Hi guys, I have a small query about power sets in naive set theory. I've learnt that if a set has $P$ number of elements, then the number of subsets the set has is $2^P$ . Does that number include the set itself?
@PedroTamaroff the reason is very subtle: power series compositions implicitly regroup terms in ways that can break apart and reassemble the partial sum valuations. For example, $$1+(1+1)+(1+1+1+1)+\cdots=1+2+4+\cdots=\frac{1}{1-2}=-1$$ converges in the $2$-adics but of course $1+1+1+\cdots$ does not.
@PedroTamaroff: That's a weird symbol for power sets. I've never seen that before. But from what your saying, the number of proper subsets of a set is $2^P - 1$ ?
note that infinite series in local fields are impervious to rearrangement: all infinite series are unconditional in their convergence or divergence. but both the archimedean and non-archimedean worlds are susceptible to regrouping. power series compositions are not enough to break sums though in the archimedean world.
@PedroTamaroff the field of p-adic numbers is the fraction field of the ring of p-adic integers. it can be thought of as the set of formal infinite sums $a_{-r}p^{-r}+\cdots+a_{-1}p^{-1}+a_0+a_1p+\cdots$ with the obvious notion of addition and subtraction
@PedroTamaroff I do the stuff I'm doing now not to impress people but because I'm interested. If I were alone on a desert island I'd still be doing it.
@PedroTamaroff I tell you man there's no point knowing what a quasi-coherent sheaf is if you don't really know the basics. Sometimes I say that to myself
I am impressed by the math you're studying. You made a big final assignment on "Weyl Schur Duality" (which I have no idea what is) that even Mariano complimented!
So, in the general case, if $B_2=\{w_1,\ldots,w_s\}$, $B_2^{\ast}=\{\psi_1,\ldots,\psi_s\}$ are the bases of $W,W^\ast$ and $B_1=\{v_1,\ldots,v_n\};B_1^\ast=\{\varphi_1,\ldots,\varphi_n\}$ are bases of $V,V^\ast$ then $$(|\hat f|_{B_2^\ast B_1})_{ij}=\psi_if(\varphi_j)$$
Guys, would it be wrong to post a question asking about proofs for $sin^2 (x) + cos^2 (x) = 1$ or is that just stupid. (I could make it a community wiki) How does it sound?
I know the whole division by zero thing is kinda undefined but can't A be zero?
Oh wait, I see the graph. Not even close
@Tobias: I was asking Peter this earlier but I'd like to ask you as well. Would it be ok to start a community wiki for the proofs of trigonometric identities or is that sort of question frowned upon?
@TobiasKildetoft: Kind of like the Pythagoras theorem certain trig identities have different ways of being proved. A sort of thread to capture all that variety would be nice to see.
@tobias:Also, there is a clear lack of written prrofs on the documents in internet unless your reading a book or something. And you know how books are, very narrow when it comes to such topics.
this is actually a really good example of why it is so important to check if the solution you have found by using various manipulations really is a solution
since it is not obvious which step is not reversible until you think a bit more about it
Yeah, I'm not even close to the answer. It's just something I have encountered in some old papers. I've been searching for an answer everywhere. I'm only asking because I really can't get it.
given some element on the right, you should try to match it with an element on the left, but picking some powers of $r$ to multiply, such that when you then further multiply by $P^2$, you get the term you had on the right
@Nick I think a possible way to do this is to pick a bijection between $n$-tuples of naturals with certain properties and itself
and picking that bijection in the right way
such that if you take the term corresponding to one tuples, if you add twice that exponent of $r$ you just found to all the entries gives you the tuples it is sent to by the bijection
