disappointed in me. ಠ_ಠ Forced to Quit PhD although the hard and proved work I am a first year Ph. D student, and I have been working on a new topic of research.  I worked day and night to prove something new and verified it. I began my Ph. D while my supervisor was on travel, but he returned before I had been here for long.  When he returned, he told me we have to stop and not continue without any reasonable reasons.  After a few months he began treating me poorly---always belittling and underestimating---although I got two grants from a top conference, made videos and podcasts, and constantly received praise for my good work and personality.    It seems that I have been treated as a competitor, not a collaborator.  Although agreeing with my approaches in private, my supervisor said false words to the committee to take the decision and stop my Ph. D work!  It is not surprising, as there was another former senior Ph. D student who I met on campus, who was not allowed to defend his Ph. D: this same supervisor claimed he was not able to finish his dissertation although he had published in a peer reviewed journal and having citation as well, it was scary and not familiar.  In front of committee, he lied, and said something without proving. I asked if he can make a detailed constructive criticism, but he could not.  All I can say that they are a gang literally, and he said words that show that he is jealous and doesn't want to see me having self confidence in my self. I know no one will believe me, but this is the truth.  Honestly, I am distressed and I'm losing sleep.  I had this dream where there were many people who saw me fail.  All things considered, I don't know whether what happened is good or bad, and I definitely don't know what I should do next!  I would be grateful for your advice and whether there is someone has encountered any similar situation. Rotation Path Not Shown in Graph Editor I created a couple of LocRot keyframes but the rotation path and keyframes aren't shown in the graph editor (no, they aren't hidden). Number of subrings of a Field Can anyone help me with this problem?     Let $f:[0,\infty)\longrightarrow \mathbb R$ be a continuous function such that for each $x&gt;0$, we have $\lim_{n\to \infty}f(nx)=0$. Then prove that $\lim_{x\to \infty}f(x)=0$.   Our teacher told first to prove Baire's theorem, and then show that this is a consequence of that theorem. I proved Baire's theorem, and I spend a few hours thinking on how Baire's theorem is related to this problem, but I couldn't find anything. I'd really appreciate your help. Electric circuits in TeX, LaTeX, and Friends What are some ways to draw electric circuits in TeX systems?  I'm making this community wiki since I haven't found a question related to this.  I was thinking along the lines of . What do you think of that? Other examples are more than welcome, of course. Deploy Community Builder audience page assignments with ant migration tool? Is the 5-foot range of the Help action my range to the ally, the enemy, or both? How to copy the contents of a file in UNIX without displaying the file contents. I don't want to cat or vi to see the contents.  I want to copy them to clipboard so that I can paste it back on my windows notepad.  I can't copy the file from that server to another due to access restrictions. Is there a difference between "Joe said" and "said Joe"? Does the subject/verb order make a difference when writing a dialog tag?  &quot;The sky is blue,&quot; Joe said. &quot;The sky is blue,&quot; said Joe.  Is one preferable over the other? Does one emphasize the speaker (or the method of speech, such as shout) more than the other? (Personally, I would use the second form to emphasize that Joe made the statement) Proving the summation formula using induction: $\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$ I am trying to prove the summation formula using induction:  $$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$$  So far I have...  Base case:   Let n=1 and test   $\frac{1}{k(k+1)} = 1-\frac{1}{n+1}$  $\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$    $\frac{1}{2} = \frac{1}{2}$   True for n=1   Induction Hypothesis:   Assume the statement is true for the n-th case   $\sum_{k=1}^n  \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$  Inductive Step:   Prove, using the Inductive Hypothesis as a premise, that   $$\sum_{k=1}^{n+1}  \frac{1}{k(k+1)} = \sum_{k=1}^{n}  \frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)} = 1-\frac1{n+1} + \frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)}{(n+1)(n+2)}+\frac{-2-n}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)-2-n+1)}{(n+1)(n+2)} = \frac{(n+1)(n+2)-n-1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)(n+1)}{(n+1)(n+2)} = \frac{n+1}{n+2}$$  To prove $$ 1-\frac{1}{n+2} = \frac{n+1}{n+2} $$ Multiply both sides by $n+2$ to get an equivalent expression. $$ (1-\frac{1}{n+2}) * (n+2) = (\frac{n+1}{n+2}) * (n+2)  $$ $$ n+1=n+2−1 $$  Does this all make sense? How can this be improved upon? SQL query that concatenates values from duplicate rows in a single table Let's say I have a table of employees, like this:  Name     Sex     Role  Bob      M       Developer Joe      M       QA   Now, I have a problem with duplicated rows in this table. I will fix it sometime but it's not the issue. What I currently need is something that queries such a table:  Name     Sex     Role  Bob      M       Developer Bob      M       Janitor Joe      M       QA Joe      M       CEO   And will output the following rows:  Name     Role  Bob      Developer, Janitor Joe      QA, CEO   It doesn't matter to me if the "Role" will be split to separate columns or be a single column with multiple values.  Using SQL Server 2008 if it matters. Is there local file manipulation that's been done with JavaScript? I'm looking for a solution that can be accomplished with no install footprint like requiring .  Specifically, I'd like to read the contents from a file and write those contents to another file. At this point I'm not worried about gaining permissions and am just assuming I already have full permissions to these files.