Can someone explain why $f(x)=\sin{1/x}$ is not uniformly continuous on (0,1]? I'm thinking intuitevly, since sine never gets bigger than 1, there must be a delta for which all epsilon are satisfied for, no?
Is the transitive closure of set membership describable in ZF? That is, does there exist a formula $\phi$ in the language of ZF such that $\phi(x,y) \leftrightarrow x \in^* y$?
You're over-interpreting these sketches - they are only sketches, and their specific details can't really be used to make any real predictions.
Here is a more accurate version of those sketches, with a proper underpinning on a solid model of the experiment's behaviour:
Mathematica source via ...
What infinite sets tend to look like to me: $R_{01}, R_{02}$
They tend to give me that impression of having a brush like structure
Given a profinite group $G$ I can define an inverse system on the sets of $p$-Sylow subgroups in each finite quotient of $G$ by just restricting the system on $G$ to those sets... does that make sense? In that case the inverse limit of such a system is a pro-$p$-Sylow subgroup of $G$, in the sense that its image in each finite quotient of $G$ has index prime to $p$ (since the image in each case is a $p$-Sylow subgroup of the relevant factor)
(And by restrict I mean that the morphisms in this system are just given by the restrictions of those for the factors of $G$)
Suppose $f(X),g(X) \neq 0$ and $f(X)/g(X) \in E$. If either $\deg f = 0$ or $\deg g = 0$ (but not both), then obviously $E$ contains a polynomial
Hmm...localization...possibly...I hadn't thought of that.
Possible way to construct counterexample: if I take a proper rational polynomial and look at the subfield generated by it, will it contain a polynomial?
E.g., the subfield generated by $f(X) = \frac{X}{X+1}$
How about this: suppose $f(X) \in F\left( \frac{X}{X+1} \right) \cap F[X]$. Then $$f(X) = \sum_{n=0}^k a_n \left(\frac{X}{X+1}\right)^n$$ for some $a_n \in F$. We can turn this into a relation in $F[X]$ by noting that $f(X)(X+1)^k = \sum_{n=0}^k a_n(X+1)^{k-n}X^n$. Let $\deg f = m$. Then $\deg f(X)(X+1)^k = m + k$ while $\deg \sum_{n = 0}^k a_n(X+1)^{k-n}X^n \leq k$, so $m+k \leq k$ ?
In fact, by Luroth's theorem, any proper extension squished between $F(X)$ and $F$ is $F(g(X))$ for some rational function $g$. You should be able to prove using that that $E$ always contains a pure polynomial.
That's the same thing as X -> 1/X isn't it, if you do a linear change of coordinates, X = Y + 1.
Let's say F = C. This automorphism, X -> 1/X, corresponds to the biholomorphic automorphism of P^1 switching the two poles. Any function which has a pole at 0 and \infty are fixed, eg X + 1/X. That's what E entails
No polynomial is there certainly; they don't have a pole at 0
Cool
E = F(X + 1/X) I am sure, because X + 1/X is "degree 1" and Luroth
Basically the problem becomes immediate if you think of $F(X)$ as the field of functions on $\Bbb P^1_F$, you're looking for a subfield containing no function which just has a pole at infinity (these are precisely the polynomials).
@user76284 If you go through the proof that $\mathrm{trcl}(x)$ exists in $\mathsf{ZF}$ (show that $\bigcup^nx$ exists via the recursion theorem, use replacement on $\omega$ to show that $\{\bigcup^n x\mid n\in\omega\}$ exists and then take the union) it looks like this should be unfoldable to some $\in$-formula, even though it's going to be kinda painful
Way easier than that $\phi(x,y)=\forall z(z \text{ is transitive})\land x\subseteq z\rightarrow y\in z$ should hold iff $y\in\mathrm{trcl}(x)$
Which, to be 100% pedantic, can be expanded to the $\in$-formula $\phi(x,y)=\forall z\forall a\forall b(a\in b\land b\in z\rightarrow a\in z)\land\forall w(w\in x\rightarrow w\in z)\rightarrow y\in z$
Here's a cool one: Let $k$ be a perfect field and $p$ a prime. Show that there exists an algebraic extension $k^{(p)}/k$ such that each finite subextension has degree prime to $p$ and such that $k^{(p)}/k$ has no non-trivial finite extensions of degree prime to $p$.
@AlessandroCodenotti Is there a typo in your formula? $z$ is free in the second conjunct. And is the outer $\rightarrow$ inside or outside this conjunct?
Did you mean $\forall z ((z \text{ is transitive} \land x \subseteq z) \rightarrow y \in z)$
That is, if you have a non-zero Hilbert space and a state $|\psi\rangle\in H$, you cannot find a linear $T\colon H\to H\otimes H$ such that $$T|\psi\rangle = |\psi\rangle\otimes|\psi\rangle,\quad\forall |\psi\rangle\in H$$
For such a commonly known theorem, it's quite easy.
Hmmm... I'm having a lot more trouble than I care to admit trying to prove (p ∨ q) ∨ r ↔ p ∨ (q ∨ r). In a discrete math course I would likely just write out the truth table, but I'm trying to write a program that illustrates the proof. :/ Something is not clicking.
But if it is anything like a deductive argument, then you could try showing two things: $$(p\vee q)\rightarrow (p\vee(q\vee r))\tag{1}$$ $$r\rightarrow (p\vee(q\vee r))\tag{2}$$
There is something like that, but the tutorial I have uses or.intro_{right|left} and or.elim... The or.intro* stuff seems to make sense, I'm having some trouble internalizing what or.elim does exactly.
hmm... I wonder, how can one define an amorphous metric...
probably some map $d$ such that it satisfy the axioms of a metric, but having some extra structure that scrambles the details of it just enough that $d$ is not any arbitrary metric, but otherwise cannot be pinned down exactly
one possible way is to consider an equivalence class of metrics and the topologies they indluces
user131753
Theorem. Let $R$ be a ring such that $R$ is a subring of a division ring $D$. If for all $d(\ne 0)\in D$ either $d\in R$ or $d^{-1}\in R$ then $R$ is a local ring.
user131753
5:28 PM
My proof. It suffices to prove that $R\setminus U(R)$ is an ideal of $R$ where $U(R)$ denotes the set of all units of $R$. For this we first show that if $x\in R\setminus U(R)$ then $Rx\subseteq R\setminus U(R)$. So let $x\in R\setminus U(R)$. Clearly then $x\ne 0$. Suppose that there exists $r\in R$ such that $rx\in U(R)$. Consequently, since $D$ is a division ring, we have $x\in U(R)$, a contradiction. So $Rx\subseteq R\setminus U(R)$ for all $x\in R\setminus U(R)$.
user131753
Let $x,y\in R\setminus U(R)$. Let $x^{-1},y^{-1}$ denote the inverses of $x$ and $y$ respectively in $D$ (clearly both $x$ and $y$ are non-zero so that they are invertible in $D$). Then note that by hypothesis $xy^{-1}\in R$ or $yx^{-1}\in R$. Since $1\in R$ it follows that either $1+xy^{-1}\in R$ or $1+yx^{-1}\in R$. In the former case, $$(1+xy^{-1})y\in R\setminus U(R)\Rightarrow y+x\in R\setminus U(R)$$ and in the later
Can someone give an intuition when a point moves across a circle with a unit speed why does acceleration always points towards the center of a circle no matter how itpoint travels over it, clock-wise or counter-clockwise I'm talking about f(t) = (cos(t),sin(t)) and f(t) = (-cos(t),-sin(t))
Well, @famesyasd, I told you to draw pictures. That's intuitive. But look at it this way: If the acceleration were NOT pointing inwards, the particle would fly off the circle.
I would switch $y$ and $z$, @GFauxPas, but that's fine.
@TedShifrin I have a question for you. Say $X \subset M$ is a Whitney stratified subset. Denote $TX \subset TM$ to be the union of the tangent bundles to each strata in $X$.
Whitney condition (a) basically says something like $TX$ is "topologically stratified", i.e., if $S < L$ are a pair of strata in $X$, $TL$ contains $TS$ in the frontier in $TX \subset M$.
@TobiasKildetoft on my part it probably helps that i didn't go to parties etc when i was in high school, so drinking didn't get normalized for me until college
and at that point i still stuck to "leave it be" until i graduated
i mean, it depends on how much you do, but cigarettes seem like a far worse lifestyle choice than alcohol
As long as you don't over-drink, yes, alcohol is less damaging. However, people who over-drink are not only damaging to themselves but often to others as well.
This summer I hope to get around to reading those characteristic classes notes you gave me, Ted. I had written a super condensed set of notes last summer as a speedy introduction to forms and integration (assuming only the definition of manifolds) for some friends, and I think I will do something similar for characteristic classes and other stuff in that vein this summer.
You're over-interpreting these sketches - they are only sketches, and their specific details can't really be used to make any real predictions.
Here is a more accurate version of those sketches, with a proper underpinning on a solid model of the experiment's behaviour:
Mathematica source via ...
