Fix $n > 1$ and let's say I have a space $X$ and an Eilenberg-Maclane space $K(\pi_n(X), n)$, I want to show that there exists a continuous map $f : X \to K(\pi_n(X), n)$ such that the induced map $f_* : \pi_n(X) \to \pi_n(K(\pi_n(X), n)) \cong \pi_n(X)$ is (equivalent to) the identity map $1_{\pi_n(X)}$. (Assuming that this is true in general of course)
Now if the functors $\pi_n$ were full functors then this would be a trivial category theoretic proof. However I'm not sure if the functors $\pi_n$ are full functors and if they are not I guess I would have to construct such a map by hand.
What's an arbitrary domain? Even in integral domains gcd does not always make sense, you need a gcd domain (or a Bezout domain if you also want Bezout's identity to work)
more specifically domain $Z[\sqrt{-5}]$ has elements $a= 3, b= 1+2\sqrt{-5}$, I have shown that these two don't have a common divisor now there is the element $c=7(1+2\sqrt{-5})$ Now question asks to prove $ca, cb$ have gcd$=1$ but i feel this is incorrect.
You're correct, but you're missing that this is not enough to conclude that the gcd is not one, you're not in a UFD, I don't see why there should be a greatest common divisor just because there is some common divisor
Morning guys, I have a simple question, is it correct to write that $\mathbb{R} \in \mathcal{B}(\mathbb{R})$? where $\mathcal{B}$ represents the Borel sigma-algebra?
I know that the sigma algebra contains all the open intervals, and in particular $(-\infty, \infty)$ right?
Hi there, in the theorem found in this question, i.e. $AX=BX \iff A=B$ for any $n \times 1$ matrix $X$ and $m \times n$ matrices $A,B$, why does $X$ specifically need to be a $n \times 1$ matrix?
@rapasite those sequences correspond to A(x)=1+2x+4x^2+8x^3+...=1/(1-2x) and B(x)=0+1x+1x^2+1x^3+..,=x/(1-x). Hence A(x)B(x)=x/((1-x)(1-2x)). Computing the first few terms, we get A(x)B(x)=0+x+3x^2+7x^3+...
So the coefficient sequence associated to A(x)B(x) in that case is {0,1,3,7,...}
But the term-by-term product of those sequences is {1*0,2*1,4*1,8*1,...}={0,2,4,8,...}
is the use of a semi-colon appropriate in this sentence?
So in the interest of preventing notational clutter any reference to $\sigma$-algebras will be dropped from now on; they are always the Borel $\sigma$-algebras.
oh, I don't know anything about Iwasawa theory apart from $SL_2(\Bbb R) \cong \Bbb R \times \Bbb R_{>0}\times S^1$, so I might attend a few of the lectures/seminars
basically generalising the result that there are no compact riemannian manifolds with non-compact isometry group to a context where that result is no longer true
@MatheinBoulomenos ok, so I guess the statement should be "I don't know anyhting about Iwasawa theory apart from the fact that $SL_2(\Bbb R)\cong \Bbb R\times \Bbb R_{>0}\times S^1$ is not Iwasawa theory" :P
consider sierpinksi's gasket. let the x-axis be defined along the bottom edge and the y-axis be defined along the left-edge of the equillateral triangle (i.e. - take regular coordinate axis and rotate the y axis clockwise by 30-degrees)
if we express all coordinates in base 2
then any coordinate where the 'x' has a 1 in the same position as the 'y' does is not part of the triangle
(e.g. --- (.101, .001) is not part of sierpinski's gasket)
also important: any coordinate where 'x' and 'y' have 1 in different positions IS part of the triangle
but like... it looks to me like this is just not true. like if i draw the first iteration of sierpinski's gasket (w/ only the middle triangle cut out) it seems easy to come up with points inside that center triangle that have 1 in different coordinates
I would advice brushing up on basic set theory before diving deeper into group theory. It needs to be completely ingrained or the rest will be too hard to follow.
@rapasite that’s a finite geometric series (n terms). This is an infinite series, which converges so long as -1<x<1. Hence x^n goes to zero as n goes to infinity
it is definitely impossible for the coefficients of A(x)B(x) to be equal to the term-by-term product of A(x) and B(x),lets take the 3 first one $a_0$,$a_1$ and $a_2$ the equation for $a_1a_1$ implies $a_2=2a_1$ and the equation for $a_2a_2$ has no solution(in $\mathbb{R}$)
Hmm. Working through the algebra, you may be right.
Doing it with fresh eyes, I seem to need a0^2>a1^2 but a1=2a0
well, it'd need to satisfy $$(a_0+a_1 x+a_2x^2+\cdots)^2=a_0^2+2a_1 a_0 x+(a_1^2+2a_0 a_2)x^2+\cdots = a_0^2+a_1^2 x+a_2^2 x^2+\cdots$$
so $a_0^2=a_0^2$, $a_1^2=2a_0 a_1$, $a_2^2=2a_0 a_2+a_1^2$
first is trivial and can be ignored. if I assume that $a_1\neq 0$, then I get $a_1=2a_0$ and therefore $a_2^2=2a_0 a_2+4a_0^2\implies a_2^2-2a_0 a_2+a_0^2=(a_2-a_0)^2=5a_0^2$
So $a_2=(1\pm \sqrt{5})a_0$. That seems perfectly sound so far.
Sometimes we can "augment knowledge" with an unusual or different approach. I would like this reply to be accessible to kindergartners and also have some fun, so everybody get out your crayons!
Given paired $(x,y)$ data, draw their scatterplot. (The younger students may need a teacher to produ...
Let $A$ and $B$ be matrices $n\times n$.
Suppose $AX=BX$ for every $n\times 1$ column matrix $X$.
How can I prove this implies $A=B$?
