I am thinking about a non hausedoff topology which is so messy that it has the following properties:
Imagine an open ball in $\Bbb{R}^n$
except that for each point in the ball, you adjoin uncountably many copies of said open ball
and then for each point in each of those open balls, you repeat the same thing
repeat the above process uncountably many times
Now the point is, if this is a region of spacetime, it is uncountably branching at every event uncountably deep so that basically determinism is completely thrown out of the window
Anyway, the above visual is inspired from this paper:
Actually, suppose the dynamics is probabilistic, meaning that given a particle at one of the points $p$ in this space at depth k and and copy x, there is some probability distribution that it will end up at the same point p, but a different copy at depth k+1 or k
Now consider a system of n particles, when these particles entered the space, some time t later, they will quickly end up in copies so far away from each other and thus they are no longer next to each other anymore
Now, take the limit as $t \to 0$ to model the scenario for uncountable depth and uncountable copies (hence uncountable branching points at each depth)
then the result is that as soon the n particles entered that region of space, they immediately being separated "far away" so they can no longer interact with each other
Therefore, an observable effect of such scenario is that everything disintegrates as soon they enter the region
The 21st century, is the first ever century where there exists interfinite elements $c$ such that $c+c+c+c+c+c+c+c+...+c = \aleph_0$
Put it in another way, the internet have empowered people enough that individuals (finite) finally have the power to affect politics at the state level (infinite)
@ForeverMozart hmm.. In that case I might have never came across it, for I am only vaguely familiar with something called the horseshoe map in chaos theory which I briefly stumbled upon in one of the vsauce or numerphile videos
@TobiasKildetoft don't you really need firsr countability more than Hausdorfness? $[0,\omega_1] is Hausdorff and $\omega_1$ is in the closure of $[0,\omega_1)$ but there is no sequence in it converging to $\omega_1$
yeah, that brieft exposure to the horseshoe map is what caused me to wonder whether the invariant quantity in all chaotic systems are some kind of strange attractor that has a fractal or self similar structure, untl one day acuriousmind showed me otherwise by pointing out there exists non chaotic strange attractor (and also the logistic map, which has no attractors of any kind)
@AlessandroCodenotti I wonder if there exists spaces that are not first countable without involving uncountable ordinals... might need to check that topology database...
I don't see how that implies that it gives a functor
but I agree that it should be natural for finite-dimensional vector spaces, but probably not in general, since the way we want to lift to morphisms is using the map from $V$ to its double dual, and the image of this map only determines a map between the double duals when everything is finite dimensional
well if f\in GL(V), f acts on x \in V** as f(x)(y)= x(f*y)
and f acts on x \in V as f(x) of course
and my brain is too fried to see if those actions are equivalent or not.
in the first line y\in V*
again iirc they aren't.
wikipedia says I recall incorrectly.
I guess its pretty simple if you assume x \in V** is an evaluation (say at x for bonus abuse of notation). Then x(f*y)= x(y \circ f) = f(x)y which is what you want.
Hello. Can someone help me in proving that if $(X,\sigma,\mu)$ is a finite measure space, and if $f_{n}$ is a real valued sequence in $M^{+}(X,\sigma)$, which converges uniformly to function $f$ belongs to $M^{+}(X,\sigma)$, and $\int f d \mu = lim \int f_{n} d \mu$
i have studied monotone convergence theorom
if i could show that $f_{n}$ are increasing, the question is done
but i do not know how, or maybe that is not needed and there is some other way?
@WilliamOliver Well, if you flip the part with the left $W$ that should do it. You get two arrows from $W$ to $X\times Y$, the $X$ on the inside of the left triangle and the $Y$ on the left of $W$
that should avoid any arrows crossing so it should still be readable
@TobiasKildetoft Awesome thanks! Forgot about parallel arrows. A separate question is (for my own sanity check), does that make sense? Would that diagram with the parallel arrows still commute?
Usually if a commutative diagram has two parallel arrows, then I will read it essentially as two different commutative diagrams. But technically I suppose this is not the way it is defined
So if you call that $T \in \text{Mat}_{n \times n}(\text{End}_A(M))$, and $X = (x_i) \in M^n$, $TX = 0$ (matrix multiplication works the same way except by feeding each entry of $T$ the elements of $X$)
Now multiply by the adjugate: $\text{adj}(T) T X = 0$ (the adjugate is defined formally as thet transpose of the cofactor matrix)
guys, would anyone here like "graph theory and algorithms"? I didn't choose the course, but I'll be doing a project in graph theory, so I was wondering if it's something that's worth considering, since I might choose to do it after all
@AkivaWeinberger The M to a power modulo problem you presented yesterday in chat, isn't M the message that is encrypted by raising it to a prime number?
there's this theorem in literature called the Serre-Swan theorem which says if $X$ is compact Hausdorff, let $C(X)$ denote the space of continuous functions on $X$; there's a natural correspondence between f.g. projective modules over $C(X)$ and vector bundles over $X$
what you should take from the two facts above is f.g. modules over $A$ should be thought as uh finite rank vector bundles on $\text{Spec}\, A$ such that the rank of the fibers can vary
the projective condition is the one which restricts rank to be constant globally, which we throw out
if you believe this philosophy, $\Bbb Z/n$ as a $\Bbb Z$-module is like the bundle over $\text{Spec} \Bbb Z$ which has a one dimensional vector space stuck at every prime divisor of $n$, and zero dimensional fibers stuck at every other primes
$f:A \to B$ a ring homomorphism, $N$ a $B$-module, then one regards $N$ as an $A$-module by $a \cdot n := f(a)n$. $B$ is also an $A$-module by $a \cdot b := f(a)b$. Now consider $N_B := B \otimes_A N$. It comes with a map from $N$ by $g:n \mapsto 1 \otimes n$. It also maps to $N$ by $p:b \otimes n \mapsto bn$. It is clear that $p \circ g = \operatorname{id}$, so $g$ is injective and $p$ is surjective. By the latter and the first isomorphism theorem, $N_B/\ker p = N$
if you take $\Bbb Z/n \otimes \Bbb Z/m$ the fiberwise tensor is nonzero iff you tensor over a common prime divisor of $m$ and $n$
so the resulting object has a one dimensional fiber stuck at every common prime divisor of $m$ and $n$
that's exactly "the bundle corresponding to" $\Bbb Z/(m, n)$
I think something similar works for $A/\alpha \otimes_A M$... every fiber over all the prime ideals that contains $\alpha$ dies unless $M$ has some fiber at that point too...
In which case the resulting object should be fibers of $M$ that lie above the prime ideals that contains $\alpha$, and zero everywhere else
A question for local calculus gurus: If $f(x)$ is smooth and bounded, is its Taylor series expansion always guaranteed to have a nonzero radius of convergence? If not, is there a special name for functions that do have this property?
This is my problem: I have an $f(t)$, bounded and smooth, and I need to approximate it as a power series over some small time horizon $\delta t$ (i.e. a polynomial in $\delta t$). Of course, there's no limit to how well I can fit a polynomial to this segment of curve as I let the order of the polynomial go to infinity.
Hence I want to elegantly make the argument that "no matter what $f(t)$ is, so long as it's smooth and bounded, it can be represented by a convergent power series over this interval", which allows me to use the power series in place of $f$ over the interval.
Let p,q be two probability distributions on a set of size n. How can I prove the inequality \sum[(p(x)-q(x))^2/(p(x)+q(x))] \leq 2\sum [p(x) log(p(x)/q(x))] ?
I have no idea... I honestly can't parse all of those nested braces---can you please try to MathJax that up? You can use basic TeX commands without difficulty; just enclose in $ signs
The division algorithm says if $a,b$ are integers and $b$ nonzero then there exist unique $q,r$ such that $a=qb+r$ and $0\le r<|b|$. I can't find $q,r$ when $a=5,b=7$
