@Perturbative The hom-set is a monoid, not a group. But it isn't useful pedantry to point out that a category is formally also the data of the particular one-object set
I do understand that Gruber's recent closing spree on the questions is because of the revival of the "missing context" close reason, and Gruber is quite accurate in closing questions. But of course mistakes are made, and a question is closed even when there is some context given
I know. But I am asking is this getting overboard?
Of course we've got the reopen queue for closed questions, but reopening takes quite a while
I do suspect this will only be a temporary inconvenience as the main close reason is shifted back to "missing context" from "needs details or clarity". But there will be people who desperately need answers
if M is closed subspace and x is not in the span of the subspace, then x_n+M should converge to some element in H for a sequence x_n that converge to x, so x+M should be closed in H as well.
The set of ends always makes sense, it's just $\varprojlim_{K \subset X} \pi_0(X \setminus K)$. I guess you'd topologize by declaring open set of an end $e$ to be one which contains a connected component of a compact subset of $X$?
wiki says that you want a connected locally connected space for the end compactification to be compact
@BalarkaSen I don't think that's strictly needed, an infinite dimensional Banach space has only one end so you should get the one point compactification even if there is no exhaustion by compact sets
There's an exhaustion by compact sets $[0, n]$; that's a cofinal subset of the inverse system. The inverse limit of those is $\cdots \to \Bbb N \to \Bbb N$ where each map $\Bbb N \to \Bbb N$ is inclusion by translating by $n$
Ok, so let's see. Their inverse system is $\{K \cup \pi_0(X \setminus K)\}$ where if $K_1 \subset K_2$, the map $K_2 \cup \pi_0(X \setminus K_2) \to K_1 \cup \pi_0(X \setminus K_1)$ is given by: if $x \in K_2$ if it already belongs to $K_1$ send it to $x$ in $K_1$. Otherwise, $x \in K_2 \setminus K_1$ so $x$ belongs to a unique connected component of $X \setminus K_1$. Send it there. And $\pi_0(X \setminus K_2) \to \pi_0(X \setminus K_1)$ there's a natural map as well.
So I take this inverse limit, let's call it $\overline{X}$. What is the natural inclusion map $X \to \overline{X}$?
Define $X \to K \cup \pi_0(X \setminus K)$ by sending a point $x \in X$ to $x \in K$ if $x$ already belonged to $K$, otherwise $x \in X \setminus K$ so send it to the connected component in $\pi_0(X \setminus K)$ it belonged to
This should be a map of $X$ to the inverse system, and by universal property we get a map $X \to \overline{X}$
$X$ is locally compact Hausdorff iff $X$ embeds as an open subset of a compact Hausdorff space, yes
Spivak, "A Comprehensive Introduction To Differential Geometry" has a sequence of exercises where he gives a reasonable definition and states the compactification result
I will work it out after lunch, maybe you will want to join me
I have an online class in the afternoon, but I'm free after that. I was planning to read some group theory stuff but if those exercises are not too painful I might join
Probably, I'm stuck at home because of the virus so it's not like I have much to do. I'm not sure whether I can help you or whether I'm interested, both depends on what you mean by set theoretic combinatorics
I know how to prove it. I was asking you to do it as an exercise, since it didn't seem to come to mind immediately for you. But if you've just read the proof there's not much meaning to the exercise
Laplace transform problem whose solution I saw recently on an old question: $$G(s)=e^{-x\sqrt{s}}\leftrightarrow g(t)=\frac{a}{\sqrt{4\pi}}t^{-3/2}e^{-x^2/(4t)}$$
Question: a train moves from A to B there are 40 stations between a and b . train stops only at 7 stations find the ways by which train stops such that train stops at no consecutive stations??
I think there are two ways of solving it, one can be done in an intuitive way and then invoking the multiplication theorem. The other method is set theoretic method which is as follows
We need to find 7 out of 40 Where the sample is ordered in a sense that it has the form $( s_1, s_2 , ... s_7)$ with $s_1 + 1 \lt s_2 .... s_6 +1 \lt s_7 $ Where $s_i$ denotes the number of station.
@Semiclassical Are you interested in a classical rigid body mechanics ? I got a problem and have solved it, want to know if the science behind the solution is correct
There are four homogeneous rods. Each rod has mass mmm and length bbb. The ends of the rods are connected by frictionless hinges such that the rods form a rhombic frame (see the picture). This frame is shaped as square $ABCD $and put at rest on a smooth horizontal table. Then one applies a force $F$ to the hinge $A$ along the diagonal $AC$. Find the acceleration of the point $C$ right after the force was applied
I did a very simple solution, my claim is that the COM of the system will lie at the Center of the square, and angle between position vector and force vector is zero therefore no torque
The only motion that will be there is translation, so by Newton’s Law the acceleration of COM is $ a = \frac{F}{4m}$. This acceleration is possessed by each point of the body, hence the point $C$ will have an acceleration of $\frac{F}{4m}$
@topologicalorientablesurface Right, and by a canonical isomorphism too (the one that sends 1 to 1). So they're indistinguishable no matter how someone defines $\Bbb F_p$ (for instance, as "the unique field with $p$ elements")
@MikeMiller now, this is just carrying intuition , but in $\mathbb{F}_p[x]$, if $g(x)$ $g(x)$ is monic with solutions $x_1,x_2....,x_n$ then $g(x)=(x-x_1)(x-x_2)...(x-x_n)$, right?
Let $F$ be a field, and consider $F[x]$. Fix $f\in F[x]$. $F[x]/(f)$ is a field if and only if $f$ is irreducible in $F[x]$. This is because $F$ is euclidean domain and so $PID$ that is not a field, and so $(f)$ is maximal, so f ought to be irreducible in pids that are not fields, right?
I am trying to prove the following:
Let $G$ be a finite group and let $\pi : G \to U(\mathcal{H})$ and $\rho : G \to U(\mathcal{K})$ be irreducible unitary representations. If $\pi$ is weakly contained in $\rho$, then $\pi$ is contained in $\rho$.
I am allowed to use the following fact:
...
Forbid this arithmetic question, but does "E(X^2)" mean getting the average of every squared unit in a set? E({1, 4, 9, 16}) rather than E({1, 2, 3, 4}), I suppose?
Let $c:I\to M$ be a $C^{\infty}$ curve on smooth manifold $M$ of dimension $n$ and $X:I\to TM$ be a vector field along $c$,
$$\forall t\in I\qquad X(t)\in T_{c(t)}M.$$
does there exist a vector field $\bar{X}:M \to TM$ such that $X=\bar{X}\circ c$?
(@Alessandro) It follows from a push-pull argument that neighborhood of a locally tame loop in a topological manifold is always a $D^n$-bundle over $S^1$
Local tameness is important, otherwise you have wild knots
he also references "modulation equations" at one point, which I believe is in reference to Whitham's modulation theory
which is something i -really- wanted to understand at ont point and failed entirely
for context from a random ref: "The Whitham modulation theory provides an asymptotic method for studying slowly varying periodic waves, and is essentially a nonlinear WKB theory."
@LeakyNun Looking through the contents it seems to have everything I need. Would you say it's sufficiently pedagogic and accessible today? I see it's from the 60's
@Knight i think i'm skeptical of that approach. suppose we applied it to the following variation on that exercise: while applying the force to point A, apply the same force in the opposite direction to point C
the net force is zero, so the center of mass will certainly stay where it started (center of the square)
and therefore there's no torque about the center of mass
however, it's not the case that the four points have equal accelerations
points A and C will accelerate away from the center, while points B and D will accelerate towards the center (in each case, the accelerations of A/C and B/D are equal in magnitude but opposite in direction)
so as to ensure that the acceleration of the COM is zero
I am trying to prove the following:
Let $G$ be a finite group and let $\pi : G \to U(\mathcal{H})$ and $\rho : G \to U(\mathcal{K})$ be irreducible unitary representations. If $\pi$ is weakly contained in $\rho$, then $\pi$ is contained in $\rho$.
I am allowed to use the following fact:
...
Let $c:I\to M$ be a $C^{\infty}$ curve on smooth manifold $M$ of dimension $n$ and $X:I\to TM$ be a vector field along $c$,
$$\forall t\in I\qquad X(t)\in T_{c(t)}M.$$
does there exist a vector field $\bar{X}:M \to TM$ such that $X=\bar{X}\circ c$?
