I've already asked a similar question yesterday (is there a well-order such that $X$ has a largest element). Can one prove the existence of a well-order on an infinite set $X$ such that $\forall x \in X \exists y \in X: x \leq y$?
According to the well ordering theorem "". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element. My question is:     For a given infinite set, can we find a well ordering on it such that there is no largest element?
I am struggling to find such languages $L_{1}$, $L_{2}$, and $L_{3}$ such that  $$ L_{1} \subseteq L_{2} \subseteq L_{3} $$ where $L_{1}, L_{3} \notin  \mathbb{R}$ and $L_{2} \in  \mathbb{R}$.  I know they exist, I need help finding them. We have a few cases of chat accounts remaining without valid parent account:  and  for example. What should mods do when that happens:  delete the chat account notify the SE team so they can reattach or clean it up leave it as is: it's a normal situation Why isn't the 'len' function inherited by dictionaries and lists in Python example:  a_list = [1, 2, 3] a_list.len() # doesn't work len(a_list) # works   Python being (very) object oriented, I don't understand why the 'len' function isn't inherited by the object.  Plus I keep trying the wrong solution since it appears as the logical one to me Enabling CORS in GeoServer (jetty)? This is not a duplicate question because I am looking for an explanation directed to a general audience as to the mistakes (if any) in  (reproduced below). (Numberphile is a YouTube channel devoted to pop-math and this particular video has garnered over 3m views.)   By general audience, I mean the same sort of audience as the millions who watch Numberphile. (Which would mean, ideally, making little or no mention of things that a general audience will never have heard of - e.g. Riemann zeta functions, analytic continuations, Casimir forces; and avoiding tactics like appealing to the fact that physicists and other clever people use it in string theory, so therefore it must be correct.)  Numberphile's Proof.  The proof proceeds by evaluating each of the following:  $S_1 = 1 - 1 + 1 - 1 + 1 - 1 + \ldots$   $S_2 = 1 - 2 + 3 - 4 + \ldots $   $S = 1 + 2 + 3 + 4 + \ldots $   "Now the first one is really easy to evaluate ... You stop this at any point. If you stop it at an odd point, you're going to get the answer $1$. If you stop it at an even point, you get the answer $0$. Clearly, that's obvious, right? ... So what number are we going to attach to this infinite sum? Do we stop at an odd or an even point? We don't know, so we take the average of the two. So the answer's a half."   Next:  $S_2 \ \ = 1 - 2 + 3 - 4 + \cdots$  $S_2 \ \ = \ \ \ \ \ \ \ 1 - 2 + 3 - 4 + \cdots$  Adding the above two lines, we get:  $2S_2 = 1 - 1 + 1 - 1 + \cdots$  Therefore, $2S_2=S_1=\frac{1}{2}$ and so $S_2=\frac{1}{4}$.  Finally, take  \begin{align} S - S_2 &amp; = 1 + 2 + 3 + 4 + \cdots           \\ &amp; - (1 - 2 + 3 - 4 + \cdots)          \\ &amp; =   0 + 4 + 0 + 8 + \cdots           \\ &amp; = 4 + 8 + 12 + \cdots           \\ &amp; = 4S \end{align}  Hence $-S_2=3S$ or $-\frac{1}{4}=3S$.  And so $S=-\frac{1}{12}$. $\blacksquare$ Dr. StrangeNumbers or: How I Learned to Stop Worrying and Love Floating Point Arithmetic General formula for evaluating integrals of the form $\displaystyle\int_0^\infty x^ae^{-bx}dx$ I've recently been trying to introduce myself to formal logic, and I've hit a stumbling point:  In the text I'm reading, a valid argument is defined as "an argument in which it is impossible for the premises to be true and at the same time the conclusion false."  The text then later prompts me to declare whether it's possible for an argument to be valid with a contradiction for a conclusion. The answer key then claims that yes, it is in fact possible, but offers no further explanation. I'm also supposed to create an example.  I just can't wrap my head around this one.  Here's the example I've worked up, is this a valid argument?   If grass is green then it is raining.  If water is wet then it is not raining.  Therefore: It is raining and not raining.   My head hurts. In Fantastic Beasts: The Crimes of Grindelwald Dumbledore teaches DADA. But from the books we always heard about him as of a Transfiguration teacher (say, in Tom Riddle's days). It is stated in the movie that Dumbledore was prohibited to teach DADA by the Ministry, though the limitations seemed to be canceled at the end of the movie together with the removal of his handcuffs (not completely clear about the teaching though). The strange thing is, however, that Dumbledore was a Transfiguration teacher and even the Head of Transfiguration Department in Hogwarts before Newt's days, as we learn from Pottermore:  Under the guidance of her [McGonagall] inspirational Transfiguration teacher, Albus Dumbledore, she had managed to become an Animagus She [McGonagall] sent an owl to Hogwarts, asking whether she might be considered for a teaching post. The owl returned within hours, offering her a job in the Transfiguration department, under Head of Department, Albus Dumbledore.  That is of course if we accept that Minerva McGonagall in the movie is the same Minerva we know. So it seems that he taught Transfiguration, then for some reason switched to DADA and then to Transfiguration again. While it is possible in general, is there any explanation for the reasons? How to merge records for query set We have a question related to a DB query :   CREATE TABLE `Action` (   `id` int(10) unsigned NOT NULL AUTO_INCREMENT,   `emp_Id` int(10) unsigned NOT NULL,   `name` varchar(60) NOT NULL,   `updated_action_at` datetime(3) DEFAULT NULL,   `created_At` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,   `notes` varchar(400) DEFAULT NULL,   PRIMARY KEY (`id`),    KEY `action_empId_fk` (`emp_Id`),    CONSTRAINT `action_empId_fk` FOREIGN KEY (`emp_Id`) REFERENCES `Employee` (`id`) ON DELETE CASCADE,  ) ENGINE=InnoDB AUTO_INCREMENT=502004 DEFAULT CHARSET=latin1   CREATE TABLE `ActionAssignedTo` (   `action_Id` int(10) unsigned DEFAULT NULL,   `emp_Id` int(10) unsigned DEFAULT NULL,   KEY `actionassignedto_emp_id_foreign` (`emp_Id`),   KEY `actionassignedto_action_id_foreign` (`action_Id`),   CONSTRAINT `ActionAssignedTo_ibfk_1` FOREIGN KEY (`emp_Id`) REFERENCES `Employee` (`id`) ON DELETE CASCADE,   CONSTRAINT `ActionAssignedTo_ibfk_2` FOREIGN KEY (`action_Id`) REFERENCES `Action` (`id`) ON DELETE CASCADE ) ENGINE=InnoDB DEFAULT CHARSET=latin1  CREATE TABLE `Employee` (   `id` int(10) unsigned NOT NULL AUTO_INCREMENT,   `vendor_Id` int(10) unsigned DEFAULT NULL,   `name` varchar(40) NOT NULL,   `mobile_Number` varchar(15) NOT NULL,   `active` tinyint(1) DEFAULT '1',   `updated_At` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,   `created_At` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,   PRIMARY KEY (`id`),   KEY `employee_vendor_id_foreign` (`vendor_Id`),   CONSTRAINT `employee_vendor_id_foreign` FOREIGN KEY (`vendor_Id`) REFERENCES `Vendor` (`vendor_Id`) ) ENGINE=InnoDB AUTO_INCREMENT=511 DEFAULT CHARSET=latin1   We want to find the action created by employee with Action.empId = 4 along with the list of Assignee (an action can be assigned to multiple people)  We have the following query :   Select     notes,             Action.id,             AssigneeName.name,             ActionAssignedTo.emp_Id  from       Employee inner join Action  on         (Action.emp_Id = Employee.id  and        Action.emp_Id in (select Employee.id                               from Employee                               where Employee.vendor_Id = 1)) inner      join ActionAssignedTo  on         Action.id =  ActionAssignedTo.action_Id inner join Employee as AssigneeName  on         ActionAssignedTo.emp_Id =  AssigneeName.id where      Action.emp_Id = 4   Now, lets say if Action with id = 10 was assigned to 2 people : John (empId : 2) And Jack (empId : 5) --> the above query returns following results :  (columns from the query)  notes                       | Action.id | AssigneeName.name | ActionAssignedTo.emp_Id 1. Finish Designing Queries | 10        | John              | 2 2. Finish Designing Queries | 10        | Jack              | 5   If we observe the above results the first 2 columns get repeated for the Assignees for SAME ACTION ID  --> is there a way we can merge them something like   1. Finish Designing Queries |10         | { John : 2  , Jack  : 5}   Basically records being merged into one. Roll back from OS X Mavericks ( or any newer OS) to Mountain Lion I installed the developer preview of OS X Mavericks, &amp; too many things broke. I can't stay on it &amp; need to roll back to Mountain Lion. Any ways of doing that? Please help!  I originally installed Mountain Lion &amp; Mavericks through the Mac App Store. Can a creature with blindsight that is not immune to the Blinded condition suffer the effects of Blindness? Show that any bounded linear functional on a normed linear space is continuous.   Can we say that it is uniformly continuous ?   Also, is it true, if we reverse the statement    any continuous  linear functional on a normed linear space is bounded ?? line 56 syntax error: unexpected end of file I have typed out the code for this mychoice.sh bash example and I keep getting the syntax error: unexpected end of file.  Am I missing something?  mychoice.sh  #!/bin/bash  E='echo -e';e='echo -en';trap ";exit" 2 ESC=$( $e "\e") TPUT(){ $e "\e[${1};${2}H";} CLEAR(){ $e "\ec";} CIVIS(){ $e "\e[25l"} DRAW(){ $e "\e%\e(0";} WRITE(){ $e "\e(B";} MARK(){ $e "\e[7m";} UNMARK(){ $e "\e[27m";} R(){ CLEAR ;stty sane;$e "\ec\e[37;44m[J";}; HEAD(){ DRAW     for each in $(seq 1 13);do     $E "   x                    x"     done     WRITE;MARK;TPUT; 1 5     $E "BASH SELECTION MENU         ";UNMARK;}     i=0; CLEAR; CIVIS;NULL=/dev/null FOOT(){ MARK;TPUT 13 5     printf "ENTER - SELECT,NEXT         ";UNMARK;} ARROW(){ read -s -n3 key 2&gt;/dev/null &gt;&amp;2      if [[$key = $ESC[A ]];then echo up;fi      if [[$key = $ESC[B ]];then echo dn;fi;} M0(){ TPUT  4 20; $e "Login info";} M1(){ TPUT  5 20; $e "Network";} M2(){ TPUT  6 20; $e "Disk";} M3(){ TPUT  7 20; $e "Routing";} M4(){ TPUT  8 20; $e "Time";} M5(){ TPUT  9 20; $e "About  ";} M6(){ TPUT 10 20; $e "Exit   ";} LM=6 MENU(){ for each in $(seq 0 $LM);do M${each};done;} POS(){ if [[ $cur == up ]];then ((i--));fi     if [[ $cur == dn ]];then ((i++));fi     if [[ $i -lt 0 ]];then i=$LM;fi     if [[ $i -gt $LM ]];then i=0;fi;} REFRESH(){ after=$((i+1)); before=$((i-!))     if [[ $before -lt 0 ]];then before=$LM;fi     if [[ $after -gt 0 ]];then after=0;fi     if [[ $j -lt $i ]];then UNMARK;M$before;else UNMARK;M$after;fi     if [[ $after -eq 0 ]] || [$before -eq $LM ];then          UNMARK; M$before; M$after;fi;j=$i;UNMARK;M$before;M$after;} INIT(){ R;HEAD;FOOT;MENU;} SC(){ REFRESH;MARK;$S;$b;cur='ARROW';} ES(){ MARK;$e "ENTER = main menu ";$b;read;INIT;};INIT while [[ "$0" != " " ]]; do case $i in     0) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS(w     )\n";ES;fi;;     1) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS(ifconfig  )\n";ES;fi;;     2) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS(df -h )\n";ES;fi;;     3) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS(route -n  )\n";ES;fi;;     4) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS(date  )\n";ES;fi;;     5) S=M2;SC;if [[ $cur == "" ]];then R;$e "\nS($e by oTo )\n";ES;fi;;     6) S=M2;SC;if [[ $cur == "" ]];then R;exit 0;fi;;  esac;POS;done; In Rey's vision, whose voice says "These are your first steps"? In the case of a non-inverting amplifier, if I DC couple the input signal with a capacitor only, \$C_1\$ (i.e. no resistor to ground), sources (as well as an experiment in the lab) show me that the op amp input will be saturated as the input bias current deposits charges on the terminal without having a return path to ground. This is solved by using connecting a resistor (\$R_1\$, as shown in the image below) to ground to create the return path. [The image is reproduced from an answer by "Neil_UK" on one of my previous questions, thanks Neil!]    My questions:   1) How is the input bias current able to continuously deposit charges and saturate the input if I don't use the resistor? The resistor should not allow the DC bias current through, so where is this current flowing from/into?  2) Why is this not a problem for the feedback path to the "-" terminal? There is no resistor to ground in that case, and capacitor \$C_2\$ seems to be causing no issues. The PDFs I've been reading suggest that the feedback behaves similar to when I have a resistor to ground by acting as the return path, how is this? get the Eyes of the Dark feature (XGtE, p. 51):     When you reach 3rd level in this class you learn the darkness spell [...] If you cast it with sorcery points, you can see through the darkness created by the spell.   Does this mean that if a Shadow Sorcerer casts darkness using sorcery points and then moves into a different area of magical darkness (not the one s/he created), the Sorcerer would not be able to see?  Or does the ability mean that the Sorcerer would possess the ability to see through the magical darkness created by the darkness spell in general (provided they were maintaining their own casting of the spell)? What are the limitations of a typical kit lens as a general purpose lens? I have a Canon EOS 550d with the 18-55mm IS kit lens. I have done  (portraits, nature, night shots etc), and find the image quality to be pretty good. In fact most of the reviews have also been positive (at least in terms of image quality):   Quote:     indeed Canon's main concern may   ultimately become whether users have   as much incentive to upgrade to more   expensive optics as they did before       So, what are the key limitations of this lens (or any other similar kit lens) that would compel one to upgrade to a more expensive general purpose zoom lens (apart from the wanting to upgrade the zoom range)? How to configure an exposed filter in a view using a date picker? I would like to install Office 2015/365 on Ubuntu 15.04. I tried to install Office using Wine, but that did not work. Please note that I do not want virtualisation unless it has a -like behaviour, where application windows are treated like regular Ubuntu application windows by the system.