I am writing an affirmation to motivate myself to be more assertive.  Choice 1: "I am assertive in any situation."  Choice 2: "I am assertive in any situations."  The question is: Should I use the word situation in singular or plural form?
I always thought with "any" I should use the plural, but on the internet I can find both:   It can be found in any book. It can be found in any books Do you have any books? It can be said in any language.  This can be understood by anyone.  It has been used in any form.   So, what's correct?  Is there any rule?
So, to explain the title, I'm referring to the necessity of the axiom of choice in the existence of a well ordering on reals, or any uncountable set.  Now, while tweaking some sets, I came across this :  We start with the natural numbers, $N$. We take the power set of the naturals, $P(N)$.  Then we remove all the finite subsets of $N$ from $P(N)$. Let us call this new set $S$. This set is the set of all infinite subsets of $N$.  It is easy to show that $S$ has uncountable cardinality, same as that of real numbers. This is because the removal of finite subsets only removes a countable number of elements. (Haven't posted this deduction, for it is very easy, but I may post it if it is not so evident)  Now, we seek to find an ordering on the set $S$. Every set in this set is an infinite subset of natural numbers, so each of these sets are well ordered by the natural ordering of $N$.  Taking any two sets in $S$, say $A$, and $B$, we seek to order them by checking their elements lexicographically. We compare the first two elements in $A$, and $B$. Let them be $a_1$, and $b_1$ respectively. If $a_1 = b_1$, then we move on to the second elements in the sets, $a_2$, and $b_2$, and so on. If, at any point, $a_n &lt; b_n$, then $A &lt; B$, or if $b_n &lt; a_n$, then $B &lt; A$.  This order seems to be a well ordering of the uncountable infinity of reals. I don't seem to have invoked the axiom of choice anywhere in the construction of this set $S$.  So, why isn't this a well ordering on the uncountable of reals? As I understand it, electromagnetic waves have two components which are the result of each other, i.e., when a moving electric charge creates a changing magnetic field at point X then a changing electric field is created at point Y and this repeating process is what creates EM waves, so therefore, it requires no medium. Is my understanding correct?  One thing that I'm surprised to know is that light is also called an electromagnetic wave.  Does this include light of any kind, for example: light from a bulb, a tube, and also from the Sun? How do they contain electric and magnetic fields? Autocovariance of an ARMA(2,1) process - derivation of analytical model for $\gamma( k)$ This is the problem I am currently having trouble with:  If $\lambda_1$ and $\lambda_2$ are eigenvalues of matrix  $$         \begin{bmatrix}         a &amp; b\\         c &amp; d\\         \end{bmatrix}$$prove that $\lambda_1^2$, $\lambda_1\lambda_2$ and $\lambda_2^2$ are eigenvalues of matrix $A$.     $$A=\begin{bmatrix}     a^2 &amp; ab &amp; b^2\\     2ac &amp; ad+bc &amp;2bd\\     c^2 &amp; cd &amp;d^2\\     \end{bmatrix}$$  What have I considered so far:   I tried to find characteristic polynomial of first matrix and then find the roots which would be $\lambda_1$ and $\lambda_2$. I was hoping I can make a connection with characteristic polynomial of matrix $A$, but it got too complicated and did not seem like the right idea. Since there are two eigenvalues for first matrix, there are two eigenvectors. I thought about somehow using $A\vec v_i=\lambda_i\vec v_i$, where $i=1,2$  Also, I thought about finding a connection between these two matrices. What I see is that the sum of elements of first and second column, of first matrix, are $(a+c)$ and $(b+d)$, respectively.  Sum of first, second and third column of matrix $A$ are $(a+c)^2, (a+c)(b+d)$ and $(b+d)^2$. Now, if I could assign $\lambda_1$ and $\lambda_2$ to $(a+c)$ and $(b+d)$, then I would get that sum of first, second and third column of matrix $A$ are $\lambda_1^2$, $\lambda_1\lambda_2$ and $\lambda_2^2$, but what would be next?   I thought it would be good to combine 2. and 3rd idea, but I am not making any progress.  I appreciate any help. Prove that the union of the interior of a set and the boundary of the set is the closure of the set I'll denote closure of A with $\overline{A}$, $A^\circ$ as the interior of A, $\partial A$ as the boundary of $A$ and $A'$ as the accumulation points. I'm trying to prove the following:   $$\overline{A}=A^\circ \cup\partial A$$  Is this a valid proof?  Take $x \in A^\circ \cup \partial A$ then $x \in A^\circ$ or $x \in \partial A$   if $x \in A^\circ$ then $x \in \overline{A}$  if $x \in \partial A$ then $x \in \overline{A} \cap\overline{(X\setminus A)}$ thus $x \in\overline{A} $ so $A^\circ\cup\partial A\subset\overline{A}$  Take $x \in \overline{A}$ then $x \in A' \cup A$ thus $x \in A'\setminus A$ or $x \in A^\circ$  if $x \in A'\setminus A$ then $x \in \overline{(X\setminus A)}$ so $x \in \overline{A}\cap\overline{(X\setminus A)}$ and $x \in\partial A$ so $x\in A^\circ\cup\partial A$  if $x \in A^\circ$ then $x \in A^\circ\cup \partial A$ so $\overline{A}\subset A^\circ\cup\partial A$ Maybe this question is too obvious to everyone.      In Noetherian ring $R$, ideal $I\subset R$, $P\in\mathrm{Ass}(I)$, then $P=I:a$ for some $a\in R$ where $\mathrm{Ass}(I)$ denotes associated primes of $I$.   This is an exercise 8.29 in Gathmann's Commutative Algebra notes. I am supposed to find this statement obvious. However, up to now, I still did not find this statement obvious. Can someone explain intuitively why noetherianness here allows me removing radical $\sqrt{I:a}$ without going into details?  If $Q$ is a primary ideal of $I$, then certainly $\sqrt{Q:a}=P$ for $a\not\in P$. It seems taking $(p_1,\dots, p_n)=P$, $p_i^{n_i}a\in Q$. Since $Q$ primary, I must have $p_i\in Q$ and $p_i^{n_i}\in Q$. (This argument seems very wrong as this deduces the same conclusion without noetherian property.) Custom pagebreak in align equation My supervisor asked me to find out which distribution represents a particular situation.  I have a VoIP generator that generates calls "uniformly" distributed between callers. This means that the volume per caller distribution is "almost" uniformly distributed between a minimum and maximum. So by running a test with 10000 users and a min value equal to 30 calls per week and a max value equal to 90 calls/week i obtain that not all the users respect this limits: we have some users that generate &lt;30 calls and some other users that generate >90 calls. It is clear that the obtained distribution is not uniform.  The situation is this:    He said that i have to perform a sort of numerical process in order to find some formulas that could define this distribution. Initially, as wrote before,we wanted to obtain a uniform (min,max) distribution (the green area in the figure) but this is not the case as proved with chi-square test.  Moreover the curve in figure is not symmetric, the probability of call generation below 30 call/week and greater that 90 call/week are not identical (it is high for 90calls/week). The variability of the number of generated call increases with the increasing call generation rates.   "Actually implementation of this distribution is nothing but assigning different call rates in a range for users in domain which indicates implementation of several delta functions. As the call rates increases the variability of the generated calls also increases with the average call rate and this leads to the asymmetric behavior of the curve." [cit. from the Voipgenerator documentation]  Someone can help me?I think that now i cannot use Q-Q plot because i don't know which theoretical distribution i have to use in order to compare it with my empirical data.  Sorry if I have stressed with a similar problem a few weeks ago, but initially we thought we could change the implementation, but now we cannot. Hence i have to discover the type of the distribution i obtained and i don't know how can i do this. I have a process I can't kill with kill -9 &lt;pid&gt;. What's the problem in such a case, especially since I am the owner of that process. I thought nothing could evade that kill option. Highlight whole TreeViewItem line in WPF I hope this isn't a silly question but I've been watching TNG and this bothers me.  The transporter 'pad' area is relatively small.  As a guess, maybe 12 people can fit on it.  But the vast majority of times it's being used by 2-3 people.  However, the transporter is able to beam people from anywhere on the ship to anywhere (within some distance I presume).  It's fairly common for them to perform a transport directly to the bridge.  So, why is there an actual room?    If a three man away team is to beam down to the surface of a planet - why do they have to walk all the way to the transporter room, then stand on the pad to get down there?  Why not simply beam them from where they are, to the surface.  The ship's computer is able to locate anyone on board. How do you take profit from stock trading while keeping capital invested? How can there be a directory named `~`? I was trying to test out this buggy/new tool to see how it works, and it prompted me to choose a directory to use for stuff I am doing. From the directory ~/Code/indep, I typed in ~/Code/indep/tarbell. (Stupid, I could have just typed 'tarbell', but whatever.) It prompted me to create it because it didn't exist, I said yes. When I inspected the contents of this directory, here is what I see:  brian@navis:~/Code/indep$ ls ~ project1 project2 project3 ...   I thought "oh, weird, it created a symlink to ~". However:  brian@navis:~/Code/indep$ file ./~ ./~: directory   file thinks it's a directory.   brian@navis:~/Code/indep$ ls -l drwxr-xr-x 3 brian brian 4096 Nov  8 16:50 ~ drwxr-xr-x 8 brian brian 4096 Oct 14 18:42 project1 drwxr-xr-x 3 brian brian 4096 Sep 10 21:15 project2 drwxr-xr-x 5 brian brian 4096 Oct 14 19:33 project3 ...   And so does ls -l.   I thought, "oh, ... is it a hard link? I've never really used those." But I tried to reproduce, and it's impossible to create a hard link to a directory. So... I guess I've narrowed this down to things I don't understand. Did I mount my home directory in this other directory somehow?   My main goal at this point is to remove the reference to my home directory in this folder (rm doesn't seem like the answer), but I'm also curious to know what's going on.  EDIT:   In retrospect, I understand exactly what happened. As was answered below, the program created a directory called "~" inside my current directory. The only really obfuscating fact about this is that I wasn't consistent with how I was actually viewing what the directory was. Sometimes I referred to it as  "~" and sometimes as "./~". It's only if you do the latter that you can actually see what's going on, but until I understood this, I was conflating what I was seeing from using the absolute path of "~" to to what was in fact a local directory. Give an example of a sequence of continuous functions $f_n:\mathbb{R} \to \mathbb{R}$ such that the pointwise limit $f:\mathbb{R} \to \mathbb{R}$ exists but is discontinuous. My idea is $f_n(x)=\displaystyle\frac{x}{n}.$ Then the pointwise limit is $0$ but $f$ is discontinuous on $\mathbb{R}.$ Is that correct? I'm having a hard time understanding the definition of pointwise convergence. I have this definition: $\displaystyle\lim_{n\to \infty} f_n=f$ pointwise iff $\displaystyle\lim_{n\to \infty} f_n(x)=f(x)$ But then what is $f(x)$? Where does the $n$ go? In the example I gave, for instance, $\displaystyle\lim_{n\to \infty} f_n(x)=\displaystyle\lim_{n\to \infty} \frac{x}{n}=0.$ But does this equal $f(x)$? Is $f(x)$=0 a discontinuous limit? Is it possible to have an ASP.NET MVC route that uses subdomain information to determine its route?  For example:   user1.domain.com goes to one place user2.domain.com goes to another?   Or, can I make it so both of these go to the same controller/action with a username parameter? I tried to follow the new Blender 2.8 tutorial from Blender Guru.   When entering "Sculpting" with the icing there were suddenly alot of holes in it. Why is this? "Out of pocket"? I'm increasingly hearing the phrase "out of pocket" used in America as a colloquialism to mean "away from the office", "unavailable", or "incommunicado".     I apologize for not replying sooner; I have been out of pocket.    What is the etymology of this usage? It always sounds odd to me, since I had previously always associated the phrase with expenses, e.g., "I had to pay for the service out of pocket," meaning, "I had to pay for the service using my own money." Should either of these usages be avoided? What is the extent of their acceptance? Number Theory: Find all solutions of $\phi(n)=16$ and $\phi(n)=24$ Java volatile reference vs. AtomicReference HTML form readonly SELECT tag/input According to HTML specs, the select tag in HTML doesn't have a readonly attribute, only a disabled attribute. So if you want to keep the user from changing the dropdown, you have to use disabled.  The only problem is that disabled HTML form inputs don't get included in the POST / GET data.  What's the best way to emulate the readonly attribute for a select tag, and still get the POST data?