Let $V$ be a real vector space equipped with a linear involution. Let $V^h := \{v \in V \mid v^* = v \}$ be the set of hermitian elements. I am reading a paper and the author claims that there is a distinguished cone $V^+ \subset V^h$. What exactly is this distinguished cone? How does one define ...
Let $V$ be a $\Bbb{R}$-vector space. In the paper I am reading, the author wants to
"...equip $V$ with a locally convex topology, called the 'algebraic topology' (also known as the finest locally convex topology), by declaring that any convex set $C$ such that $int(C) = C$ is open."
Why is this...
How can I prove: no matter how small interval I take containing $0$, there will always be a number of the form $\frac{1}{\pi/2 + 2\pi n}$ in it ($n \in \mathbb N_0$)
??
What I want to know is if I take a small enough interval and say “$~|\sin (1/x)| \lt 1/100$” in that interval, I can disapprove it by saying “that interval” contains a number of the form $\frac{1}{\pi/2 + 2\pi n}$ and hence $\sin (1/x)$ is 1 for a value in that interval.
I know that we can make $\frac{1}{\pi/2 +2\pi n}$ as small as we like by taking a sufficiently large $n$, but what I don’t get is that it will always occur in every interval no matter how small.
Context: I’m trying to prove that limit of $\sin(1/x)$ as $x$ goes to zero doesn’t exist.
@CalvinKhor Dostoyevsky wrote hard and complicated novels, for example if you read Crime and Punishment (its one of the greatest works in Literature) you will find difficulty in understanding the condition of Raskolnikov, the novel starts very strangely (I think by soliloquy or self-talking/thinking).
Leo Tolstoy is very simple to read and understand and I must say Tolstoy’s writings are of different domain.
@CalvinKhor When I take an interval containing $0$ do I need to take the number to both the sides of zero? Won’t just the positive ones gonna work? For example taking $0$ as one of the end points and taking all number with $\delta$ of it?
@Knight so is $\mathbb R \setminus \{ 0\}$ an interval?
an interval is a set that looks like $\{x\in\mathbb R: a < x < b \}$, $a\le b$, ( $b< a$ also allowed but then the set is empty), the other sets that are also intervals come from replacing one or two $<$s with $\le$, or by only having a one-sided condition $\{ x: a\le x\} = [a,\infty)$ or no restrictions (so the whole line $(-\infty,\infty)$ is an interval)
but yes, what you are trying to write is what I wrote
this is a bother so you can just restrict your question earlier to just open intervals containing 0 (these are the ones like $(a,b)$) or closed intervals (these are $[a,b]$) but for the first kind, there are no open intervals containing 0 where 0 is an endpoint
Yes, but its not 100% the same as subtraction because you can have $A - B = A$ even when $B$ is not a trivial set, so I will refuse to write it using a minus sign
This is because you chose to ask about a limit, whose definition includes an arbitrary open neighhbourhood (with one point deleted) around the point of consideration
as mentioned an open neighbourhood of a point cannot have that point as one of the end points
if you want to only look at one side, then ask for a one-sided limit
The mapping class group of a manifold $M$ is interpreted as the group of isotopy-classes of automorphisms of $M.$ How can I convince myself that $MCG(M)$ is always a discrete group? And is there a continuous analog?
@geocalc33 This is not true in general, at least with natural topologies, but is true for compact manifolds. The natural topology to put on the space of homeomorphism of a manifold is the compact open topology and since for compact manifolds you get uniform convergence you get sequences of functions converging to something isotopic will eventually be uniformly close to something isotopic to the identity and just perturb to get the sequence was eventually just functions isotopic to the identity
@geocalc33 Also what do you mean by continuous analogs? I guess the space of homeomorphisms is a sort of continuous analog.
One of the things I have been working on is coming up with an independent "elementary" proof that quasiisometries of the curve complex of a surface is close to curve graph automorphisms. Curve graph is a graph where vertex is isotopy classes of simple closed curves and you get edges if there are representative that are disjoint.
@geocalc33 An example where it is not true: consider a surface where you keep on taking connect sums of tori going out one directions ( "looks like" 0000000....) then take a sequence, $f_n$, of mapping classes where $f_n$ is a dehn twist on the "nth tori" then the sequence converges to the identity
@PaulPlummer what I mean by continuous analogs...and I probably don't understand enough about this I'm just curious..but from what I understand there's discrete groups and then there's Lie groups, and I'm just wondering if there's a decent way to understand why the MCG is discrete and why it can or cannot be made into a Lie group I guess
I have heard people say the the space of homemorphisms or diffeomorphisms is sort of an infinite dimensional lie group, but I don't really know how useful or "true" that is
I wonder how difficult it is to prove that every finite group is a subgroup of the mapping class group of a closed orientable surface. It sounds like an incredible result!
@Paul I should probably become more familiar with manifolds and Riemannian geometry first before I embark on that, but please give me the hint and I'll save it for later
Sort of. The mapping class group acts on simple closed curves and it turns out that the automorphism group of this simplex is the mapping class group (including orientation reversing)
It is a "nice" geometric space to act on. Remarkably it is delta hyperbolic
and infinite diameter outside of some trivial cases (and there is a slightly different def for torus, once punctured torus, and annulus)
Hi everyone today our teacher gave a question on p and c
Question state as I have cube of six surfaces and I have the six different colour and we have to paint the six faces such that the two particular color should always paint on the sides whose edges are marked with the rest of the four colors I need to find all the possible cases
@moteutsch Yes, switch the order of integration. Draw the region $\{(r,s): 0\le s\le t, 0\le r\le s\}$. This is a right triangle. Now ask. What are the smallest, largest values of $r$ in this region? For fixed $r$, how does $s$ vary? Draw the picture.
My book writes : A stationary point $x^{\ast}$ is stable if given any $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that $$ | x_0 - x^{\ast}| \lt \delta \implies |x(t) - x^{\ast}|\lt \epsilon$$
In Dostoyevsky's time his writings were heavily critiqued because everyone felt his characters were completely unrepresentative of the Russian folk, heavily caricatured and dramatized to a frenzy. The reason for this is that in context of pre-Soviet, Dostoyevsky doesn't make sense.
@Yuvraj Not sure I understand what you've written. What edges are marked? If an edge is marked then either adjacent side should be painted one color or the other?
@moteutsch Your change of order was correct, but I don't follow the last step. $\int_0^t (t-r)f(r)\,dr$ can't be simplified the way you did it. You can pull out the $t$ from the integral but you still have $f(r)\,dr$. The best you can do is $\int_0^t uf(t-u)\,du$ if that's any easier.
My book writes : A stationary point $x^{\ast}$ is stable if given any $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that $$ | x_0 - x^{\ast}| \lt \delta \implies |x(t) - x^{\ast}|\lt \epsilon$$
But I remember when I was young there was a huge outcry when wearing seat belts in cars became required. This is like the 2nd amendment maniacal behavior.
@TedShifrin I wonder how all those people who feel that they are being imposed upon by needing to wear a mask would like to be put onto the Apocalypse at Magic Mountain without seatbelts, being told they shouldn't be restricted by a seatbelt.
I've seen various ironic letters to the editor. Among things listed: Wouldn't want to infringe on the rights of cooks in restaurants to cough in food or serve chicken cooked only to 120º.
What's interesting to me is that prior to Achebe no-one saw the problem with the book (also apparently Achebe hated that the book went out of favour as a celebrated work after his criticism).
Let $a_{1}=3$ and $a_{n+1}=\frac{a_{n}}{2}+\frac{5}{2 a_{n}}$ for all $n \geq 1$
So we need to check whether it is converging sequence or not
SO what I think is if I am able to show this sequence is decreasing then it is already bounded by 0 so it becomes convergent.
But not sure how to show it i...
So we need to take limit n tend infinity then a(n+1) is a(n) converge to same limit say L. Then we just need to find value by some algebraic manipulation.
In limit form n tends to infinity a(n+1) is same as a(n)
I'm not, I'm helping you prove that $\lim_{n\to\infty}a_n$ exists. We can do that by any means necessary, as long as our logic is sound. So for example, it's helpful for us to know what the limit is before we actually prove it, after all the definition of the limit is $n>\delta\implies|a_n-x|<\epsilon$ so it's harder to prove the implication holds if you don't know x
