Though this one probably grows slower than all of the already submitted programs at my challenge above.
@SimplyBeautifulArt I suppose the restriction of $k\in\Bbb N$ is unnecessary, no need to get ahead of yourself.
In this one, $C(0,n)$ and $\psi(0,n)$ are the ones in my question. $C(1,n)$ is then much faster growing, since it includes $\psi(0,k)$ for any $k\in C(1,n)$, so it kinda has the check points to grow off of. And ofc, $\psi(1,n)$ is the first natural missed in $C(1,n)$, etc.
Hm, I wanna try computing a few values...
Hm, not spectacularly faster yet it seems.
I wonder what happens when we include exponentiation...
With exponentiation: C(0)={0,1} C(1)={0,1,2} C(2)={0,1,2,3,4} C(3)={0,1,2,3,4,5,6,7,8,9,12,16,27,64,256} C(4)={0,...,25}U{27,...,36}U{40,42,45,48,54}U{63,...,73}U{...?}
Seems like we have to expand pretty far out before exponentiation has an effect on $\psi(n)$.
O.o perhaps we could have a question: Find the first $n$ such that this $\psi$ differs from the previous.
12² = 144, so C(5) will contain everything from 0 to 144+25 = 169.
Yes. For example let $\mathcal{F}$ be a sheaf of sets on $X$. Then basically it could be $\mathcal{F}(X)=\emptyset$, couldn't it?
The point is, I am doing some exercise where I need to show surjectivity of a certain map, so I would like to say "let $s\in\mathcal{F}(X)$", but I am not sure whether I need to consider this special case $\mathcal{F}(X)=\emptyset$.
Let $X=S^1$ be the unit circle and let $\pi:X\to X, x\mapsto x^2$. Let $\mathcal{F}(U)$ be the set of all continuous functions $s:U\to\pi^{-1}(U)$ such that $\pi\circ s=id$.
This (together with the usual restrictions) defines a sheaf on $X$
But $\mathcal{F}(X)$ is empty, since there is no "continuous root function" on $S^1$.
If you set $\mathcal{G}$ to be the final sheaf on $X$, then the unique morphism $\varphi:\mathcal{F}\to\mathcal{G}$ is surjective, but $\varphi_X:\mathcal{F}(X)\to\mathcal{G}(X)$ is not.
So this exercise was supposed to motivate sheaf cohomology, somewhat measuring the defect of surjectivity.
I suspect it might be similar to the notion used in orthonormal basis in hilbert spaces, that the set produces by this basis are dense in function space
@user685252 well, the dream does not say what the endpoints of those intervals are, only the trend of their lengths forming a bounded sequence of reals. However if they are really disjointed as shown, then they cannot form any function other than $y=0$ with values restricted to some intervals.
It is not clear what happens if they are allowed to overlap and add together though
@SimplyBeautifulArt That's a central focus in game theory. One would say white has a "winning strategy" in that instance. According to Wikipedia, it is currently unknown
Given a problem $P$ and solution steps $S_0,S_1,S_2,...$
Define a solution strategy $K$ to be an ordered list of $S_i$
What is the sufficient and necessary criteria for $P$ to ensure all solution strategy to be equivalent (meaning that it does not matter what angle you approach the problem, you will guarantee to get the solution)?
This is a hard metamathematics question, but it is worth to wonder about since almost all our problem solving methods relies on simplifying the problem by replacing with a simpler concept. The big why is then, why does it work, and when it will fail to work
If $P$ is taken from a smaller domain of discourse, then we can easily find examples where it does not work. For example in functions of multiple variables, it is not always true that $f(x,y,z)=f(x)+f(y)+f(z)$
@SimplyBeautifulArt Microincrement games like these also illustrates the notions of limit ordinals well. Note how as the actual number and the growth rate differ by some orders of magnitude, it will seems like for sufficiently large but finite number, the actual number does not look like it is changing very much for a long time. The notion of limit ordinals take this phenomenon up to eleven, by having the time you need to wait for the number to change to shoot to countable
What contradiction can we get here? What comes to my mind is that the argument in that answer says that $[0,1]=f^{-1}([0,x))\cup f^{-1}((x,1])$, and since $f$ continuous, this says $[0,1]$ is union of two open sets, so is open, is this correct way to derive contradiction?
Imagine we have a linear or affine space of real symmetric $n\times n$ matrices:
$$\mathcal{A}=\{A_0+c_1 A_1+\ldots+c_m A_m: c\in \mathbb{R}^m\}$$
The question is how to efficiently test if it contains matrix of given eigenspectrum? Equivalently: of given characteristic polynomial?
Trying to exp...
Any non-empty open subset of $\Bbb R$ contains rationals...... And your statement that no uncountable subset of irrationals is closed is FALSE. Let $S$ be open with $S\supset \Bbb Q$ and $\lambda (S)<1.$ Then $T=\Bbb R \setminus S$ is closed and is disjoint from $\Bbb Q$ and has non-zero measure so it cannot be countable. — DanielWainfleet24 mins ago
Illustrates very nicely why I want to understand measure theory: I don't even know such sets exists!
We have that $A=\begin{pmatrix}1& 0 \\ -1 & 0\end{pmatrix}$ is an element of the $\mathbb{R}$-vector space $V=\mathbb{R}^{2\times 2}$.
We consider the linear map $l:V\rightarrow V$ with $l(X):=AX-XA$. I have show that $\text{ker }l=\left \{ \begin{pmatrix}x& 0 \\ 0 & 0\end{pmatrix} : x\in \mathbb{R}\right \}$.
I want to check if $l$ is injective or/and surjective.
We have that the kernel is not the zero matrix, $\text{ker }l\ne \{0\}$, does it mean that $l$ is not invertible, so it is not bijective, that means that teh map is neither injective nor sujective?
I wonder how I can get an open set with an uncountable union of singletons without intervals... Aren't all open sets in the open interval topology must be a union of open intervals?
Ah ok! Why do we have to check if $A$ can generate all elements of $V$ ? Surjectivity means that for every $Y\in V$ there is a $X\in V$ such that $l(X)=Y$, right? @Secret @TastyRomeo
Do we need here the image of the map? I have found that $\text{im }l=\left \{\begin{pmatrix}x_{12} & x_{12} \\ -2x_{21} +x_{22}& -x_{22}\end{pmatrix} : x_{12}, x_{21}, x_{22}\in \mathbb{R}\right \}$ So, do we take a matrix that is not in the image? @Secret
@MaryStar So if I understood correctly, you want to check whether $\ell$ is bijective. If so, then $\ell$ is surjective if its image is all of $V$, otherwise it is not surjective and there will be matrices in $V$ not in the image of $\ell$
@TastyRomeo Hmm, I think I can understood it (since the measure will be given by a geometric series $\sum_{i=1}^{\infty} \frac{1}{n^i} = \frac{1}{1-\frac{1}{n}}$), but wow this set is so hard to visualise
So, since at the image there are only the matrices where at the first row they have the same element, we have that for example the matrix $\begin{pmatrix}1 & 0 \\0 & 0\end{pmatrix}\in V$ is not in the image, and so it is th emap is not surjective. Is this correct? @Secret
@MaryStar yeah. You can also see when compared to a generic matrix, the entry $x_{11}$ is constrained by the equation $x_{11}=x_{12}$ hence all matrices where $x_{11}\neq x_{12}$ will not be in the image
hmm... so as we continue to enumerate the rationals, eventually all the rationals will be picked, but only "a small portion of (and occasionally overlapping)" uncountably many irrationals were included, leaving behind uncountably (large) many of them as the closed set
(Dang, it is so hard to talk about how much stuff left behind when cardinals don't really change much whenever you take away or add smaller cardinal number of elements to it)
such behavior cannot be reproduced by having a union of sets consists of rationals and coutably many irrationals, because then countable subadditivity kicks in and turn the Lebesgue outer measure to zero
hmm... so the key thing that controls Lebesgue outer measure is the nature of the uncountable set that is included ... :?
It is known that to compute the Lebesgue outer measure $\lambda^*$ of irrationals $\Bbb{I}$ we make use of the following property:
$$\lambda^*(\Bbb{I}) +\lambda^* (\Bbb{Q}) = \lambda^* (\Bbb{R})$$
and noting that since $\Bbb{Q}$ has zero measure being a countable union of singletons and using c...
what I am constructing here will no longer be intervals at the countable limit, thus it is no longer an open cover. While yes such non open cover will contain every irrational and excluding every rational, since it is not an open cover and is not countable, Lebesgue outer measure failed to be applied, thus we have to calculate the measure of the irrationals indirectly...
Meanwhile, going into the opposite direction on attempting to construct an uncountable open cover by first using axiom of choice to enumerate a well ordering of the irrationals, and then exclude them in the same manner as outlined in this MSE, we will end up with a non open (and also not closed) cover consists entirely of singletons. However, since in this cover only countably many singletons are there, countable subadditivity kicks in and we get the zero measure of rationals as required
so one small consequence of not having some version of axiom of choice that holds at least at the continuum, is that this direct route of proving the rationals having nonzero measure will become nonconstructive since an enumeration is needed to construct such cover
Later on, I will check if I understood this correctly by computing the measure of the cantor and fat cantor sets...
seems so, take a countable union of intervals centred at rationals. This is an open dense set. Meanwhile, the reals are locally compact since for every point there exists compact neighbourhoods in the form of closed intervals and singletons. The reals are also hausedoff since given any two points a,b, we can always find some real c between them. The neightbourhoods (a-e,c) and (c,b+e) will then be disjoint. Therefore the criteria for Baire category theorem holds. Thus by Baire category theorem,
the intersection of that open collection will be dense
... but yeah, I don't really know Baire category theorem.
Hypothetically, combining with what we discussed earlier on why we need countable subadditivity, had the measure been defined to have continuum subadditivity, invariant under translation and singletons having zero measure, then the reals and all intervals would have zero measure, thus such notion will not be useful as a measure
I am now starting to wonder, whether Lebesgue measure is not just a generalisation of length, but actually is a tool that give us more idea on how "uncountable" an uncountable set is...
In mathematics, the Smith–Volterra–Cantor set (SVC), fat Cantor set, or ε-Cantor set is an example of a set of points on the real line ℝ that is nowhere dense (in particular it contains no intervals), yet has positive measure. The Smith–Volterra–Cantor set is named after the mathematicians Henry Smith, Vito Volterra and Georg Cantor. The Smith-Volterra-Cantor set is topologically equivalent to the middle-thirds Cantor set.
== Construction ==
Similar to the construction of the Cantor set, the Smith–Volterra–Cantor set is constructed by removing certain intervals from the unit interval [0, 1].
The...
So once again, we cannot attack this problem directy with covers, as the cover is not open.
Using the usual method to prove that, it does seems given a Lebesgue measurable set, whether the set has a nonzero outer measure is all controlled by the value of the infinite sum as the total length of the cover is summed up
now, what if we construct a cantor like set by removing the middle halves...
@AlessandroCodenotti how to handle a cantor like set with $\frac{2}{3}$ of the interval [0,1] removed each time, since summing up the total amount of lengths removed will be the geometric series $\frac{2}{3}\sum_{n=1}^{\infty}(\frac{2}{3})^n = 2 > 1$. Do we just resolve this and higher cases by noting whenever the total length removed is $> 1$ we just say the resulting cantor like set will be of measure zero?
@Kenshin, if the relation is not fixed is it possible to infer a more precise relation than $r = AO*\frac{BO}{AO+BO+AB}$ and is that relation correct at all? If you can, help me, please.
since $p < q$ and $\frac{p}{q-1} > 1$ only when $p+1 > q$, the best can happen is $p+1=q$ which then give us the upper bound of 1 as required and no longer, hence no issues
I feel that given the position of the third circle isn't clearly defined on your diagram (could be anywhere along the BD line based on your diagram) then the radius can't be determined
In mathematics, a Vitali set is an elementary example of a set of real numbers that is not Lebesgue measurable, found by Giuseppe Vitali. The Vitali theorem is the existence theorem that there are such sets. There are uncountably many Vitali sets, and their existence depends on the axiom of choice. Assuming the existence of an inaccessible cardinal, Solovay constructed a model of Zermelo–Fraenkel set theory (ZF) without the axiom of choice, where all sets of real numbers are Lebesgue measurable.
== Measurable sets ==
Certain sets have a definite 'length' or 'mass'. For instance, the interval [0...
Bernstein Sets are even harder to comprehend than Vitali sets even at the construction. My brain go BSOD when there are too many intersection symbols in the expression
currently at beginning and heading to 6th dimension
I am teaching a measure theory class, where we are in the process of constructing Lebesgue measure on $\mathbb{R}$ via the usual Caratheodory outer measure construction.
As motivation, we began by constructing a Vitali set $V \subset [0,1)$ which has the property that $\bigcup_{q \in \mathbb{Q} ...
But given today's progress, soon enough I will be able to count to nonmeasurability, and thus one step closer in comprehending infinity in a foundation free system
and once infinity is down, everything except number theory will become a lot easier
@SimplyBeautifulArt i found couple of conjectures, in aim to be all wraped in form of a partial answer iff i could design a polynomial time program to check high levels of n>10.
"I have already introduced the formal point of view that point of $\text{Spec} A$ are in one-one correspondence with homomorphisms of $A$ to fields, with $P$ corresponding to the composite $A\to A/P\hookrightarrow{Frac}(A/P)$." Quoting from Reid's commutative algebra book. I'm not sure what that means, every morphism from $A$ to a field has a prime ideal as kernel and from a prime ideal I can costruct such a morphism, but why is that one-one? Can't I have different morphisms with the same ker?