$y=\text{Percentage}(25,10,40)$ doesn't mean that $y=\text{Percentage}(g(25),g(10),g(40))$, where $g(x)$ is an arbitrary function, in this case $g(x)=\sqrt{\frac{x}{10}}$. It means so if $g(x)=a\times x$ where $a$ is a constant @Dragneel
@TedShifrin $f_n(x)=e^{-nx^2}$ is uniformly convergent on $[-1,1]\backslash\{0\}$. This is because $e^{-n0^2}=1$ regardless of what $n$ is. Right? In my case, with $$f_n(x)=\frac{x}{1+nx^2}$$, $x=0$ isn't a problem because it equals $f(x)$ there. For all other $x\in\Bbb{R}$ it's fine as well. So the subset of $\Bbb{R}$ is $\Bbb{R}$. No?
@TedShifrin Help me remember the result of Weyl on tubes again: the polynomial in r you get is supposed to be about the intrinsic geometry of the submanifold, but not the extrinsics of the normal bundle?
(You can show that Mahlo cardinals are inaccessible, inaccessible limits of inaccessibles, inaccessible limits of inaccessible limits of inaccessibles, etc.)
Well, going back to measurable cardinals. It's easy to prove that if $\kappa$ is a cardinal there is no $\kappa$-complete ultrafilter on $\kappa$ (exercise), so the next best we can ask is that there is an ultrafilter on $\kappa$ that is $\lambda$-complete for all $\lambda<\kappa$. Such a $\kappa$ is called measurable
Interestingly ZFC can't prove that there are measurable cardinals, but it can prove that if there exist an uncountable cardinal admitting an $\omega$-complete ultrafilter (also known as an $\omega$-measurable cardinal) then there is a measurable cardinal
(The uncountable in my previous sentence is completely superfluous)
Oh, and all ultrafilters are non principal, principal ultrafilters are clearly $\kappa$-complete for all $\kappa$
@Oskar: Nope, not uniform on $[-1,1]-\{0\}$. Think pictorially. Draw an $\epsilon$-"fence" around the graph of the limit function. You need all the graphs of $f_n$ for $n\ge N$ to fit everywhere inside that $\epsilon$-fence.
There is a general cofinality issue with "least" fixed points. Think about fixed points of the aleph function $\alpha\mapsto\aleph_\alpha$, you can climb to the first fixed point by defining $\gamma_0=0$, $\gamma_{n+1}=\aleph_{\gamma_n}$ and taking the sup of the $\gamma_n$, so it has countable cofinality, hence ZFC proves that it isn't inaccessible
@MikeM: The coefficients are in fact integrals of various invariant polynomials in the curvature tensor. But they arise extrinsically (and can be interpreted by taking an actual $\epsilon$ tube and looking at its geometry, then averaging appropriately). This is what Lipschitz-Killing curvatures are.
The cool thing, @MikeM, is that for complex manifolds (say in $\Bbb C^n$ or $\Bbb CP^n$), all the coefficients are topologically invariant. This is because they're all built out of Chern forms.
@Alessandro: A special party arranged with the boar and zebras?
In fact you can keep iterating that construction by defining $\gamma_\beta=\sup\{\gamma_\alpha\mid\alpha<\beta\}$ when $\beta$ is a limit, then you have $\aleph_{\gamma_\beta}=\gamma_\beta$ for all limit $\beta$
So not only the least fixed point can't be inaccessible, but every fixed point which can be proved to be the least fixed point above some given ordinal will suffer from the same cofinality issue
Well, it starts out with the second fundamental form, but then because of the invariance it turns into stuff in curvature. Remember that the generalization of the Gauss equation actually relates intrinsic curvature to second fundamental form (and in Euclidean space there's no ambient curvature term). @MikeM
(It's very common to only consider uncountable cardinals anyway in this kind of discussions because otherwise you have to specify it to avoid having $\omega$ strongly inaccessible or other weird stuff)
Similar to before, you can show that these are much larger than things such as $\sup\{I(0), I(I(0)), I(I(I(0))), \dots\}$
Similar to the aleph fixed-point
Likewise you can show that recursively generalizing this will always yield smaller than the first 1-inaccessible (on inputs smaller than the first 1-inaccessible)
These will also satisfy $\kappa=I(\kappa)$, just like how inaccessibles satisfy $\kappa=\kappa$-th regular
In general, consider an ordinal of the form "limit of ... which is regular" and the corresponding club set who's elements are of the form "limit of ... which may or may not be regular" and you'll see the Mahlo ordinals must be inaccessible, 1-inaccessible, (1,0)-inaccessible, etc. with every sort of "inaccessible extension".
Hi, I'm wondering if any intersection of unions can be expressed as a union of intersections. I want to know if this is correct so I don't duplicate a question: In math.stackexchange.com/questions/1576633/… there is a line going from $\forall i\in I\exists j\in J,x\in X_{ij}$ to $\exists j_i\in J^I\forall i\in I,x\in X_{ij_i}$.
Is that legitimate? The function seems like it should be in $J^I$, defined coordinatewise, but I'm not sure.
instead of $j_i\in J^I$, lets write $f\in J^I$, and the function $f:I\to J$ i think they have in mind is defined by $f(i)=j$ for some $j$ that exists by the previous line $\exists j\in J$. Then the $x\in X_{ij_i}$ should say $x\in X_{if(i)}$. but i'm not sure this function actually exists
Why not, @quallenjäger? It's the simplest thing you could possibly have.
There's one chart for it. If it's an affine $k$-dimensional subspace through the point $a\in\Bbb R^n$, it's diffeomorphic to the linear subspace, which is in turn diffeomorphic to $\Bbb R^k$.
Or, more directly, if $v_1,\dots,v_k$ is a basis for the linear space parallel to your set $S$, you get a global parametrization $\phi\colon\Bbb R^k\to S\subset\Bbb R^n$ by taking $\phi(t_1,\dots,t_k) = a + \sum t_i v_i$.
There are various technical reasons smooth is better than just $C^1$. You need smooth to say that the tangent space comes from derivations, for example. With $C^1$ (or finite $C^k$) there are infinitely many independent derivations.
Hey. Anyone wanna take a stab at this measure theory problem with me? It should be easy lol, somehow missing the idea, so a hint would be cool: Show that if f: R->R is (once) everywhere differentiable then f' is Borel measurable.
(You may assume that f is exactly once differentiable, otherwise it is trivial)
cause im certainly not white (despite some portuguese ancestry), but i also dont qualify as amerindian by most US standards cause no tribal affiliation despite the fact that most of my ancestry is amerindian
Suppose we have the space of tensor algebra $T(E):=\oplus_{n=1}^{\infty}(\Bbb R^d)^{\otimes n}$, the log series of $a \in T(E)$ is defined as $log a =log(a_0)+\sum_{n \geq 1}\frac{(-1)^n}{n}(\boldsymbol{1}-\frac{a}{a_0})^n$
and because we have to deal with the problem of convergence of such series, one says log(a) is locally finite because it produces only finite many terms for a given degree
And hence we do not need any norms on $T(E)$ to make the series convergent and the log(a) is purely algebraic.
I don't understand why this is enough to see this series as an algebraic object.
