Axiomatic set theory is basically a first order logic theory, so I believe we must certainly begin with a treatment of first order logic. However, to define a first order logic vocabulary we need to define a set of constants, a set of variables, a set of relation symbols, etc, whereby we are already using set theory.  Is this avoidable? What would be the cleanest way to define these two subjects?
Suppose we want to define a first-order language to do set theory (so we can formalize mathematics). One such construction can be found . What makes me uneasy about this definition is that words such as "set", "countable", "function", and "number" are used in somewhat non-trivial manners. For instance, behind the word "countable" rests an immense amount of mathematical knowledge: one needs the notion of a bijection, which requires functions and sets. One also needs the set of natural numbers (or something with equal cardinality), in order to say that countable sets have a bijection with the set of natural numbers.  Also, in set theory one uses the relation of belonging "$\in$". But relation seems to require the notion an ordered pair, which requires sets, whose properties are described using belonging...  I found the following in Kevin Klement's,  (pages 2-3).  "You have to use logic to study logic. There’s no getting away from it. However, I’m not going to bother stating all the logical rules that are valid in the metalanguage, since I’d need to do that in the metametalanguage, and that would just get me started on an infinite regress. The rule of thumb is: if it’s OK in the object language, it’s OK in the metalanguage too."  So it seems that, if one proves a fact about the object language, then one can also use it in the metalanguage. In the case of set theory, one may not start out knowing what sets really are, but after one proves some fact about them (e.g., that there are uncountable sets) then one implicitly "adds" this fact also to the metalanguage.  This seems like cheating: one is using the object language to conduct proofs regarding the metalanguage, when it should strictly be the other way round.  To give an example of avoiding circularity, consider the definition of the integers. We can define a binary relation $R\subseteq(\mathbf{N}\times\mathbf{N})\times(\mathbf{N}\times\mathbf{N})$, where for any $a,b,c,d\in\mathbf{N}$, $((a,b),(c,d))\in R$ iff $a+d=b+c$, and then defining $\mathbf{Z}:= \{[(a,b)]:a,b\in\mathbf{N}\}$, where $[a,b]=\{x\in \mathbf{N}\times\mathbf{N}: xR(a,b)\}$, as in  question or  on Wikipedia. In this definition if set theory and natural numbers are assumed, then there is no circularity because one did not depend on the notion of "subtraction" in defining the integers.  So my question is:     Question Is the definition of first-order logic circular?   If not, please explain why.   If the definitions are circular, is there an alternative definition which avoids the circularity?   Some thoughts:   Perhaps there is the distinction between what sets are (anything that obeys the axioms) and how sets are expressed (using a formal language). In other words, the notion of a set may not be circular, but to talk of sets using a formal language requires the notion of a set in a metalanguage. In foundational mathematics there also seems to be the idea of first defining something, and then coming back with better machinery to analyse that thing. For instance, one can define the natural numbers using the Peano axioms, then later come back to say that all structures satisfying the axioms are isomorphic. (I don't know any algebra, but that seems right.) Maybe sets, functions, etc., are too basic? Is it possible to avoid these terms when defining a formal language?
ReadList can't read from file containing high-precision Reals Did Snape ever commit murder? It's well known that Severus Snape was, in one way or another, a Death Eater since he was at Hogwarts. He hung out with the likes of Avery and Mulciber who became well known Death Eaters, he joined Voldemort right after he left Hogwarts, and he basically became Voldemort's right hand man until the very end.   However, in his time as a Death Eater did he ever kill anyone? We know that he was a Death Eater for around 3 years before Voldemort fell; Lily and James died at 21 and they were the same age as Snape, so it would have been 3 years since they left Hogwarts. Then after Voldemort came back in book 4, Snape would have been a Death Eater again for around 3 years. That's a total of around 6 years of Death Eaterhood with Voldemort.  So, in either of these periods, did Snape commit murder as part of his job as Death Eater? I'm not counting a certain death at the end of book 6 for reasons which are obvious if you've read book 7.   Clearly it's much more likely he killed before Voldemort fell since that roughly correlated with the period when he pledged allegiance only to Voldemort, before he became a teacher at Hogwarts. But did he ever do it? Image, pre-image, and injection proof verification Just wanted to verify this proof was correct. Trying to make sure I have a handle on these subjects before the exam. Any style comments also more than welcome.    Statement: Let $f : A \mapsto B$ be a function. $f$ is injective if and only if for all $X \subseteq A$, $f^{-1}(f(X)) \subseteq X$.  Proof:  $\Rightarrow$ Suppose that $f$ is injective. Let $X \subseteq A$, and let $x \in X$. Since $f$ is injective, $f^{-1}(f(X)) = \{x\} \in X$, so $f^{-1}(f(X)) \subseteq X$.   $\Leftarrow$ Suppose that for all $X \subseteq A$, $f^{-1}(f(X)) \subseteq X$. Let $a_1, a_2 \in A$ and $X = \{a_1\}$. Suppose $f(a_1) = f(a_2)$. If $a_1 \neq a_2$, then $f^{-1}(f(X)) = \{a_1, a_2\} \not \subseteq X$, a contradiction. Thus $a_1 = a_2$, so $f$ is injective. Tikz: setting options for pin I just found the pin option to drawing, and it looks useful in some of the things I want to draw. How do I format it though? I've tried pin=[red]60:{Hey!} and I get the error Cannot parse this coordinate.  \documentclass{standalone} \usepackage{tikz} \usetikzlibrary{shapes,arrows,positioning} \begin{document} \begin{tikzpicture}[node distance=2cm,&gt;=latex, every node/.style={     font=\sffamily\scriptsize     },       circtext/.style={draw,circle,minimum size=8pt,inner sep=2pt},     dot/.style={draw,circle,fill=black,minimum size=0.6mm,inner sep=0pt} ] { \node[rectangle, fill=green!10!white!90!black, minimum width=6cm, minimum height=3cm] at (2,0){}; \node[circtext, fill=yellow](A) at (0,0) {A}; \node[circtext, fill=yellow, right=of A](B){B}; \draw[-&gt;] (A) -- node[dot, pos=0.5, pin=[red]60:{Hey!}](C){}(B); } \end{tikzpicture} \end{document}     The default style looks like it's a gray thin arrow, and I want to change it to a black thin dotted line with no arrow.    update: I figured out how to add style to the pin node:  pin={[red]60:Hey!}]  But I can't figure out how to change the linestyle though. Number of homomorphisms between two cyclic groups. Convert LaTeX to HTML Is there a converter that will convert LaTeX to HTML? I have tried several services but the LaTeX diagrams and some of the maths came out weird.  I actually converted the Latex to PDF without a problem. Then I used  to convert from PDF to HTML. The square root signs did not appear properly and some of the Cartesian coordinate plane diagrams had nothing on them except the grid.   \documentclass[12pt,a4paper]{report} \usepackage{amssymb} \usepackage{amsthm} \usepackage{amsmath} \usepackage{pstricks,pst-plot} \usepackage{pst-eucl} \addtolength{\hoffset}{0.2cm} \addtolength{\textwidth}{1.1cm} \addtolength{\textheight}{1.0cm} \addtolength{\voffset}{-0.7cm} \linespread{1.3} \begin{document}  If $x$ is non-positive then $x \leq -\sqrt{\frac{1}{2}}$     $\hspace{30 mm} $     (6)\\  (4) and (6) tell us that\\  (4)  $-1 \leq x \leq 0$  \\  (6)  If $x$ is non-positive then $x \leq -\sqrt{\frac{1}{2}}$\\  Therefore $-1 \leq x \leq -\sqrt{\frac{1}{2}}$\\  \vspace{50 mm}  \begin{align*} \psgrid[unit=2cm](0,0)(-2,-2)(2,2) \pnode(-1.8,.87){b}\uput[50](b){b} \pnode(-1.8,1.8){a}\uput[50](a){a} \psset{linecolor=green} \psline[linewidth=1pt](-1.8, 0)(-1.8,4) \psset{linecolor=green} \psline[linewidth=1pt](-1.8, 0)(-1.8,-4) \psset{linecolor=blue} \psellipticarc(0,0)(2,2){0}{180} \psset{linecolor=red} \psline[linewidth=1pt,linestyle=dashed,dash=2pt](-2,2)(-1.41,1.41) \psset{linecolor=red} \psline[linewidth=1pt](-1.41,1.41) \psset{linecolor=red} \psline[linewidth=1pt]{-&gt;}(4,-4) \psset{linecolor=red} \psline[linewidth=1pt]{-&gt;}(-2,2)(-4,4) \end{align*}  \end{document}   PDF output:    PDF->HTML output: Update  This has been confirmed resolved in a later version 376.33 and possibly the update prior.     Okay, I recently did a computer update. I used to run Blender 2.77 on a Dell desktop (5 GB RAM, 2.7 GHZ, Windows 8, build in graphics card). Since that computer wasn't fast enough for my preference, I saved up and got a new gaming laptop - the Asus ROG GL552 (Windows 10, 16 GB DDR4 RAM, Nvidia GeForce GTX 960M with 4 GB DDR5 RAM, 2.6-3.5 Ghz, and quad-core intel processor). That laptop has been known to have outdated drivers, so when I got it, I updated it to up to date Windows 10 drivers.  As soon as I started Blender, I immediately noticed its speed. Even doing simple things, like clicking buttons, or rotating the view with the middle mouse button, took too long to register in the program. I tried changing some User Preference System settings, but that didn't really help much. And the hardware seems to be fine - the render is lightning fast (renders the startup scene in under .1 second; it's just laggy when loading the uv screen), so I don't know why it's like this, or what to do. Also, I ran Blender 2.77 on this computer, and it was also slow (although it was fine on my previous computer).  It could be an issue with drivers perhaps (or maybe OpenGL?), but I ran Unreal Engine 4, and had none of the same issues - the speed was pretty decent on that; so I'm not sure why Blender would be so slow. Any help is appreciated - thanks guys! Using a comma before "but" Who is / was the "Lord of the Rings"? Who is the Lord, as referred to in the title? Frodo? Sauron? Someone else? Or is it more of an abstract title, not referring to a single person, like "He who wields the One Ring"? Center of a group ring? Let $R$ be a ring with unity and let $G$ be a group. Then what is the center of the group ring $R(G)$? I feel that the center $Z(R(G))$ is $R(Z(G))$. I assumed if $x=\sum_{i=1}^{n} r_i g_i$, then it should commute with every element of $R(G)$. In particular it should commute with $g=1 \cdot g \in R(G)$ Thus we get $$\big(\sum_{i=1}^{n} r_i g_i\big)g=g\big(\sum_{i=1}^{n} r_i g_i\big)$$ Thus I get $$g_i \cdot g=g_i \cdot g \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \forall g \in G$$ Hence $x \in Z(R(G)) $. Is this correct? '\r' added end of the script command This is my script:  #!/bin/bash hd='pwd' cd $hd   When I try to run it, it fails with:  line 3: cd: $'pwd\r\r': No such file or directory   What is going on? Modern Linux commands that parallel uuencode/uudecode functionality I need to get binary files into text format.  Years ago, I used uuencode/uudecode, and it was easy:  uuencode file.xls &gt; file.txt # Delete file.xls # Do whatever with file.txt # Now I need file.xls back uudecode file.txt &gt; file.xls   But uuencode/uudecode is not on my system, and as I look around, the yum/rpm process is ugly--this gives me the clue that maybe no one uses it any more and instead there is something more modern to do this.  I looked at xxd and hexdump but couldn't get them to do it.  Could just be dumb user issues.  What can I use on Linux to do this?  Would like it to be something that tends to be in the native install so I don't need to yum/rpm it if possible.  Just for your own understanding, what I'm specifically trying to do now is source control an *.xls file but the corporate tool doesn't allow *.xls in.  I respect that constraint normally, but in my context, an *.xls is a configuration table for my code so It is logically like source code, but the corporate SVN trigger is physical--if it's *.xls, it gets rejected, so I'll encode it and source control the *.txt file.  Thanks! What are good advanced LaTeX books? The tablet functionality stops working after I wake my Lenovo Tablet from sleep mode (closing &amp; opening lit). The error message is "Tablet Serial Wacom Tablet (Lenovo X201) removed" (screenshot: ). In the system settings under input devices it states that "No tablet device is detected". Only a restart helps and the touch screen works again.  This error appeared only after upgrading from Kubuntu 12.04 to 14.04. In 12.04 the same error message appeared after waking from sleep mode, but then another followed that the tablet was activated again and it worked properly. Any ideas how this could be fixed? Show $f$ can't be irreducible over a finite field if $f^\prime$ is the zero polynomial. Unbounded continuous function on non-compact metric space Suppose $(S,d)$ is a non-compact metric space. Is it possible to construct an unbounded continuous function from $S$ to $\mathbb{R}$? If it is possible, please show the construction method :)  Here's my attempt to solve this problem: Because $S$ is not compact, we can find a neighborhood assignment function $N(x)$ so that a finite cover is not possible. Then each time we can find an uncovered point from the union of neighborhoods before, and change the function by $f(x):=max\{f(x),\text{a function which equals to the indice of this point at that point and gradually vanishes at further points from the selected point}\}$. However I'm concerned that while this function is unbounded, there may be some invalid points for this function (goes to infinity)...  EDIT: As I found out there is a duplicate question, however the elementary solution is not presented in that question, so I still wish to know about this problem. :) How can I copy location info from geotagged phone images to images taken with my other camera around the same time? So i did quite a big trip recently, and took lots of pictures with my DSLR and occasionally Android phone. Only after coming back did i read about the whole GPX file, tracklog options in Lightroom. meaning i don't have such a tracklog file.  The android phone did geotag most photos it took. (at least, when it got a gps fix i guess)  So my question is, is there some software or way that will automatically copy the geotag from the android's photos to photos taken on the same time with my DSLR?  Or alternatively, generate a tracklog from the geotagged data in the android pictures? Should we write “allow” or ”allows” in “require that the class allow[s]”? Can you explain to me whether I should or should not use ‑s at the end of the  verb allow in this sentence?     This function requires that the container class allow[s] random access   The sentence in the book I read wrote allow without ‑s. However, the container class is a third-person singular, so the ‑s must be required. Is that right? Realtek RTL8723BE Wi-Fi incredibly weak Verbatim in beamer showing error: "File ended while scanning use of \@xverbatim." I am using Texmaker and MiKTeX on Windows 7. I have the following beamer code,  \documentclass{beamer} \theme{AnnArbor} \begin{document} \begin{frame} \begin{verbatim}     \begin{frame}[&lt;alignment&gt;]    \frametitle{Frame Title Goes Here}    Frame body text and/or LATEX code    \end{frame}  \end{verbatim} \end{frame} \end{document}   I was expecting the following output,         \begin{frame}[&lt;alignment&gt;]        \frametitle{Frame Title Goes Here}        Frame body text and/or LATEX code        \end{frame}   But instead I got the following Error Message,  File ended while scanning use of \@xverbatim.   Can any one please point out the error here. As far as I know verbatim can have anything inside it.  Also when I tried the following code,  \begin{verbatim}  \ begin{frame}[&lt;alignment&gt;]  \frametitle{Frame Title Goes Here}  Frame body text and/or LATEX code  \ end{frame} \end{verbatim}   I got the following output,      \ begin{frame}[&lt;alignment&gt;]      \frametitle{Frame Title Goes Here}      Frame body text and/or LATEX code     \ end{frame}   But I don't want space between \ and begin and \ and end.