TypeError: 'set' object is not subscriptable? (tkinter)
What does it mean if a Python object is "subscriptable" or not?
Is the "live" in "He only lived a few days after the accident" intrasitive or transitive? An example sentence from the :  [I] He only lived a few days after the accident.  [I] means &quot;intransitive verb&quot;. He (subject) + lived (verb) + a few days (noun). What part of speech is a few days? From the :  number 5  way of life[intransitive always + adverb/preposition, transitive] to have a particular type of life, or live in a particular way   used 'transitive verb' like 'She lives a very busy life.' From the ELU question :  She is 16 year old.  '16 years' as acting as an adverb. My questions are:  In &quot;He lived a few days.&quot;, is a few days an object (noun) or an adverb? How is it different from &quot;He lived for a few days.&quot;? &quot;I walked ten miles.&quot; Number + measurement — how can I deal with 'part of speech'? In a field that is the direct product of two fields, why isn't it a field? Suppose that we have the direct product of two fields, defined as: $\mathbb{F}^2 = \mathbb{F} \ \times\ \mathbb{F}$. Then, this is obviously not a field because we can take $(1,0)$, and this doesn't have a multiplicative inverse. However, I saw a mention to this today in a book, and the reasoning was:     "$\mathbb{F} \ \times\ \mathbb{F}$ has the zero-divisor $(1,0) \neq  (0,0)$, hence it is not a field".   (Lectures on Rings and Modules by Joachim Lambek, pg. 17).  I am not sure what they mean by zero-divisor here nor what $(1,0) \neq  (0,0)$ has anything to do with the direct product of fields not being a field. I always assumed there wasn't a multiplicative inverse, and that was it. Am I missing something deeper? Are static fields open for garbage collection? Given an hypothetical utility class that is used only in program setup:  class MyUtils {    private static MyObject myObject = new MyObject();    /*package*/static boolean doStuff(Params... params) {        // do stuff with myObject and params...    } }   will myObject be garbage collected when it is no longer being used, or will it stick around for the life of the program? I have been reading Hamilton Chapter 13 and he has the following state space representation for an ARMA(p,q). Let $r = \max(p,q+1)$.Then the ARMA (p,q) process is as follows: $$ \begin{aligned}  y_t -\mu &amp;= \phi_1(y_{t-1} -\mu) + \phi_2(y_{t-2} -\mu) + ... + \phi_3(y_{t-3} -\mu) \\ &amp;+ \epsilon_t + \theta_1\epsilon_{t-1} + ... + \theta_{r-1}\epsilon_{t-r+1}.  \end{aligned} $$ Then, he defines the State Equation as follows:   $$  \xi_{t+1} = \begin{bmatrix}      \phi_1 &amp; \phi_2 &amp; \dots &amp; \phi_{r-1} &amp; \phi_r \\      1 &amp; 0  &amp; \dots &amp; 0 &amp; 0  \\      \vdots &amp; \vdots &amp; \dots &amp; 0 &amp; 0 \\      0 &amp; 0 &amp; \dots &amp;1 &amp;0 \end{bmatrix} \xi_t  +  \begin{bmatrix}      \epsilon_{t+1} \\     0\\\vdots  \\     0       \end{bmatrix}  $$  and the observation equation as:  $$y_t = \mu + \begin{bmatrix}      1 &amp; \theta_1 &amp; \theta_2 &amp; \dots &amp; \theta_{r-1} \\     \end{bmatrix} \xi_t. $$  I do not understand what the $\xi_t$ is in this case. Because in his AR(p) representation it is $\begin{bmatrix}      y_{t} - \mu \\     y_{t-1} - \mu\\\vdots  \\     y_{t-p+1} - \mu       \end{bmatrix}  $ and in his MA (1) representation it is $\begin{bmatrix}      \epsilon_{t} \\     \epsilon_{t-1}      \end{bmatrix}  $.  Could someone explain this to me a little bit better? I'd like to create a Categorized style in QGIS, classify with it data in a shapefile and then save the shapefile with the new categorized style. Is it possible? Possible Duplicate:          In  the following claim was mentioned:  $R/I \otimes_R M \cong M / IM$  So I tried to prove it. Can you help me finish my proof? Thanks!    We recall that $M \otimes_R -$ is a covariant right-exact functor that it is exact if $M$ is flat and we observe that the following is an exact sequence: $$ 0 \to IM \xrightarrow{i} M \xrightarrow{\pi} M / IM \to 0$$  $R/I$ is an $R$ module since $R/I$ is a subring of $R$ closed with respect to multiplication from $R$ but it is not necessarily flat hence we only get exactness on one side:  $$  (R/I) \otimes_R IM \xrightarrow{id \otimes i} (R/I) \otimes_R M \xrightarrow{id \otimes \pi} (R/I) \otimes_R M / IM \to 0$$  Then $$ \mathrm{Im(id \otimes \pi)} = (R/I) \otimes_R M / IM  \cong (R/I) \otimes_R M / \mathrm{Ker} (id \otimes i) $$  so we want to show two things:   (i) that $\mathrm{Ker} (id \otimes i) = \{0\}$. To this end let $r + I \otimes im \in (R/I) \otimes_R  IM $ and assume $ id \otimes i(r + I \otimes im ) = r + I \otimes im = 0 + I \otimes 0$. I'm not sure how to proceed from here.   (ii) and that $M /IM \cong (R/I) \otimes_R M / IM $  Can you help me? Thanks. Easy way to determine the virtualization technology of a Linux machine? How do I uninstall Pandoc in Ubuntu 19.10 (and install most recent version) I have an old version of Pandoc installed, but I'd like to install the latest version of Pandoc.   To begin I've run the following to get rid of the old version:  sudo apt-get autoremove --purge pandoc  But when I check that everything is uninstalled by running: pandoc --version Then pandoc 2.2.3.2 pops up.  I can't seem to get rid of this old version of pandoc so that I can install a new one.  Could you advise me on how I can get rid of the old version and install the new one please. A problem about an $R$-module that is both injective and projective. Let $R$ be a domain that is not a field, and let $M$ be an $R$-module that is both injective and projective. Prove that $M= \left \{ 0 \right \}$.   This is exercise 7.52 of Rotman's Advanced Modern Algebra. Using theorems before exercises, because $M$ is injective and $R$ is a domain, I conclude that $$\forall m\in M ,\forall r\in R\ (r\neq 0) ,\exists {m}'\in M \Rightarrow m=r{m}'$$ and also because $M$ is projective there is a surjective $\psi$ from free $R$-module $F$ with basis $\left \{ e_{i} \right \}_{i\in I}$ to $M$ and thus we can conclude that for every $m\in M$ we have $$m=\sum r_{i}\Psi (e_{i})$$ now I don't know how should I use these together.  The idea of what is happening or a suggestion or a hint will be great. What tool can I use to manage the configuration of my Windows Server environments I help manage environments for an application that uses cloud-based Windows server VMs. The application stack consists of Windows Server 2008 and Windows Server 2012, SQL Server 2008, IIS, and SharePoint 2010. I have several environments Dev/Test/Stage/Prod. I am looking for a way to do configuration management of those environments to make sure any environmental changes are applied in a consisent way and there is a way to verify that no changes have been made to one environment that have not been made to other environments.  I have been reading up on Puppet, Chef, etc and I like the idea of doing delcarative configuration management. What is a good method or tool to use to help manage the configuration of these environments. I know a little  about SCCM but it is too expensive for this particular situation although I would still be interested in knowing if it has the capabiltiy to do those things. Say $R$ is a commutative ring and $I\in R$ is an ideal. Let us consider the quotient $R/I$. It is created by taking every element $a\in R$, and adding all the elements of $I$ to it. The elements of $R/I$ are of the form $a+I$ and $b+I$, $\forall a,b\in R$.   Now $$(a+I).(b+I)=(a+i_{1}).(b+i_{2}),\forall (i_{1},i_{2})\in I \times I$$  $$(a+i_{1}).(b+i_{2})=ab+a.i_{2}+b.i_{1}+i_{1}i_{2}=ab+I$$  We know that $a.i_{2}+b.i_{1}+i_{1}i_{2}\in I$.   If $(a+I).(b+I)=ab+I$, then taking suitable $(i_{1},i_{2})\in I\times I$, we should be able to prove $(a+i_{1}).(b+i_{2})=ab+i_{3},\forall i_{3}\in I$. However, can every element in $I$ be generated by taking suitable $i_{1},i_{2}\in I$? And if not, is that the reason why $(a+I).(b+I)$ has to be defined as equal to $ab+I$ in violation of the distributive property of ring elements? This has confused me for a long time.   EDIT: I figured addition does not have to be defined as such because $(a+i_{1})+(b+i_{2})=a+b+i_{1}+i_{2}$, where $i_{1}+i_{2}\in I$. In fact, every element $i\in I$ can be constructed by taking $i_{1}=0$ and $i_{2}=i$. Hence, $(a+I) + (b+I)=a+b+I$ naturally. This is the logic I followed to determine that $(a+I).(b+I)$ doesn't quite work as nicey. I'm not sure if this logic is flawed or not as I haven't had the opportunity to ask anybody.   Thanks for your help in advance! Suppose $f:[0,1] \to (0,\infty)$ is a Riemann integrable function. Prove that the integral of the function from $0$ to $1$ is strictly positive.   I have been trying to do this for awhile but I can't seem to get it. Here is my thought process: If the function is Riemann integrable, then its set of discontinuities has measure zero (Lebesgue). I am not sure how to connect this to the integral being positive A thought experiment: Imagine the Sun is suddenly removed. We wouldn't notice a difference for 8 minutes, because that's how long light takes to get from the Sun's surface to Earth.   However, what about the Sun's gravitational effect? If gravity propagates at the speed of light, for 8 minutes the Earth will continue to follow an orbit around nothing. If however, gravity is due to a distortion of spacetime, this distortion will cease to exist as soon as the mass is removed, thus the Earth will leave through the orbit tangent, so we could observe the Sun's disappearance more quickly.  What is the state of the research around such a thought experiment? Can this be inferred from observation? Problem: Evaluate $$ L=\displaystyle \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left(  \left\lfloor \frac{2n}{k} \right\rfloor -2\left\lfloor \frac{n}{k}\right \rfloor \right).$$  Please help me with this one.  I tried using the sandwich theorem by first constructing the following  $$ 2 \geq \left\lfloor \frac{2n}{k} \right\rfloor -2\left\lfloor \frac{n}{k} \right\rfloor \geq -1 $$ and then summing over. But the limit of the two bounds are different so sandwich theorem doesn't help!. Is it possible to construct sharper bounds?  Also I observed that if we ignore question of continuity, etc. then we can write,  $L= \displaystyle \int ^{1}_{0}\left\lfloor \frac{2}{x} \right\rfloor -2\left\lfloor \frac{1}{x} \right\rfloor dx$  Now I am able to compute the above integral. But is the above process correct? If then how can make it rigorous? Or if any other approach is possible, please tell. Apply a string replacement on the citation as returned by \cite Can my child attend a private school whilst I'm on USA ESTA? I am British, currently staying in the USA for 6 weeks on an ESTA Visa Waiver. I am living with a local family and whilst here my child (9 years old) has been attending a private (non-state) school with the children of the local family.  Today somebody told me that by having my child attend a private school I may be in violation of the terms of the ESTA Visa Waiver.  I have found nothing about this at   Where can I find official documentation / online advice about this? We are supposed to prove that the zero vector is unique with the additive identity $x+O=x$ and then by creating another additive identity with $O'$, then $x+O'=x$.  We set them equal $x+O=x+O'$ and use the vector cancelling theorem to get $O=O'$, but how does this prove that the zero vector is unique? Infinitely many primes for quadratic residues I have a Google Apps for your domain account for my domain which we will call DomainA  I have since replaced that domain with DomainB which is set up in the Google Apps account as an alias.   Since I want to let DomainA expire at some point in the future, I would like to change my primary domain to DomainB in my Google Apps account with DomainA as an alias.   How can I do this? I am Indian passport holder living in Hungary for 4 years and I have been issued a Temporary residence permit (EU Schengen).  Can I go to visit Dublin without any visa, or do I need to apply for a UK visa? I know Ireland still in Schengen but part of United Kingdom.