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12:04 AM
Hm, I wonder if my function is dominated by n^n
C(5) is an odd case, the first gap occuring within additions
@Abra What do you mean by that?
Suppose that C(n) contained all the naturals from 0 to k.
i was extrapolating and i failed at C(5)
^ There's C(5)
until C(4), all gaps occur after performing all additions
12:08 AM
which turns the problem trickier.
By adding them all up, C(n+1) will contain everything from 0 to 2k
By multiplying them, we will also find that we get all the multiples of 3 from 0 to 3k, multiples of 4 from 0 to 4k, etc., multiples of k from 0 to k²
By repeating the addition once more, we can fill the gaps from x to x+2k.
This fills the gaps from nk to n(k+1) for any n<2k
And so we fill everything from 0 to k²
So ψ(n+2) > ψ(n)²
Or ψ(n+1) > ψ(n)^√2
Or ψ(n) > 2^(2^(n/2))
it would be better coloring the multiples to distinguish them.
12:18 AM
So right now we have something along the lines of:
$$2^{2^{n/2}} < \psi(n) \le 2^{2^{n-1}}+1$$
Not too shabby
@Abra :P Much too lazy
k' i'll scratch some js snippet.
gotta have an hour of so.
Haha, cool
So I was actually thinking of using such a function for my BigNum Bakeoff Reboot...
But like a better version of what you see over there.
$$C(\alpha,0)=\{0,1\}\\C(\alpha,n+1)= C(\alpha,n)\cup\{\gamma+\delta,\gamma\cdot\delta ,\psi(\eta,k):\gamma,\delta, \eta,k\in C(\alpha,n),\eta<\alpha, k\in\Bbb N\}\\\psi(\alpha,n)= \min\{k\notin C(\alpha,n),k>0\}$$
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:
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.
Not quite 178
12:49 AM
Hey chat
Quick question: Do (pre)sheafs need to have global sections?
C(3) contains 6, 16, and 27.
C(4) contains 16 and 6*27 = 162.
C(5) contains 16 + 162 = 178.
@lattice The space is open.
I mean, nowhere in my definition it says that if $\mathcal{F}$ is a sheaf on a topological space $X$, then $\mathcal{F}(X)$ needs to be non-empty.
12:53 AM
@lattice Yes, but it does say something about open sets.
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 $\mathcal{F}(X)=\emptyset$.
@lattice You are right, that can occur :O.
I also just figured out that I have dealt with some example before haha
@lattice What example is it?
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 $S^1$.
1:08 AM
@lattice +1
Is this in Ravi's notes?
Not that I know of
Where did you find it?
It was some exercise from last semester^^
It is actually a nice example
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.
1:18 AM
Last night dream involve the following set of intervals which is claimed to be a dense basis of functions
that doesn't look like the real number line?
no, it's not the real line, more like line segments that progressively getting smaller
I also suspect that we cannot produce some continous functions like $\sin x$ by adding intervals
What do you mean by "dense"?
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
1:34 AM
@SimplyBeautifulArt see this it may have some functional bugs i dunno yet ...
Like the set of fractions?
@leaky is a subtle bug sniffer :D
right, I sniff 100 crickets everyday
You're in England now, "cricket" means something completely different :P
oh god
1:44 AM
@Abra uh
@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
@LeakyNun hey man. Did you manage to get through it? Is it to your satisfaction?
@DavidReed yes, and yes
it's amazing
@SimplyBeautifulArt i need to sort it :S
@LeakyNun Glad to help man. Was good for me too, mental agility wise--trying to reconstruct it.
1:48 AM
@Secret You may enjoy this: ivark.github.io
It's a game to get a lot of antimatter
@Abra :P Sorry, I just can't read js and html very well
@SimplyBeautifulArt hmm, .io are making cookie clicker -esche games now?
I designed an AI game in my 20 yo no human being could settle a win against.
Now try beat AlphaZero with that >:D
2:06 AM
selflearning AI ?
yup, the latest by google. Can learn all chess in human history in 4 hours by playing with itself
google is mastering all fields of CS and even extra CS.
I want an AI that master infinities and get it to teach me what it saw
2:10 AM
If you start out as white in chess and play 'perfectly,' will you always win?
I play with white + italian opening (this is my standard winning condition)
2:31 AM
/14 -- the gap is highlignted
2:56 AM
How to find the "C" in y'=4y^2*e^(-5x) = 0, y(0)=5 ?
I have so far:
y = 5/4*e^(5x)+c
I'm doing something wrong because when I solve for C I get 5-(5/4) and I am marked as wrong
Think I screwed up solving the differential equation
Can someone just tell me where I screwed up?
separate y and x: dy/y^2 = 4e^(-5x)dx
integrate: -1/y = 4/-5 * e^(-5x) +c
Reciprocate both sides and cancel negatives: y = 5/4e^(5x) + c
What am I doing wrong? If anything...
Lol, nobody is online on fridays :/
3:21 AM
@10Replies Check this step again
You had -1/y = -4/(5e^(5x)) + c = (-4+d)/(5e^(5x))
These are all different forms of the same equation...
-1/y = 4/-5 * e^(-5x) +c => (-1/y)^(-1) = (4/-5*e^(-5x) + C)^(-1)
So how do I then compensate? y= (4/5 * e^(-5x) + C^-1) ?
Oh wait, no, it's addition
3:36 AM
I solved for C using the y = (4/-5*e^(-5x) + C)^(-1) equation and got 3/5
> Remember kids, reciprocals don't distribute over addition
I sometimes made that mistake myself when solving ODEs
I fixed my sign error and got c= -5 +(4/5)
but -1*(-4/5*e^(-5x)-5+4/5)^-1 was still marked as wrong...
3:49 AM
@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
wait nvm still screwed up
As I said:
-1/y = -4/(5e^(5x)) + c = (-4+d)/(5e^(5x))
Solve for d
"Prove that there exists a sequence of 999 consecutive integers with exactly 10 primes"
[some metamathematics]
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
4:10 AM
[short rambles before moving back to rambles]
The different levels of infinity
Picture counting numbers one by one from zero
You can imagine counting this faster and faster and faster
but no matter how fast, you still get a finite number in the end
Now imagine counting so fast that you count all the naturals in one single step: Congratulations, you have just reached $\omega$ the smallest infinity
Now, you can imagine counting up, and again faster and faster and faster
eventually you count so fast that you can no longer write an alogorithm to demonstrate how fast you count
if you continue this procedure, eventually you count so fast that everything seemed to be counted right at step 1. Now you are at $\omega_1$
So in general, as we move from one level of infinity to the next, what used to be a step by step process takes only one step to finish
5:05 AM
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?
@Silent why is $[0,1]$ the union of two open sets?
$[0,x)$ isn't open
I mean, if we take $[0,1]$ as a subspace itself , I thought. @LeakyNun
then $[0,1]$ is open in the subspace itself
Why is that wrong?
So, how we get contradiction?@LeakyNun
@Silent by arguing that $[0,1]$ is connected
btw, "For a different approach then the ones above" is really the same approach (it's how you prove ivt in the first place)
5:09 AM
Thank you so much.
6:27 AM
Counterexample: I understand maths much quicker in english than my mother tongue chinese
6:40 AM
sry wrong chatroom
6:51 AM
Given a countable sequence of computable functions $\{f_n\}_{n \in \Bbb{N}}$ on $\Bbb{R}$, show that:
$$ \cdots \circ f_2 \circ f_1 \circ f_0 (x)$$
is uncomputable
2 hours later…
8:27 AM
second missing link for testing graph isomorphism in polynomial time:
Q: Testing if given linear space of symmetric matrices contains matrix with given eigenspectrum?

Jarek DudaImagine 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...

1 hour later…
9:41 AM
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. — DanielWainfleet 24 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?
nontrival kernel always means non injective, for surjectivity, check if $A$ can generate all elements of $V$
injective and surjective implies bijective
not bijective implies not injective OR not surjective
In some cases you can upgrade this, but still, better not to assume this straight away
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?
what does the set $S$ even look like?
9:57 AM
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
@Secret It's a countable union of intervals, for example
Take any enumeration $(q_i)_{i \in \mathbb{N}}$ of the rationals
And, take the open intervals $(q_i- 2^{-i},q_i + 2^{-i})$ around every $q_i$
It's easy to check their union will have finite measure
And by taking some number larger than $2$ or higher powers or whatever, you can get the measure of the union arbitrarily small
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
The measure will be bounded by that geometric series, but won't be equal to it
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
10:06 AM
A lot of those intervals will overlap or be contained in eachother
@MaryStar that's correct, yes
@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
Also, note that in this case, you're working with a map on a vector space
You could use the rank-nullity theorem to show that not injective iff not surjective :P
(if you've seen it)
rank nullity theorem is indeed one of the most useful things in linear algebra that fails in functional analysis
@TastyRomeo Do you mean that using this theorem we see that $dim(ker) \neq 0$ and $dim(im)\neq dim(V)$ ?
@MaryStar We see that $\dim \ker l \neq 0 \iff \dim \operatorname{im} l \neq \dim V$
@TastyRomeo Yes! I see!
Thank you very much!! :-) @TastyRomeo @Secret
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 ... :?
Q: Trying to compute Lebesgue outer measure of irrationals by constructing a cover

SecretIt 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...

so that means...
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...
10:51 AM
@Secret If you know the Baire category theorem you don't need much more to construct such a set
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...
11:14 AM
how is "outlier" underlined by google chrome
underlined by dots or squiggles?
if former, it is likely got indiced by an ad
12:03 PM
red squiggles @Secret
then I have no idea why it is underlined as typo
12:24 PM
Any ideia how to obtain Lim (x->0) [ ( (1+x)^a -1 )/x ] without Lhospital ?
I hate those "without" questions
they're ill-defined
and pointless
"with" would be much better
They're not pointless at all
Before you get your "cheat sheet" of standard derivatives, all you have is the definition of a derivative, which uses limits
If you can't solve a bunch of limits without l'Hopital, you can't prove that cheat sheet which you later use when invoking l'Hopital
@TastyRomeo what I mean is they shouldn't phrase it in the negative
one might argue "can I use Taylor series then?" and if they ban it, replace it with "Mclaurin series"
a better way would be "using only the basic properties of limits"
or even "from the definition of limit" if that is what is desired
instead of "without"
Agreed. Its is. Using only basic properties of limits.
These "without using X" question is one reason I like first principle proofs and derivations
first principle proofs are always immune to pathologies and special cases
12:33 PM
my point is they shouldn't say "prove without ...", but instead they should say "prove using ..."
Thats why a friend asked me to solve that, and I applied Lhopital, but he does not know Lhospital and cant use it on his test.
We cannot really say "prove using ...", because there are at most countably many possible proof strategies
Hi guys. Is every point in $\mathbb{R}$ a limit point of $\emptyset$?
oh, it was you who said "without", not the question, lol
@philmcole no
and it would had been worse if human beings are inherently countable dimensional
and developed a countable alphabet
12:35 PM
in space X with subset S, a point x in X is a limit point of S iff every neighbourhood of x in X contains a point of S other than x itself
@LeakyNun So then it's the other way around: No point in $\mathbb{R}$ is a limit point of $\emptyset$?
@philmcole yes
@LeakyNun Thanks!
1:04 PM
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...
$$1-\frac{1}{n}\frac{1}{1-\frac{1}{n}} = \frac{1}{n-1}$$
no, I need something more powerful than this. Consider:
$$1-\frac{p}{q}\frac{1}{1-\frac{p}{q}} = \frac{p}{q-p}$$
sorry typo
$$1-\frac{1}{n}\frac{1}{1-\frac{1}{n}} = 1-\frac{1}{n-1} = \frac{n-2}{n-1}$$
$$1-\frac{p}{q}\frac{1}{1-\frac{p}{q}} = 1-\frac{p}{q-p} = \frac{q-2p}{q-p}$$
$\frac{n-2}{n-1} > 1 $ when $n-2 > n-1$ which is impossible
similarly for the p,q case
actually, what I tried to say is the following:
$\frac{1}{n-1} > 1$ when $n < 2$
1:29 PM
g'day fuckers
so constructing a cantor like set removing the $\frac{2}{3}$ of an interval each time should result in overcounting to occur
Could someone help me to tackle the problem, please?
yeah that looks fucking easy
I can help you out
where r u stuck?
I do not know how I can express it in terms of only AO and r
$\frac{p}{q-p} > 1$ when $2p > q$
1:33 PM
alright mate
let me draw a diagram myself, find the answer
so picking $\frac{2}{3}$, the total amount of stuff removed is $\frac{2}{3-2} = 2$
or actually I do not know the approach at all
@trafalgarLaww first note
you want the ratio in terms of what quantities?
because the answer isn't a number
AO and r
I thought you want to evalualte AO/r
1:35 PM
I would like to find the ralation of AO and r
I'm saying that AO/r isn't a fixed number
it depends on the other variables in teh iamge
can you show that?
show it, please
(1) Draw a line of arbitrary length AO
(2) draw to circles at the end of the line of arbitrary radius r
(3) construct an isoseles triangle containing the circles as shown in your diagram
Since the choice of AO and r were arbitrary
the ratio of AO/r isn't fixed
1:37 PM
sounds very reasonable
let me think about it for a while, please
@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, thanks a lot for the help.
If you remove 2/3 at the first step at the second step you remove 2/3 of the remaining 1/3, which is 2/9, not (2/3)^2
1:53 PM
@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.
now you fuckers are asking a lot, i'm going to have to do some equations
hang on
Ah I see
$$\frac{p}{q}\frac{1}{1-\frac{1}{q}} = \frac{p}{q-1}$$
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
@trafalgarLaww this should help mathforum.org/library/drmath/view/54670.html
Basically we get
(AO - r) + (OB - r) = AO^2 + OB^2
2:13 PM
@Kenshin, so I updated the question. So, it seems, that the relation now is fixed. But I do not know how to show that, I just feel it.
2:31 PM
@Abra Woot woot, I got a really good approximation of $\psi$.
@trafalgarLaww I think the diagram isn't quite right
That third circle off to the right
Nvm, I made a mistake
should it be touching the line CD
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
2:46 PM
and now...
perhaps I can finally comprehend this proof now...?
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...
@Kenshin yes, it should
3:06 PM
@Secret I'm almost 20% of the way to ∞
still in the 5th dimension
@SimplyBeautifulArt wth is this ?
@SimplyBeautifulArt hold on, i comitted some errors in my jfiddle.
@Secret I think I figured out how to get good at the game, for now at least
i know the git address i just wonder what is it.
3:09 PM
You basically try to grind for antimatter
1st dimension gives you antimatter.
Every 10 of the 1st dimension you buy, the antimatter it produces will double.
2nd dimension gives you first dimensions.
Every 10 of those you buy, the first dimensions produced will double.
I'm down at the 8th dimension
When you reach the 8th dimension, you can also do dimensional sacrifices.
You lose your 1st-7th dimensions (but the 10 buys multiplier stays) and your 8th dimension gets boosted
My first dimension has a multiplier of +100 million right now x'D
Lots of people seem to have difficulty tackling my question lol
@Secret Btw, infinity = 10^308 I think
And the bar at the bottom is probably logarithmic scaled.
Haha, I reset again
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
So my strategy is to always buy the "Until 10, cost ..."
As well as the tick boost
3:26 PM
Q: A Vitali set is non-measurable, direct proof, without using countable additivity

Nate EldredgeI 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} ...

ugh, my brain is definitely not in nonmeasurable set mode today, going to continue on my chemistry stuff...
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.
this where the challenge figures ....
3:29 PM
3:47 PM
Looks like this going to be our first Winter Bash without robjohn around? @anon
In the chat room.
according to his chat profile he was here at some point on 5/7 days last week
ok, there's still hope :-)
never mind, that's not a recent graph lol
but prolly just lurking
3:51 PM
"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?

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