I'd like to put other symbols than arrows, e.g. a + sign, between the nodes of a tikzcd diagram. Is this possible?
Please help me to draw the following diagram:  \begin{displaymath} \xymatrix{ H\ar[r]&amp;H .1={}_HI_Q\supset 1.Q&amp;Q\ar[l]_\epsilon^{\cong}  \\  &amp;R\ar[u]\ar[r]&amp;R\ar[u] } \end{displaymath}
I need to add a coloured line before my caption; this is what I used:  \usepackage[labelfont=bf]{caption}  \DeclareCaptionFormat{myformat}{\hrulefill #1  #2 #3} \captionsetup[figure]{format=myformat}   but the figure turns out to be line this:    and not a horinzontal line which what I want  2) I want the line to be coloured RGB [152 204 204] with thickness about the same as \top rule. Can anyone show me how to do this? Age and Graduate school If $e^{itx_n}$ converges for every $t\in\mathbb R$, then does $x_n$ converge? Programmatically selecting non adjacent Cells What is the earliest work considered to be Science Fiction? How can I detect if I'm in a subshell? Eeeek! Something stole the protection! (Why is this protected question unprotected?) asked me to report this (apparent) bug.  When he tried to protect  question someone (or something) stole the question's protection, which has since been missing:          The question shows no sign of having the protection removed, thus I believe it was abducted. I am sure justice will be done, and this poor question will receive the protection it deserves.     I recommend interrogating the  that was said to have announced this question to the world, I bet it would have some useful information. How and when is old-releases.ubuntu.com updated? Construct a Borel set on R such that it intersect every open interval with non-zero non-"full" measure This is from problem $8$, Chapter II of Rudin's Real and Complex Analysis.  The problem asks for a Borel set $M$ on $R$, such that for any interval $I$, $M \cap I$ has measure greater than $0$ and less than $m(I)$.  I was thinking of taking the Cantor approach: taking $R$ to be the union of $[a,b]$ with $a$ and $b$ rationals, and for each $[a,b]$ we construct Cantor sets inside it. During theconstruction of each Cantor set, in order to have positive measure on it, we need to take off smaller and smaller intervals from it, namely the proportion goes to $0$. As a result, these Cantor sets are extremely "dense" on their ends. If for an interval $I$ it intersects with the Cantor set on $[a,b]$ while $b-a&gt;&gt;m(I)$, we shall expect the measure of intersection to be rather close to $m(I)$ and then we lose control on these cases.  Is there any way to fix this or shall I consider other approaches?  Thank you At what point does a caster define the target of a spell? The new School of Invention tradition for Wizards () while still in playtesting (and indeed a bit silly in my opinion, more of a "chaos wizard"), got me thinking about its reckless casting ability. This allows the caster to cast magic they do not have prepared and get a random result. The random spells have different targeting criteria (Wall of Stone, Fireball, Feign Death , Mass Cure Wounds) and so it is not possible to decide the target of the spell until it has been cast, as you don't know what you are going to get until it is cast.  This suggests that the designer of the subclass has made the assumption that you don't actually target until you have cast the spell and release it, which makes sense from one point of view, strongly supported by the way a prepared action casting works, PHB p.193 where you cast the spell and hold the energy until the condition is met for the readied action at which point you target the spell.  While the Unearthed Arcana is not RAW or finalized and released, my question is not about this subclass, but about the official rules as a whole.  Is there was any direct RAW, or other supporting or opposing RAW, that states at what point a caster defines the target of a spell?  As a sub-question: are there any circumstances where it actually matters, other than the already defined Readied Action spell casting, in the official rules? Wouldn't the thermodynamic cost of creating alternate universes make the Many Worlds interpretation implausible? I was thinking about the many interpretations of quantum physics, and one thing that never made sense to me was the many world's interpretation. Basically at any given moment for which something exists in a superposition, isn't this interpretation basically saying that there would have to exist a new universe for every possible discrete outcome that could result once the measurement is made?   So let's say an electron's location is a superposition of two distinct locations. Doesn't that mean there would at that moment have to be two distinct universes for each distinct location that is superposed? If so, then clearly the amount of universes being spawned every moment must be astronomical.   There could be trillions upon trillions of electrons(for instance) having superposed states, and then every moment in time (I suppose every planck time?), it seems there would have to be yet another set of universes spawning for every superposition.  How could this be? Surely there must be some kind of thermodynamic cost to replicate an entire universe. What would fuel or supply the additional mass and energy for every universe at every moment for every superposed state?  To my lay person's mind, this seems to be a very big weakness of this interpretation, but I needed to ask people with the physics background required to answer this question. Could there be some kind of loophole that could allow alternate universes to spawn so rapidly? Otherwise, doesn't this present a bit of a problem for taking this interpretation seriously? $(\Bbb R,+)$ is a topological group. Is there any other group structure on $\Bbb R$ such that it is still a topological group and this group is not isomorphic to $(\Bbb R,+)$ ?  Refer to  for the general problem. In that problem I have conjectured that there is no such group however I couldn't prove it. Transform the NFSA with transition table Transform the NFSA with transition table  $$\begin{array}{ccc} X&amp;a&amp;b\\ A&amp;\{\}&amp;\{F\}\\ B&amp;\{\}&amp;\{\}\\ F&amp;\{\}&amp;\{A,S\}\\ S&amp;\{F\}&amp;\{A,F\} \end{array}$$  and starting state $S$ inta a DFSA and give the transition table of that DFSA. Place the letter $X$ (uppercase) in the top left corner of the matrix. Your states in the first column and inputs in the first row can be in the order of your choice.  The question then offers a $7\times 3$ matrix to be filled in.  Unsure how to transform the NFSA with transition table.  This is my answer but it is incorrect.  $$\begin{array}{ccc} X&amp;a&amp;b\\ \{S\}&amp;\{F\}&amp;\{A,F\}\\ \{F\}&amp;\{\}&amp;\{A,S\}\\ \{A\}&amp;\{\}&amp;\{F\}\\ \{A,F\}&amp;\{\}&amp;\{A,F,S\}\\ \{A,S\}&amp;\{F\}&amp;\{A,F\}\\ \{A,F,S\}&amp;\{F\}&amp;\{A,F,S\} \end{array}$$ To infinitive used after adjective This question is relatively simple. I don't understand why we never use passive form of to infinitive after the adjective unless the subject is "it".  For example:     He is difficult to please.   People consider the example above a correct sentence. But:     He is difficult to be pleased.    People, for example, my teacher, said it sounds wrong and told me not to write this type of sentence. Is it grammatically flawed, or is it just a stylistic choice? Proving $\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$ without induction I was looking at: $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$  It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation? I have a database which contains 5 tables. Each table contains 24 rows and each row contains 4 columns.  I want to display these records in Excel sheet. The heading of each table is the name of the table, but I am unable to merge the columns for heading.  Please help me. I'm new to this website. I've checked the other topics and couldn't find the answer to my question. I've also tried to find the result with TI-CAS but the only thing it does is simplify the inequality which really doesn't help at all.  the equation is : $1000n^3 &lt; 2^n$ Now I've been trying to use log but it gets me nowhere (I haven't seen this form before), the answer is supposed to be 24 but I'm not sure. So my question is how to solve step by step this inequality for n. Could you guys help me with this one? Real Numbers Raised to Imaginary Powers? What is a real number to the power of an imaginary or complex number? e.g. 3i. I have searched through sites about imaginary numbers, but none seem to say anything about imaginary indices. Examples and explanations would be appreciated. How do I login to the Gamecenter on PewDiePie's Tuber Simulator, if you have not done it when they asked? number of edges induction proof Proof by induction that the complete graph $K_{n}$ has $n(n-1)/2$ edges.   I know how to do the induction step I'm just a little confused on what the left side of my equation should be. $E = n(n-1)/2$  It's been a while since I've done induction. I just need help determining both sides of the equation.