Assuming that $a \neq 0$ and $a d-b c \neq 0,$ perform the progression of row operations (in the order
indicated) on the matrix $A=\left[\begin{array}{cc}{a} & {b} \\ {c} & {d}\end{array}\right]$.
(a) $E_{1}\left(\frac{1}{a}\right)$
(b) $E_{21}(-c)$
(c) $E_{2}\left(\frac{a}{a d-b c}\right)$
(d...
Consider any sequence (x_n) in the open disk that converges to a point on the unit circle. Then every subsequence of (x_n) converges to the same point on the unit circle. Hence no subsequence of (x_n) converges to a point in the open disk.
Is $\aleph_{\aleph_1}$ singular under a model where $\aleph_1$ is singular?
It seems naively one can find that nonconstructive countable sequence that get to $\omega_1$ and then use replacement to map alephs to this sequence to go from $\aleph_0$ to $\aleph_{\aleph_1}$
btw, it seems the finite cardinals are club wrt $\omega$ because $\sup (\omega \cap n) = n \in \omega$ and for every $n \in \omega$ there is an $m > n$
finite cardinals are not club wrt any cardinals because they each have a maximum meaning they are bounded despite closed. w is also not club wrt any cardinal because it is bounded (there are no elements in w that are larger than any countable ordinal in w1). The simplest example of a club set wrt w1 is the set of all countable limit ordinals
Lets say we have a linear subspace $F$ of a Hilbert space $\mathfrak H$. How could I prove that $\bar F\subset (F^{\perp})^{\perp}$ (the closure of $F$ is a subset of the "double" orthogonal complement) implies $\bar F^{\perp} \subset F^{\perp}$?
I've made a type
actually the condition is $F\subset \bar F$ implies implies $\bar F^{\perp}\subset F^{\perp}$
@AlessandroCodenotti Will in singular models $\text{cof}(\aleph_{\aleph_1})=\omega$ ?
because I should be able to find the cofinal sequence of $\aleph_1$ and then apply replacement $x \mapsto \aleph_x$ to turn them into alephs and hence producing a sequence for $\aleph_{\aleph_1}$?
You are basically mapping the euclidean plane onto the surface of a sphere while preserving many things such as right angles, hence straight lines become their noneuclidean counterpart: Geodesics
Hiya @Alessandro, it's going alright, haven't really done anything yet lol. First 2 weeks of ANT 1 are going to be elementary number theory, and for modular forms we're reviewing Möbius transformations so that's a bit slow too
Thinking about the philosophy of infinity again (don't roast me, I have actually read the prerequiste text on this topic months ago) :
Let a be an object
Define some partial ordering so we can talk about $\leq$
Let b be an object
Then a potential infinite of $a$ is if for every $c \geq a$, there exists a $d$ such that $d \geq c$
(need to think about how to phrase this better, because $\forall c (c \geq a) \implies \exists d (d \geq c)$ is supposed to be a schema that is iterated until the condition is no longer met)
Thus a potential infinity $p$ is any mathematical object where it is preserved under the operation of upward extension.
Now the program can give a couple of possibilities:
We knew from the above definition, a potential infinity $p$ is something that is left invariant by the program and the program does not terminate in any part of its run
@LeakyNun I don't know why you feel like that, but it's not necessarily the case :P
In fact ZF cannot prove that there is a single uncountable regular cardinal (but this is a result way harder than showing that $\aleph_1$ might be singular)
I have a following integral: $$\int_0^1 dx\int_0^{\sqrt{1-x^2}}dy \int_\sqrt{x^2+y^2}^\sqrt{1-x^2-y^2}z^2dz$$
Which i have to solve by introducing polar coordinates, which is, by itself, relatively simple:
$$x=\rho\cos\theta\sin\phi \\ y=\rho\sin\theta\sin\phi \\ z=\rho\cos\phi$$
Besides this,...
5. step 1-2 revists an element: Output "cyclic order of period <number of steps 1-2 for that c> detected"
6. step 1-2 never revists an element, but the elements obtained are distributed between a pair of lower and upper bound, Output "aperiodic bounded order of length <number of steps 1-2> detected"
ok screw that again, should have put a different order
Checklist:
1. If $Pc$ maintain its run all the way through without switching steps, output "potential infinity detected, $C$ is not strictly finite". Pick another $c$ and recompute $Pc$
2. If there is no $d \in C$ for a given $Pc$, output "maximal element detected". Pick another $c$ and recompute $Pc$
3. If Condition 2 occur for all $c \in C$, output "$C$ is tarski finite"
4. If Condition 3 occurs and the value that triggers Condition 2 are the same, output "maximum detected"
5. If $Pc$ maintain its run all the way through without switching steps, and some elements in that run are revisted, output "cyclic order of length <number of steps ran for $Pc$>" detected
6. If $Pc$ maintain its run all the way through without switching steps, no elements are revisited, and the whole output of $Pc$ has both a upper and lower bound, output "aperiodic cyclic order detected"
I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876). In the fourth chapter, he deals with different definitions of finite set. Here is the classical one:
Definition 1. A set $X$ is called finite if there exists a bijection $n \to X$ for some $n \in \ma...
I always found it redundant when the same concept is given many different names
and I always thought tarski finite is weaker than finite because "names are cool"
hmm...
Let $C$ be a partial order of objects, with elements $c \in C$
Let $P$ be a program that does the following when $PC$ is executed:
1. Randomly pick a $c \in C$ that is not picked in the previous iteration, compute $Pc$ 2. If $d \in C$ such that $c \leq d$, compute $Pd$ 3. After a complete run for each $Pc$, refer to checklist. 4. Repeat Steps 1-3
(Very important assumption here. $P$ is not a turing machine. The best way to understand it is it can run for absolute infinite amount of time and steps, thus it can terminate for any transfinite value of time and steps thus there is no infinite loop in the usual sense of the term. I have yet to figure out a way to drop this)
Checklist:
1. If the current $Pc$ maintain its run without switching steps, output "potential infinity detected, $C$ is not finite"
2. If the current $Pc$ terminates due to inability to fulfill step 2, output "maximal element detected"
3. If all $Pc$ fulfills Condition 2, output "$C$ is bounded by maximal elements"
4. If all $Pc$ fulfills Condition 3 and the element that triggers termination is identical, output "$C$ has a maximum"
5. If the current $Pc$ fulfills Condition 1 but some elements are revisited, output "cyclic order of period <number of steps before each revisiting> detected"
6. If the current $Pc$ fulfill Condition 1, some elements are revisited irregularly, output "quasiperiodic cyclic order detected"
7. If the current $Pc$ fulfills Condition 1 and no elements are revisited, and that the output has both a upper and lower bound, output "aperiodic cyclic order detected"
8. If all $Pc$ fulfills Condition 1, and none of these have length equal to $C$, output "$C$ is dedekind finite". Otherwise output, "$C$ is dedekind infinite"
(actually no, I need to rule out countable subsets)
8. If all $Pc$ fulfills Condition 1 or 2, and none of these have the same length if one is a subsegment of another, output "$C$ is dedekind finite", else output "$C$ is dedekind infinite"
9. If all $Pc$ and its subsegments fulfills Condition 3, output "$C$ is tarski finite"
10. If all $Pc$ fulfills Condition 9 and each subsegment contain subsegment of length 1, output "$C$ is finite"
(Here I introduce a new kind of set call an indecomposable set. They are sets of some cardinality that taking its powerset only gives itself. Still need to work out how it will mess up set algebra. Bijections between these sets with other sets are done in an abstract fashion. Bijection can occur if they have the same cardinality, despite no element of such set can be demonstrated)
Anyway, it appears that starting with a computer that can compute indefinitely (and hence all transfinite steps will terminate), it is much easier to check for an infinite thing compared to a finite thing. Likewise, in real life computers, where the number of steps are bounded by a finite number, it is much easier to check for finite thing compared to an infinite thing
My suspicion is that, given any proof that start with an infinite thing, it is much easier to check whether something is infinite more than whether something is finite, and the converse is true if the proof has some notion of finiteness in it
hence there seemed to be some kind of correspondence between an infinite and a finite prover
NB An infinite prover is a program that proves things without assuming any in built notion of finite. In particular, it does not have a notion of natural numbers, only whether the job is completed with yes or no
In contrast, a finite prover is your typical turing machine, which has bounded memory and time, thus if it get stuck in an infinite loop, it cannot output anything
So it seems, even if in an alternate history, we humans knew more about infinity than finite mathematical objects, it seems to take quite an effort to define finiteness
analogous to how in this history where we are more familiar with the finite, it is very hard to prove most theorems involving infinity
well you said your context is bose gas, so it seems more likely that statistical mechanics related users will knew it better. Otherwise I don't know of anyone right here who deal with summation identities
Consider the equation $0=x+2y+z+e^{2z}-1$. If one solves this equation $z=z(x,y)$ at a point $P$ and tries to find the taylor expansion of $z$ at $P$, which will look like $z=Ax+Bx+Cx^2+Dxy+...$ (1), how does one solve for $x,y$ when plugging (1) in the equation? The $e$-term seems to cause some trouble here.
Why is the "basis vector field" $d/(dx_i)$ in Minkowski space constant and covariant (I assumed the basis vectors in a tangent space are contravariant)? Or is there such a thing as “covariant constant”?
@eigenvalue: Yes, vector fields are contravariant tensors. "Covariant constant" is the phrase to say that the covariant derivative (i.e., the derivative using the connection) is zero. Nothing to do with co- and contra-variant.
I just wanted to work something out in words with someone familiar with the topic.
In particular if $\Delta = [0,1]\times [0,1]$ is the diagonal, then I want to compute $\int 1_\Delta(x,y)\,d\mu(x)$ where $\mu$ is the Lebesgue measure.
Basically that integral is only being done across the x, so that means that $1_\Delta(x,y)$ is fixed in $y$. Meaning that it is basically the indicator function $1_{\{y\}}$, right?
Incidentally some set theory which is somewhat measure theoretic. I'm going to give a seminar talk about a strong form of Fubini's theorem which is independent of ZFC
Suppose that $F\colon\Bbb R\times\Bbb R\to\Bbb R$ is such that $F_x$ and $F^x$ are measurable for almost all $x$ (those are $F$ with the first and second coordinate fixed) and such that $G(x)=\int F_xdx$ and $H(y)=\int F^ydy$ are both measurable. Must we have $\int G(x)=\int H(y)$?
Basically we don't ask that $F$ is measurable, but instead minimal assumptions to be able to write a double integral as in Fubini
There are models where it is false (this is easy to show, it fails in every model of CH for example) and there are models where it's true (this is hard to show, it's a result published in 1980 by Friedman)
@BalarkaSen In set theoretic topology an $L$-space is a regular, hereditarily Lindelöf but not separable space, but I guess there's another concept with the same name lol
2) The hat version of Heegaard Floer has $\chi(\widehat{HF}(Y)) = |H_1(Y;\Bbb Z)|$, where if that cardinality is infinite we return the number 0. So in particular we have the rank requirement $\dim \widehat{HF}(Y) \geq |H_1(Y;\Bbb Z)|$. An L-space is a rational homology sphere for which you have equality there.
Because $S^3$ is so simple its Heegaard Floer complex has 1 generator and no differential, so it is an L-space. Same with any $L(p,q)$, whence the name; but also connected sums of L-spaces are L-spaces.
The L-space conjecture provides a conjectural description of which 3-manifolds are L-spaces
I do want to know more but it might be out of my league; I'll think and ask some questions later. Do you know a description of the Poincare homology sphere as a surgery along some knot in $S^3$?
@MikeMiller Oh, ok. $x = aba$ and $y = ab$ works, because $x^2 = aba\cdot bab = (ab)^3 = y^3$. There should be a way to interpret this in terms of the $(2, 3)$-symmetry of the trefoil?
@Semiclassical Okay. Would there be a way to integrate f’(x) (as given in the question)? That is, f’(x) with the $e^{-x^3}$ terms. The range is supposed to be (0,ln 2), so maybe there is some power series solution to it as well...
@Semiclassical Cool. How does one evaluate, say Ei(x) for any x? It is somewhat unfamiliar territory, although the gamma function is a familiar function. Are they related in some way?
@Semiclassical Okay, so if one doesn’t have Wolfram, one would look Ei(-x^3) up in a table? There is no way that the gamma function could be of any help?
I have a following integral: $$\int_0^1 dx\int_0^{\sqrt{1-x^2}}dy \int_\sqrt{x^2+y^2}^\sqrt{1-x^2-y^2}z^2dz$$
Which i have to solve by introducing polar coordinates, which is, by itself, relatively simple:
$$x=\rho\cos\theta\sin\phi \\ y=\rho\sin\theta\sin\phi \\ z=\rho\cos\phi$$
Besides this,...
I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876). In the fourth chapter, he deals with different definitions of finite set. Here is the classical one:
Definition 1. A set $X$ is called finite if there exists a bijection $n \to X$ for some $n \in \ma...
