Recent content by Infrared

While this is the definition I was taught and think makes the most sense, the other perspective is also out there. From https://en.wikipedia.org/wiki/Continuous_function#Real_functions "A partial function is discontinuous at a point if the point belongs to the topological closure of its...

Any differentiable function is automatically continuous, so that's not an additional assumption. The question was whether ##f'## must also be continuous (it doesn't need to be). The point of explicitly mentioning continuity in the wikipedia article is that the function doesn't need to be...

No, MVT is valid for any differentiable function.

I don't think this is correct without the additional assumption that ##f'## is integrable. In general ##f## being differentiable does not imply that ##f'## is integrable. See https://en.wikipedia.org/wiki/Volterra%27s_function

Abelian subgroups do not need to be normal. For example, if ##\tau## is a transposition in ##S_n## with ##n>2,## then ##\{1,\tau\}## is abelian but not normal in ##S_n.## On the other hand, index 2 subgroups, like ##SO(3,1)\subset O(3,1),## are always normal.

In a metric space ##(X,d)##, a set ##E\subseteq X## is bounded if there is a constant ##C## such that ##d(x,y)\leq C## for all ##x,y\in E.## In this case ##E## is the unit ball in the ##1## norm and ##X=\mathbb{R}^n## and ##d## is the usual metric on ##\mathbb{R}^n## (induced by the 2-norm). So...

The unit sphere in the 1-norm is the set of points ##(x_1,\ldots,x_n)\in\mathbb{R}^n## satisfying ##|x_1|+\ldots+|x_n|=1.## This set is bounded since ##|x_i|\leq 1## for each ##i##. It is also closed, because the map ##f:\mathbb{R}^n\to\mathbb{R}, f(x_1,\ldots,x_n)=|x_1|+\ldots+|x_n|## is...

Chapter 1 exercise 6.

No, this doesn't make sense. All that changes during Ricci flow is the metric ##g## according to the PDE ##\frac{\partial g_t}{\partial t}=-2\text{Ric}(g_t).## The actual manifold doesn't change. What is actually meant by your statement is probably that the metric becomes zero in finite time...

@mathwonk I think you have it all there! An isotropic subspace can have dimension at most half the dimension of the total space and since the first cohomology of ##M_g## over a field has dimension ##2g##, the result is immediate.

Some hints for the remaining problems :) 7)In general, for a finite group ##G##, the number of commuting pairs ##(g,h)\in G\times G## is ##|G| \cdot \left(\text{number of conjugacy classes of G}\right).## To prove this, find a formula for the number of elements which commute with a given ##g\in...

This is correct but this problem was intended for much less advanced students than you :)

A bit of googling finds a fun argument giving examples: Let ##f## be an irreducible monic degree 4 polynomial in ##\mathbb{Z}[x]## satisfying: 1) ##f## has exactly two real roots. 2) The coefficient of ##x^i## is the same as the coefficient of ##x^{4-i}.## An example of such a polynomial is ##...

This is not the case. For example, ##\frac{3}{5}+\frac{4}{5}i## is on the unit circle, but the angle ##\text{arcsin}(4/5)## is not a rational multiple of pi (the only time ##x/\pi## and ##\sin(x)## can both be rational is when ##\sin(x)=0,\pm 1/2,\pm 1##).

The solutions by @PeroK and @mfb for the first problem look correct! I'll also share how I counted. There are ##\frac{8!}{2!2!2!}=7!## ways to arrange the 8 pieces. If we choose a setup uniformly at random, there is a 4/7 probability that the two bishops have opposite colors, and then...