When light changes medium, does it decelerate, or instantaneously change speed?
Do photons have acceleration?
Is this proof of $\liminf E_k \subset \limsup E_k $ correct? I am wondering if my proof is correct? Thank you for whoever willing to take a look at it for me.     Proof $\liminf E_k \subset \limsup E_k $   If $\{E_k\}_{k=1}^\infty$ is a sequence of sets, we define \begin{align*} \limsup E_k &amp; = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right)\\ &amp;= \bigcap_{j=1}^\infty(E_j \cup E_{j+1} \cup \cdots)\\ &amp;= (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots. \end{align*}  Therefore, $\limsup E_k$ consists of those points in $\mathbb{R}^n$ which belong to infinitely many $E_k$. \begin{align*} \liminf E_k &amp; = \bigcup_{j=1}^\infty\left(\bigcap_{k=j}^\infty E_k\right)\\ &amp;= \bigcup_{j=1}^\infty(E_j \cap E_{j+1} \cap \cdots)\\ &amp;= (E_1 \cap E_2 \cap \cdots) \cup (E_2 \cap E_3 \cap \cdots) \cup \cdots. \end{align*} Therefore, $\liminf E_k$ consists of those points in $\mathbb{R}^n$ which belong to all $E_k$ for $k \geq k_0$.  Now consider $x \in \liminf E_k$, then $x \in E_1 \cap E_2 \cap \cdots$, or $x \in E_2 \cap E_3 \cap \cdots$ and so on, that is to say, $x \in E_{k_0} \cup E_{k_0+1} \cup \dots$ for some $k_0$. More specifically, $x \in E_k \forall k \geq k_0$. Therefore, $x \in (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots$, so we showed $x \in \limsup E_k$. When would you use the Builder Pattern? What are some common, real world examples of using the Builder Pattern?  What does it buy you?  Why not just use a Factory Pattern? Proof Check: Every Cauchy Sequence is Bounded Sorry if I keep asking for proof checks. I'll try to keep it to a minimum after this. I know this has a well-known proof. I understand that proof as well but I thought I'd do a proof that made sense to me and seemed, in some ways, simpler. Trouble is I'm not sure if it's totally correct. It's quite short though. I was just hoping someone could look it over and see if it is a valid proof. Thank you!  Lemma: Every Cauchy sequence is bounded.  Proof: Let $(a_{n})$ be Cauchy. We choose $ 0&lt;\epsilon_{0}$. So $ \forall \; n&gt;m\geq N_{0}$ we have that $\vert a_{n}-a_{m} \vert &lt; \epsilon_{0}$. Therefore $(a_{n})$ is bounded for all $ m \geq N_{0} $ by $ \epsilon_{0} $. Since $ \mathbb{N}_{N_{0}}$ is finite, it is bounded. So, for all $ m&lt;N_{0} $, $ (a_{n})$ is bounded. Therefore $(a_{n})$ is bounded.  I realize I haven't said what the bounds are but I think that's sort of irrelevant. So long as we know it's bounded. Any help is much appreciated! change folder icons in ubuntu-mono-dark theme Let $a$ and $b$ be natural numbers with $ab&gt;2$. Suppose that the sum of their highest common factor and least common multiple is divisible by $a+b$. Prove that the quotient is at most $\frac{a+b}4$. When is this quotient exactly equal to $\frac{a+b}4$. Round a double to x significant figures How to draw circle tangents and removing the inner lines? I'm new to Adobe Illustrator and I wanted to accomplish this task. How do I remove the inner lines in my drawings? I tried with all the pathfinders effect but it keeps removing my tangent lines (the lines connecting both my circles).    It works like a charm as I followed the selected's answer guide:  Here it is. I would like my sections to start on a new page if they do not fit on the current page. In other words:  if a section fits on the current page print it, if it does not fit start printing it on the next page.  I would like this to be done automatically.  \usepackage{titlesec} seems to be the solution, but I can not get it to do the job.   Should it be done with titlesec? how ? Is there a better approach? Question title overlap in new design when title is longer Today I noticed that in new design view count is getting overlapped with question title on scifi.se for longer title:    But it's not site specific as same problem happen in Beta site too ( Hinduism.se):    That's why asked in main meta. Approximating a $\sigma$-algebra by a generating algebra Theorem. Let $(X,\mathcal B,\mu)$ a finite measure space, where $\mu$ is a positive measure. Let $\mathcal A\subset \mathcal B$ an algebra generating $\cal B$. Then for all $B\in\cal B$ and $\varepsilon&gt;0$, we can find $A\in\cal A$ such that $$\mu(A\Delta B)=\mu(A\cup B)-\mu(A\cap B)&lt;\varepsilon.$$  I don't think there is a proof in this site. It's a useful result for several reasons:  We know what the algebra generated by a collection of sets is, but not what the generated $\sigma$-algebra is. The map $\rho\colon \cal B\times\cal B\to \Bbb R_+$, $\rho(A,A')=\mu(A\Delta A')$ gives a pseudo-metric on $\cal B$. This makes a link between generating for an algebra and dense for the pseudo-metric. We say that a $\sigma$-algebra is separable if it's generated by a countable class of sets. In this case, the algebra generated by this class is countable. An with the mentioned result, we can show that $L^p(\mu)$ is separable for $1\leq p&lt;\infty$, which makes a link between the two notions. In ergodic theory, we have to test mixing conditions only an a generating algebra, not on all the $\sigma$-algebra. Create subdomains on the fly with .htaccess (PHP) I am looking to create a system which on signup will create a subdomain on my website for the users account area.   e.g. johndoe.website.com  I think it would be something to do with the .htaccess file and possibly redirecting to another location on the website? I don't actually know. But any information to start me off would be greatly appreciated.  Creating a sign up area is not the problem - I have done this many a time. I am just unsure where to start with the subdomain. How can the petrification from the Flesh to Stone spell be removed? Poisson Distribution of sum of two random independent variables $X$, $Y$ $X \sim \mathcal{P}( \lambda) $ and $Y \sim \mathcal{P}( \mu)$ meaning that $X$ and $Y$ are Poisson distributions. What is the probability distribution law of $X + Y$. I know it is $X+Y \sim \mathcal{P}( \lambda + \mu)$ but I don't understand how to derive it. When Ron broke his wand, why didn't he apply for Hogwarts financial aid? We know, from Book Six, that Hogwarts has a financial aid for students in need. It's supposed to cover all expenses regarding materials and school supplies, so it certainly covers wands. So why didn't Ron simply apply for this fund when he broke his wand in Book Two, since his family could not afford a new one?  Seems absurd that no teacher would notice that Ron had a broken wand and inform him there's a school fund. Plot hole?  Obs.: Even if Ron could afford a new wand, one could ask how we would purchase it (you must experiment personally to find a wand that chooses you). It turns out that there's a local branch of Ollivander in Hogsmeade. I'm sure that, under those circumstances, a second-year student like Ron would be allowed to visit Hogsmeade to purchase a new wand. How did Vader saving Luke turn him to the good side? Couldn't Vader save his son and continue being a Sith Lord? Couldn't he have let Luke go and basically become the Emperor and have the Galaxy to himself? I don't think killing the Emperor to save your son will automatically convert you to the light side. Enlighten me. When signing in with Google (OAuth), it shows a list of accounts it already knows about. How can I remove some accounts from this list?    I know I can nuke all of them by deleting all cookies, but I'd prefer to remove one account that I don't want there, rather than all of them.  There used to be an ability to edit this list in the older version of this screen, but in 2018 redesign Google has removed [×] buttons next to the accounts. Run cURL commands from Windows console Is there a way to install cURL in windows in order to run cURL commands from the command prompt? Computing the Galois group of $x^4+ax^2+b \in \mathbb{Q}[x] $ I want to compute the Galois group of some polynomials, but I want to see some examples first. For example this proposition could be helpful. I don't know how to prove it &lt;.&lt;  Let's consider a polynomial $f(x)= x^4+ax^2+b \in \mathbb{Z} $. Let $ \pm\alpha , \pm\beta$ denote the roots of $f(x)$. Note that $f$ is irreducible iff $\alpha^2, \alpha\pm\beta \notin \mathbb{Q}$. So let's suppose that $f(x)$ is irreducible. Let's denote $G$ the galois group of $f$ (the Galois group of the splitting field $\mathbb{Q}(\alpha,\beta)/\mathbb{Q}$)  $i)$ $G\cong V$ the Klein 4-group iff $\alpha\beta \in \mathbb{Q}$  $ii)$ $G\cong \mathbb{Z}_4$ iff $\mathbb{Q}(\alpha\beta)=\mathbb{Q}(\alpha^2)$  $iii)$ $G\cong D_8$ the dihedral group of order 8, iff $\alpha\beta \notin \mathbb{Q}(\alpha^2)$  I need help with the proofs of this propositions... Well I'm very lost with the computation of Galois group, some care must be exercised, I don't have it. For example I know that Galois group acts transitively, in the sense that $u\in K/F$ a galois extension, the other roots of the minimal polynomial of $m_u(x)\in F[x]$ are precisely the elements $ \sigma(u)$ , where $\sigma \in Gal(K/F)$. And clearly any automorphism, is determined by only the value of a basis. In this case we are working on $\mathbb{Q}(\alpha,\beta)$, so I have to know the values of both. But I have to respect the algebraic relations involving $\alpha , \beta$ I proved that in $i),ii)$ $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\alpha) $ So I have to know the values on $\alpha$.   My first question is: If in both cases $i),ii)$, I have an extension of degree 4 (so $|G|=4$) I have to map $\alpha$ to determine the automorphism, but I have also to map $\alpha$ to all of it's four roots. Why the galois group in this case are different?  My second question is about $iii)$ I have no idea How to attack it. Prove that if $A$ is transitive, then $\mathcal{P}(A)$ is transitive too. My book on set theory says     Prove that is a set $A$ is transitive, then $\mathcal{P}(A)$ is also transitive, where $\mathcal{P}(A)$ is the power set of $A$.   I don't know understand how. Let $A=\{a,b\}$. Clearly it is transitive. Now $\mathcal{P}(A)=\{\{a\},\{b\},\{a,b\},\emptyset\}$. This is clearly not transitive. Gauss's Lemma for polynomials claims that a non-constant polynomial in $\mathbb{Z}[X]$ is irreducible in $\mathbb{Z}[X]$ if and only if it is both irreducible in $\mathbb{Q}[X]$ and primitive in $\mathbb{Z}[X]$.   I wonder if this holds for multivariable case.      Is it true that a non-constant polynomial in $\mathbb{Z}[X_1,\dots,X_n]$ is irreducible in $\mathbb{Z}[X_1,\dots,X_n]$ if and only if it is both irreducible in $\mathbb{Q}[X_1,\dots,X_n]$ and primitive in $\mathbb{Z}[X_1,\dots,X_n]$?    Thank you for your help.