How can I select all instead of one by one in gmail Android version application. ?
I've previously used the default Email app on my CyanogenMod 7 EVO because I had an Exchange account at work. However, I switched jobs recently and my new place just uses GMail corporate accounts (which I enjoy). Since I don't have the Exchange account anymore, I'd like to move over to the GMail app because I prefer the UI and whatnot, but I can't for the life of me find any kind of "Select All" functionality in it.  The Email app allows me to do this from the options menu (Menu-&gt;Select All selects everything in the current folder). In GMail I noticed that if I select one or more items, then "Deselect All" becomes available in the options menu, but not "Select All". I've also tried various forms of long-pressing on messages, accounts, and folders to no avail.  Is there any way to do this in the GMail app, or am I just stuck without it? I searched the issue tracker and didn't find anything relevant, which surprised me a bit, to be honest. Obviously I can do this from a computer if I need to empty a folder/label, but I'd kind of like to retain the functionality on my phone, too. Am I missing some non-obvious (or perhaps even obvious) menu item somewhere?
Is there a geometric reason why two merging black holes never “decay” into two separate black holes All black hole mergers seem to result in one Kerr black hole after ring-down.   Could it be possible, that two initial black holes with two separate event horizons form an “intermediate compound state” with one event horizon, which - at a later time - “decays” in two final black holes?  I know that the answer to this question is “no” when applying Bekenstein-Hawking entropy where the decay would violate the second law which postulates $dA/dt &gt; 0$.   And the answer seems to be “no” when relying on numerical relativity for black hole mergers.  My question is if there is a more elegant, geometrical proof following entirely from general relativity, which rules out such a decay.  Thanks Object referencing its own property on initialization How do I get the title of a .pdf document to appear as the page title? A question that may sound very trivial, apologies beforehand. I am wondering why $( \mathbb{Q}_{&gt;0} , \times )$ is not isomorphic to $( \mathbb{Q} , + )$. I can see for the case when $( \mathbb{Q} , \times )$, not required to be positive, one can argue the group contains elements with order 2 (namely all negatives). In the case of the requirement for all rationals to be positive this argument does not fly. What trivial fact am I missing here? Assurance that ST_Envelope works with geography as expected Using PostGIS, I have a table of data with geographic points defined as last_known_coordinates GEOGRAPHY(POINT,4326). I want to find all of these within a geographic bounding box with given top left and bottom right geographic coordinates, and be sure that the calculations are geographic, not geometric, e.g. that the results account for the curvature of the earth. The query I'm using contains the fragment: last_known_coordinates &amp;&amp; ST_MakeEnvelope(?, ?, ?, ?, 4326)::geography As you can tell I'm a bit new to this, and I know what I want, but not necessarily how to do it correctly. Other answers hint that this is correct, but I'd like to know for sure: Will this point data and SQL fragment take into account the curvature of the Earth in the bounding box? I am often logged in several SSH sessions at once. To logout from several sessions, I press CTRL+d, until I am back on my local machine. However, I occasionally press it once too many, and my terminal exits. Is there a way to make CTRL+d unable to close my terminal ? I am using terminator as my terminal emulator. What are independent parameters in Hellmann–Feynman theorem? A typical example in textbooks about the application of Hellmann–Feynman theorem is calculating $\left\langle\frac{1}{r^2}\right\rangle$ in hydrogen-like atoms.  has a nice demonstration of this. At some point in the Wikipedia derivation is used that $$\frac{\partial n}{\partial \ell}~=~1. \tag{1}$$   But why is eq. (1) true? I know that  $$n=n_r+\ell+1,$$   but $n_r$ is just another variable with different physical meaning, so why is $n_r$ independent from $\ell$, whereas $n$ is not? The Wikipedia proof for Hellmann–Feynman theorem does not address the problem of independence of different parameters. What variables are kept fixed during the differentiation $(1)$ and why?  The Wikipedia page seems to have only a vague notion of $\frac{\partial\hat{H}}{\partial\lambda}$ and $\frac{\partial E}{\partial\lambda}$, unlike in, e.g., thermodynamics, where all partial derivatives are typically written like  $$\left( \frac{\partial U}{\partial V} \right)_S \qquad ,\qquad  \left( \frac{\partial U}{\partial V} \right)_p \qquad ,\qquad  \ldots $$ so that it is clear which variables are kept fixed during the differentiations. How do I uninstall Windows 8 and install Windows 7? I have a friend, who just got a new Toshiba Satellite C series with Windows 8. I tried to uninstall Windows 8 and install Windows 7 but was unable to do so.  I did everything I could:   Boot disk - would not load (error message Media -- Fail) Boot USB - would not load (error message Media -- Fail) Loading Windows 7 disk from system, fails with different errors   Does anybody know what I can do in order to install Windows 7? Suppose $n$ is an even positive integer and $H$ is a subgroup of $\mathbb Z/n \mathbb Z$. Prove that either every element of $H$ is... What is the right first step to take when integrating $1/(2+ \sin x + \cos x)$? I want a hint on this following integral. I don't know the first step, do I do $u$-sub. or do I use trig. sub or something?  $$\int \frac{1}{2+ \sin x + \cos x} ~\mathrm{d}x$$ Can you contract the main verb '[I] have' in a sentence? Where is the operator panel in Blender 2.8? The tutorial I read is based on 2.79. So I'm not sure if it's still the same name in 2.8. But I can't find anything providing similar functionalities. This is fairly easy.  If I go to https://webserver.com/someurl?a=b, will the someurl?a=b part be protected, or just the contents of the site? Is it possible to test apex that relies on field history tracking? I have some code that relies on field history tracking by querying the History for a custom object. This works fine, but it appears that the history records aren't created during the test so none of my tests will pass. Maybe the concept, why it's used, and an example. Why is compactness in logic called compactness? In logic, a semantics is said to be compact iff if every finite subset of a set of sentences has a model, then so to does the entire set.   Most logic texts either don't explain the terminology, or allude to the topological property of compactness. I see an analogy as, given a topological space X and a subset of it S, S is compact iff for every open cover of S, there is a finite subcover of S. But, it doesn't seem strong enough to justify the terminology.   Is there more to the choice of the terminology in logic than this analogy? Elementary properties of closure Hi everyone I'd like to know if the following is correct. I really appreciate any suggestion. (Honestly the only one that matters me is the second property the others are easy, I think) Thanks.  Definition: Let $X\subset \mathbb{R}$ and let $x'\in \mathbb{R}$, we say that $x'$ is an adherent point of $X$ iff $\,\forall\varepsilon&gt;0\,\exists x\in X \text{ s.t.} \; d(x',x)\le \varepsilon$. We say that $\overline{X}$ is the closure of $X$ iff contain all the adherent points of $X$.  Lemma (Elementary properties of Closures): Let $X$ and $Y$ be arbitrary subset of $\mathbb{R}$. Then $X\subset \overline{X}, \; \overline{X\cup Y}=\overline{X}\cup\overline{Y},\; \overline{X\cap Y}\subset\overline{X}\cap\overline{Y}$. If $X\subset Y$ then $\overline{X}\subset \overline{Y}$.  Proof:  Clearly if $x\in X$,  $x$ is $\varepsilon$-adherent to $x$ (indeed is $0$-close to itself) for every $\varepsilon$, hence $x\in \overline X$.   Let $z'$ be an adherent point of $X\cup Y$, then there exists some $z\in X\cup Y$ such that $d(z',z)$ for a given $\varepsilon&gt;0$. Suppose  $z\in X$, hence $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}\subset \overline{X}\cup \overline{Y}$. Similarly when  $z\in Y$, $z'\in \overline{Y}\subset \overline{X}\cup \overline{Y}$. Thus $ \overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. We  know that $X\cup Y \subset\overline{X\cup Y}\subset\overline{X}\cup \overline{Y}$. Then $\overline{\overline{X\cup Y}}=\overline{X}\cup \overline{Y}$. Since $\overline{\overline{X\cup Y}} = \overline{X\cup Y}$ we're done.  Let $z'\in \overline{X\cap Y}$, i.e., $z'$ is an adherent point of $X\cap Y$. So, there is some $z\in X\cap Y \subset X$ such that $d(z',z)\le \varepsilon$ for any given $\varepsilon&gt;0$. It follows that $z'$ is an adherent point of $X$, i.e., $z'\in \overline{X}$. Similarly since $z\in X\cap Y \subset Y$ we have $z'\in \overline{Y}$. Hence $z'\in \overline{X}\cap\overline{Y}$.   Let $x'\in \overline{X}$ we'd like to show that $x'\in \overline{Y}$. Let $\varepsilon&gt;0$ be arbitrary then $\exists x\in X\subset Y$ such that $d(x',x)\le \varepsilon$, so $x'$ is an adherent point of $Y$, i.e., $x'\in \overline{Y}$ as desired. $\Box$     Claim 1: If $X\subset Y \subset \overline{X}$. Then $\overline{Y}=\overline{X}$.    Proof claim: Let $y'$ be an adherent point of $Y$, i.e., $y'\in \overline{Y}$ and let $\varepsilon&gt;0$ be given. Thus, $\exists y \in Y\subset \overline{X}$ s.t. $d(y,y')\le \varepsilon$. So, $y$ is an adherent point of $X$. Thus $\exists x\in X$ s.t. $d(x,y)\le \varepsilon$ and therefore  we must have $d(x,y')\le 2\varepsilon$ which shows that $y'$ is an adherent point of $X$, i.e., $\overline{Y}\subset \overline{X}$. Conversely if $x'$ is an adherent point of $X$. Then $\exists x\in X\subset Y$ s.t. $d(x,x')\le \varepsilon$, and so is adherent to $Y$. $\Box$     Claim 2: $\overline{\overline{X}} = \overline{X}$    Proof: It will suffice to show that if $x''\in \overline{\overline{X}}$, then $x''$ is an adherent point of $X$. Let $\varepsilon &gt;0$ be arbitrary and let $x''$ be an adherent point of $\overline{X}$. Then there is some $x' \in \overline{X}$ such that $d(x'',x')\le \varepsilon$. Since $x' \in \overline{X}$ by definition is an adherent point of $X$, so there is some $x\in X$ for which $d(x,x')\le \varepsilon$. Thus $d(x,x'')\le d(x,x')+d(x',x'')\le 2\varepsilon$ it follows that $x''$ is an adherent point of $X$ as desired. $\Box$ Form within a block I'm using drupal_get_form('event_signup_form') to embed my custom form inside a block. The form is being rendered as intended, but submitting the form triggers nothing.  The form function itself is called event_signup_form($form, &amp;$form_state) and the submit function is called event_signup_form_submit($form, &amp;$form_state) but it's not being called.  I am using  to show the form:  function event_block_view($delta = '') { $blocks = array(); if ($delta ...) { $blocks['content'] = drupal_render(drupal_get_form('event_signup_form')); } }   I managed to get this working - here is how (stub):  function hook_block_view($delta = '') { $blocks['content'] = drupal_render(drupal_get_form('event_signup_form')); }  function event_signup_form() { // Register form submit handler, same goes for validation handler $form['#submit'][] = 'event_signup_form_submit';  return $form; }  function event_signup_form_submit($form, &amp;$form_state) { // This is now triggered } I want to initialize a struct element, split in declaration and initialization. This is what I have:  typedef struct MY_TYPE {   bool flag;   short int value;   double stuff; } MY_TYPE;  void function(void) {   MY_TYPE a;   ...   a = { true, 15, 0.123 } }   Is this the way to declare and initialize a local variable of MY_TYPE in accordance with C programming language standards (C89, C90, C99, C11, etc.)? Or is there anything better or at least working?  Update I ended up having a static initialization element where I set every subelement according to my needs. What do I need to run a D&D 5E campaign? I have GM'd several RPGs before that are not D&amp;D (Fate, Deathwatch, Star Wars). We first tried to play D&amp;D using the 3.5 edition and it went horribly — too many books and too many options with no idea of where to start.  I have been told that 5E takes care of all that and is easier and more fun so I want to try that but I would like a definitive list of what I need.  I expect that the kinds of books we need divide up like this:   Rulebook for me as GM Rulebook for players Fluff for everyone Monster book for me as GM   Links to these would be helpful as we got very confused with the 3.5 books — who should use what books confused us.  What other things do we need to play a campaign? I expect things like:   Dice: what kinds and how many of each kind, roughly. Models and maps. Do we need these? Does 5E need models and a grid or can it be played without (which is preferred)?   Is there anything else that we need that I haven't anticipated?