Several users expressed disagreement with some of 900 sit-ups a day's actions. OTOH there is no doubt that this user devotes lot of effort and time to this site. It is not surprising that he leads participation tab in the moment. I find especially his work on improving questions impressive.
I guess these 3 threads give an image on what was going on on meta recently. But there is a lot of answers and comments there, so it would take a long time to read all of them.
@AlecTeal I would not call him (or her?) arrogant. He certainly has a point.
@MartinSleziak I don't get the point in wanting to delete things (the obvious spam ones of course....) because it's a few KB somewhere, it hurts no one to keep it.
Quote from his answer: I see such Q&As as broken windows, through which new users enter (via Google search, typically), and conclude that the site is of low standards.
@MartinSleziak Just out of curiosity, it almost seems ill-advised to pump out calculus and precalculus answers with such high volume. I don't think that math.se should be solely for 'advanced' mathematics, but the vast majority of these more 'basic' questions have millions of completely analogous (if not literally the same) answers elsewhere on the web.
Just out of curiosity, what's your opinion about: *
I think the campaign is not against precalculus, calculus and low-level questions. It is mainly against badly written questions.
If you notice his profile, he quite often improves questions: By improving titles, wording, adding context or OP's effort if it was mentioned in comments.
@MartinSleziak I wasn't even necessarily commenting on any 'campaign', I was just asking in the abstract. Personally, I think that a lot of people take this website way too seriously. When I go on meta, which granted isn't that often, it seems like a bad high-school drama.
Off topic. If I define $CS(\mathbb{Q})$ as the set of all Cauchy sequences of rational numbers, the equivalence relation ~ on CS by $(x_n)~(y_n)$ if $x_n - y_n\to 0$, then define $\mathbb{R}=CS(\mathbb{Q}$, how do I show that this identification is unambiguous?
@Chris'ssis We are kids. You are a kid reading in a higher standard than @Khallil. That's all.
As for the series, you know I am not attracted to manipulations of integrals and series much. Beauty is highly subjective and to me it comes from Number Theory.
@Sush $$\sum_{j,k=1}^n \lvert x_j\rvert\,\lvert x_k\rvert = \sum_{j=1}^n\left(\lvert x_j\rvert \sum_{k=1}^n\lvert x_k\rvert\right) \leqslant \sum_{j=1}^n \lvert x_j\rvert \sqrt{n\cdot \lVert x\rVert^2} = \sqrt{n}\sum_{j=1}^n \lvert x_j\rvert \leqslant \sqrt{n}\sqrt{n\lVert x\rVert^2} = n.$$ For $n = 1$, we have equality, for $n > 1$ we have $n < n^2$.
@Sush The intention was probably to use $\lvert x_k\rvert \leqslant 1$ for $x\in U$, however, so each of the $n^2$ terms in the sum is bounded by $1$, hence the sum is $\leqslant n^2$, and we only can have equality if $\lvert x_k\rvert = 1$ for all $k$, but for $n > 1$ that contradicts $\lVert x\rVert = 1$.
@Alizter and then, after some simpel calcualtions, you get that $$\int _1^\infty \frac{\{t\}}{t^2}dt=1-\gamma$$
@Alizter your initial integral cannot evaluate to a negative value. Why? It was the first thing I noted there, and I concluded that something was wrong.
$\varphi^c$ is closed ... but we're talking about subspace (relative) topology for all of this. Well, work out my example. I mean open intervals $(0,1)$ and $(2,3)$, not ordered pairs. I should use your notation from Europe: $]0,1[$ and $]2,3[$. Agh, hard for me to type.
I mean like Subtract integers that can be positive and negative. Otherwise if two integers have the same number then it equals 0 and it be called additive inverse.
@Alizter What is wrong? Well, it's wrong the way you do the replacements there. You should get something like $$\int _0^\infty \frac{\{x\}}{(x+1)^2}dx$$
@Shisui Oh also If you can get a book on set theory and proving things it can help when proving theorems in modules. I reccomend "Book of Proof" by hammack
@BabakS. In several ways. If the intervals are $[a,b]$ and $[c,d]$ (or open, half-open, ...) with $b < c$, we can for example define it as $(a+d)/2$. Or we can define it as the centre of mass.
Whether either of these is useful depends on what the goal is.
@BalarkaSen That seems simple enough. From that we can say that $A \cup A' = U$, where $U$ is the universal set. So the two, in a way, complement each other.
@DanielFischer I had great timing this morning. I had written a comment late last night and planned to expand into an answer this morning, but as I was writing the first sentence, 900 answered it :)
Yea, that's what I thought. @BalarkaSen - It's like comparing an empty box, $\varnothing$, with a box with nothing but an empty box inside, $\{ \varnothing \}$.
@MikeMiller Yes. It is somehow funny if one answers - or sets out to answer - a question that hasn't been answered for hours or days, and hey presto, somebody else also writes an answer. Happens surprisingly often.
@BalarkaSen I know! I know! That's why I said I'll get to tying some actual math now. All I've been doing is drawing pretty Venn diagrams =P
Hmmm, it seems difficult to start off. I'm thinking of trying to show that if $x \in (A \cup B)'$, then $x$ must be an element of $A' \cap B'$ and that if $x$ isn't an element of the former, then it can't be an element of the later.
@MikeMiller If G is a group with some subgroup H define Aut_H(G) to be automorphisms of G that fix H. The claim is that if N is a characteristic subgroup of H, a characteristic subgroup of G then there is a short exact sequence 1 --> Aut_N(G) --> Aut_H(G) --> Aut_N(H) --> 1. I'll start with the proof iff you are convinced. (Also, I am a bit hesitating about the short exact part. It might as well be just left-exact)
But then your map is trivial. If Aut_H(G) --> Aut_N(H) is just the restriction, then you're sending autos that fix H to autos of H... but they're fixing all of H already
@BalarkaSen Unfortunately, it's only exact at the start. The map you have on the far right is an injection, so it has trivial kernel. But the image of Aut_H(G) isn't trivial.
Surely you don't believe that such a group exists? Serre proved that all groups are realizable as galois groups of real algebraic number fields, assuming inverse galois.
