Mathematics

you're going to have to define those functors

sure, Sym is a functor from the category of finite sets with morphisms bijections to the category sets

it takes a set $X$ to the set of all permutations of $X$

@Rememberme that was unnecessary, but go on.

I would use this theorem which says that W is a subspace of A only and only when there are two scalars $\alpha$ and $\beta$ such that $c\alpha + \beta \epsilon W$ only then W is a subspace

sure

4:03 PM

and it takes a bijection $f:X\to Y$ to the map $\text{Sym}(f):g \mapsto f \circ g \circ f^{-1}$ where $g$ is a permutation of $X$

I don't understand @Rememberme

yes

ok

@Rememberme er, you mean $W$ is a subspace iff $a + bw$ is always in $W$ for any $w \in W$. go on.

what is episolon here ?

okay now i'll define the other functor Ord: Finite-bijective $\to$ Sets

4:04 PM

and wrong quantifications : $\forall a, b$ not $\exists a,b$ @Remember

Yup @BalarkaSen now I will try to do this for R^n-1

it takes the finite set $X$ to the set of all total orders on $X$

try to do what for $\Bbb R^{n-1}$?

@Rememberme you need a subspace to begin with

and it takes the bijection $f:X\to Y$ to the map Ord$(f): (x_1,...,x_n) \mapsto ( (f(x_1),...,f(x_n))$

4:06 PM

for example

Take some vector a in R^n-1 and b in it. These vectors will have n-1 components and I am trying to prove that R^n-1 is a subspace of R^n

@Rememberme ah, but that is not relevant to the question

we already know that R^{n-1} is a subspace of R^n in many ways (hyperplanes through the origin).

Okay ....... And you were talking about curves right @BalarkaSen

but the question was to prove that there are no other subspaces of R^n excepts k-dimensional planes through the origin for k < n

columbus8myhw

Taylor's theorem can be written as $e^{a\ d/dx}f(x)=f(x+a)$, if you define $e^x$ as a formal series.

4:08 PM

no, i was not. i was just making a point.

and i'm trying to show now that there is no natural transformation from Sym to Ord

Ok then I have to work again....

right

columbus8myhw

It turns out that if you have a linear operator, you can differentiate it with respect to $d/dx$, in a sense.

yes. classifying vector spaces are fundamental results. i bet it's in H-K, but you haven't read it.

4:10 PM

I have a moral reason it doesn't work but I'm having trouble making it precise

ok please tell me

i sort of have a moral reason too

for any two vectors in $R^n$ a + b is $R^{n - 1}$ by definition and scalar multiple of any vectors is in $R^{n - 1}$

a natural transformation would, in particular, give a canonical ordering on every set (start with the set of permutations, and sending it to the set of orders, you're sending the identity permutation to some order)

just add two vectors in $R^{n - 1}$ component wise

it seems likely that this will cause inconsistencies with the permutations

i haven't checked the details

4:12 PM

for example take x and y in $R^{n - 1}$ then x = ($a_1,....,a_{n - 1}$), y= ($b_1,....,b_{n - 1}$), so x + y = ($a_1 + b_1,....,a_{n - 1} + b_{n - 1}$) which is in $R^{n - 1}$

ok

i've been trying forever to find a rigorous reason

there is a hint saying to consider the identity permutations

@Rememberme a good excerise for you is proof that for any vector space V and any subspace W,X prove that W $\cap$ X is always a subspace of V.

oh lol

follow the diagram around for the identity permutation

yes

meh what i wanted to say doesn't work

i'm done with this one, sorry

4:16 PM

ok no prob

im giving up for now too lol

@KarimMansour that's a theorem in HK

Prooved

I would suggest proving some problems of vector space and subspace until you get comfortable with them, because those ideas or similiar ideas carry for example to proof in abstract algebra.

Proving something is a group or subgroup etc they have the same form as this kinda of proofs in linear algebra, so take your time until your comfortable with them.

@BalarkaSen how would you go about proving what you said???

well, you said you have understood vector spaces. use your knowledge.

Varun

Hi what is the definition of bias with respect to a coin flip, suppouse you have a coin whose probability of heads is 0.8 what is its bias?

4:21 PM

work with n = 2. assume, for a contradiction, that there is a subspace V of R^2 which is not vector space-isomorphic to a line through the origin. start with that.

i'm gonna leave.

ping me when you prove it.