I'm a bit confused by this definition. We have a family $G_\alpha, \alpha<\omega_1$ of sets and define $H_\alpha=\bigcup\limits_{\beta\le\alpha} G_\beta$ and $K_\alpha=H_\alpha\setminus(\bigcup\limits_{\beta<\alpha} H_\beta)$, I'm having trouble parsing the definition of $K_\alpha$, what exactly is in it?
Aha, so $K_\alpha$ is "stuff that is $H_\alpha$ but not in smaller $H_\beta$", but isn't this "stuff which is in $G_\alpha$ but not in smaller $G_\beta$?
Hm, ok, so I guess they introduced the $H_\alpha$ here because they're needed later in the proof, not because they're actually needed to define the $K_\alpha$
@Danu well-ordered set - meaning every subset has a minimum. you can label more and more of its elements starting with the minimum, call it 0, then you get 1,2,3..., then you get omega, omega+1, etc.
You know, I've found linear algebra to be quite useful for dimensional analysis via a representation of derived units as 7-dimensional vectors, where the basis vectors are the SI base units and each magnitude represents the exponent of that unit
Turns out of A is a 2x2 octonionic matrix then A(AA)=(AA)A where A={{a,b},{c,d}} (wolfram notation) iff either a,d are in a complex subalgebra of O or a,b,c,d are in a quaternionic subalgebra of O. Will try my hand at various generalizations.
@AkivaWeinberger The Erdös-Sierpinski Duality Theorem. That is "Assuming CH there exist a bijection $f:\Bbb R\to\Bbb R$ such that $f=f^{-1}$ and $f(A)$ is a nullset iff $A$ is meager"
Or in english "let $P$ be a proposition involving solely the notions of measure zero, meager and notions of pure set theory, let $P^*$ the proposition obtained from $P$ by exchanging the terms "nullset" and "meager set" whenever they appear, then assuming CH $P\implies P^*$ and $P^*\implies P$"
Yeah — "null" and "meager" are both meant to capture the idea of a set (a subset of $\Bbb R$) being "small". $\Bbb R$ itself is "large", so it's neither of those things.
The surprising thing is that $\Bbb R$ can be written as the union of a meager set and a null set.
That is, it's the union of two sets that we wanted to call "small". That's counterintuitive. @ZachHauk
@PVAL Reeb stability says that locally you're a foliated neighborhood of your leaf. If you do this and all the leaves are compact then the holonomy homomorphism must have had image in Z/2, and so your foliation is either locally trivial or locally an I-bundle over your hypersurface. In the first case you local look like R, in the second [0,1).
I'm reading over this question trying to prove the same thing math.stackexchange.com/questions/671222/… and I'm not sure what "axiom" the answer is referring to. Can anyone anyone point out what axiom this is? I don't want to post an entirely new question for something already answered...
Anyone know? I had assumed it meant the fifth incidence axiom for space consisting of at least three noncollinear points. Hence, we would have the line defined by two and one which is not on the line.
Let $f$ continuous on $(a,-a)$ and there exists a $k \in (0,1)$ such that $\displaystyle\lim_{x \to 0}{\displaystyle\frac{f(x)-f(kx)}{x}} = L$. Prove $f$ is differentiable at $x=0$ and obtain $f'(0)$.
and, because $k\in(0,1)$, fixed an $\epsilon > 0$ there exists $\delta > 0$ such that if $0<|t|<\delta$ then $|g_{n}(t)-L|<\varepsilon$ for all $n \geq 0 $
On the other hand, $\dfrac{f(t)-f(0)}{t}=g_{0}(t)+kg_{1}(t)+\dots+k^{n}g_{n+1}(t)+\dfrac{f(k^{n+1})-f(0)}{t}$
and because $L/(1-k)=L(1+k+k^{2}+\dots+k^{n})+Lk^{n+1}/(1-k)$ we can apply the triangle inequality getting: $\Big|\dfrac{f(t)-f(0)}{t}-\dfrac{L}{1-k}\Big|\leq |g_{1}(t)-L|+\dots+k^{n}|g_{n+1}(t)-L|+\Big|\dfrac{f(k^{n+1}t)-f(0)}{t}\Big|+$
If the limit isn't $L/(1-k)$, there's some $\epsilon_0>0$ and $x_n\to 0$ with $\left|\dfrac{f(x_n)-f(0)}{x_n} - \dfrac L{1-k}\right|\ge \epsilon_0$. Playing with the reverse triangle inequality gives a contradiction.
You're in more advanced analysis class, I assume ;)
Yes, I get that. And you certainly used $0<k<1$ to get $\lim k^n = 0$. But you can get the limit for $k>1$ by manipulating that one and vice versa, so it's true in general, as I thought.
