@user76284 Ordinals are actually encoded as a well ordering of the natural numbers, at least until you reach $\omega_1$
e.g. $\omega + 1$ is encoded as 1,2,3,4,...,0 algorithmically speaking. No pairs needed
(continue on unrelated note)
Last night dream introduce a very weird kind of large cardinals which is made from non well founded sets (i.e. sets that contain itself such as X={X}). These weird cardinals, are "large" and inaccessible, in the sense that they are larger than any self containing sets, which are already large because they contain themselves as elements and hence an infinitely decreasing sequence. Thus let a non well founded set be Y={Y}. Then these large cardinals are T such that $Y \subsetneq T$ for any $T > Y$
According to discussions with user21820 on this topic, an encoding is basically a function $f : \Bbb{N}^2 \to \{0,1\}$. Thus e.g. $f(x,y)$ outputs $1$ if $x > y$, otherwise outputs $0$
Hence $f$ is a program that tells you how to order the natural numbers, and hence they encode a given countable ordinal
$f : \Bbb{N}^2 \to \{0,1\}$ absolutely does not necessarily imply $f(x,y)=1$ if $x \gt y$, if I am correct in my interpretation of the former statement, $f$ is any two variable arithmetic function that is equal to either $1$ or $0$, for any $x \in \mathbb N$ and $y \in \mathbb N$
But I feel as if I should point out, this is not a medium for collaboration between the vast many of us in the sense that our thoughts are going to be respected as having come from us as individuals, it's more a sham set up to benefit a select few, much like all things are really, and providing answers isn't going to lead you anywhere but perpetually being taken for granted
In the philosophy of mathematics, ultrafinitism (also known as ultraintuitionism, strict formalism, strict finitism, actualism, predicativism, and strong finitism) is a form of finitism. There are various philosophies of mathematics that are called ultrafinitism. A major identifying property common among most of these philosophies is their objections to totality of number theoretic functions like exponentiation over natural numbers.
== Main ideas ==
Like other finitists, ultrafinitists deny the existence of the infinite set N of natural numbers.
In addition, some ultrafinitists are concerned with...
In the philosophy of mathematics, ultrafinitism (also known as ultraintuitionism, strict formalism, strict finitism, actualism, predicativism, and strong finitism) is a form of finitism. There are various philosophies of mathematics that are called ultrafinitism. A major identifying property common among most of these philosophies is their objections to totality of number theoretic functions like exponentiation over natural numbers.
== Main ideas ==
Like other finitists, ultrafinitists deny the existence of the infinite set N of natural numbers.
In addition, some ultrafinitists are concerned with...
It's very strange, I mean in that yes I have done quite a bit of work in analytic number theory, I cannot see where they have drawn the conclusion of denying the existence of an infinite natural number line, and yet, they seem to agree that there exists a limit of measurability, which is really the same thing as saying that the set is infinite lol
The notion of a proper class is only a natural consequence of how Peano axioms work but it does not prove there exists a proper class, only that you can define such in logic
It remains open whether the existence of infinite set can be a theorem
But I feel as if I should point out, this is not a medium for collaboration between the vast many of us in the sense that our thoughts are going to be respected as having come from us as individuals, it's more a sham set up to benefit a select few, much like all things are really, and providing answers isn't going to lead you anywhere but perpetually being taken for granted
and providing answers isn't going to lead you anywhere but perpetually being taken for granted
I guess combining with the 76284 data, that kinda justify it
Well at least 76284 is not as JEE machine as gateprep, but it really give me the urge to spam about large cardinals
What’s the order type of the set of all integer polynomials, where $f \leq g \leftrightarrow \exists x \in \mathbb{Z} : \forall y \in \mathbb{Z} : f(y) \leq g(y)$?
Turns out that RudyRucker book I have been reading in the past weeks have an answer to this question
@Secret come on, don't go and think that a username composed of any alpha numeric characters awards any kind of legitimate identity, everyone is just a unit of a certain monetary fraction of the website owner's profits
we are all Stack Exchange lol
@RyanUnger ok that's quite a number of rhetoric @BalarkaSen compliments yes he knows he is good too I can only assume, therefore probably doesn't get any satisfaction except for the first one you give if he is impressed by you, the rest id keep to references to specific proofs he has shown you if I were to give social advice there
> One last picture of alef-one. Go back to Figure 42. In this picture we saw how various ordinals can be represented as sequences of functions ordered according to steepness (the <bep ordering). How long a sequence of functions can be found with each function steeper than all the ones before? At least alef-one. That is, if S is a set of functions so that for every g (no matter how steep), there will be an f in S that is steeper, then S must have at least alef-one members.
Ok nvm, I made a read error. They are saying all eventually dominanting functions, not just polynomials
They are talking about aleph 1 as the first number that cannot be injected into $\omega$, so they are talking about $\omega_1$. However, when I first read this I do felt strange. Problem is I do not have as good an intuition nor experience to handle the set of all increasing functions ordered by eventual dominance, thus I have not tried to (dis)prove myself on that yet
@Slereah lol no I am just saying it probably only feels rewarding when he impresses himself, if I show my math to my family or my cat and the miracle of them saying something nice occurs, I don't think I would be any happier with the math, so if he is the same in that regard, it only feels good when there is mutual appreciation for something he has done
@LeakyNun translate it in English? I'm not that familiar with lean yet, still learning it
Hi @ÍgjøgnumMeg
Actually let me translate it
Suppose that $\alpha$ and $\beta$ are measurable spacces, $\nu$ is the measure on $\beta$ and $s\subseteq \alpha\times\beta$. Then $s$ measurable implies that for all $b\in\alpha$, $\{y\in\beta\mid (b,y)\in s\}$ is $\nu$-measurable?
(If my translation is correct the theorem is false by the way)
Specifically let $V\subseteq\Bbb R$ be the Vitaly set and consider $V\times\{0\}\subseteq\Bbb R^2$. It is measurable having measure $0$, but obviously not all slices are measurable
The correct version is "let $(X,\mu)$ and $(Y,\nu)$ be spaces with measure. Suppose that $S\subseteq X\times Y$ is $\mu\times\nu$-measurable. let $S_x=\{y\in Y\mid (x,y)\in S\}$ and $S^y=\{x\in X\mid (x,y)\in S\}$. Then $S_x$ is measurable for $\mu$-almost every $x$ and $S^y$ is measurable for $\nu$-almost every $y$."
The converse can fail badly, you can ask all slices to be measurable and yet $S$ can be nonmeasurable, you can even ask all slices to be open while keeping $S$ nonmeasurable
Which is why some people define the product algebra as the completion of the product, because you really want to say that every subset of a measure zero set is measurable with measure zero
If you use the product $\sigma$-algebra on $X\times Y$ then $S\subseteq X\times Y$ measurable implies that $S_x$ and $S^y$ are measurable for all $x,y$
@Slereah well no I didn't say that at all, that interpretation indicates it isn't worth me paraphrasing, but no I never implied he was from a royal family either. I don't know what the social environment is like in India, I doubt they have a kings etc, if they do they are probably like ours, totally unimportant super rich reality tv stars
also everyone should look at the way @LeakyNun writes questions this makes it far easier for people not necessarily in that specific field of mathematics to attempt it and or learn from the experience, asking in natural language with a f%$# ton of field specific terminology is just silly
It's a little strange that some people can have a career and declare themselves to be ultrafinitists. Sure ok academics is in all likelihood like any other industry now, it isn't what you know it's who you know. Actually it was always like that wasn't it
psychiatry is by far the most corrupt industry by a long shot tho
in my own personal experience I mean, of course there are banks that have stolen billions from people and never spent a single day in a cell, so round that up to all authorities and we have a winner
I am aware of that but I'm just curious as to how Doron can have a career and still go in opposition to classical view, sure yes his algorithms are very impressive we study very similar areas of number theory, that's the only reason I came across them, but we live in a world where majority rules, popular opinion is equal to correct opinion in an institutionalized mind set, so, I don't make any bold assertions like that, only state what i know to be true, its just interesting that's all
I guess already being important might explain it so if i were to guess i would have to go with nepotism. Yes it shouldn't have anything to do with it, but it does. \
well no I wasn't talking about the success and already stated I hold his work in very high regard, I only implied this for the career of the individual
do you truly believe we are all given equal opportunity @anakhro ? That's very amusing and defensive of you
no they are not the same thing. So in the hypothetical scenario of what I implied being true, which you have agreed, does happen, you find this amusing?
in a 2-category for example, you have objects, which are 0-morphisms and 1-morphisms going from objects to objects and you have 2-morphisms going between 1-morphisms
well, a career is a good measure of how comfortably the individual lives, and how much appreciation for their work they have received from others. This is not in any sense a measure of success
The category of small categories is such a 2-category, with objects being small categories, 1-morphisms being functors, and 2-morphisms being natural transformations
another example is the category of topological spaces with 2-morphisms being homotopies between continuous maps
I see. I just had this thought because it will look more like an interpolation if $\eta$ literally translate the top row of arrows to the bottom row, and then for some reason I started to imagine the whole thing to behave like a homotopy where the top row of the commutative diagram looks like a directed curve to me
yeah i think it was Venice? I wasn't really listening but yeah her brother's wife is Italian so they have family over there enabling her to stay for over a month which was great
Oh ok yeah I did plan at one stage to do one of those semesters abroad things in Budapest, but I literally don't want to travel unless I can live in that place for an extended period, I just don't see the point in vacations personally
My ancestors are French lol apparently they fled from the French revolution to England and then well by no wrong doings ill assume one generation ended up here in Australia
Let $X$ be some compact Hausdorff space, let $C(X)$ denote all of the continuous complex valued functions on $X$, and let $A \subseteq C(X)$ be a subalgebra. If $f \in A$, does it follow that $|f| \in A$?
$|f| = \max \{f,-f\}$, but I don't think that is helpful...
Let $X$ be some compact Hausdorff space, let $C(X)$ denote all of the continuous complex valued functions on $X$, and let $A \subseteq C(X)$ be a subalgebra. If $f \in A$, does it follow that $|f| \in A$?
Please recommend me good book to learn undergrad algebra. I took algebra course at uni which covered speedily topics of basic structures, vector fields, transformations, polynomials, characteristic polynomials, residue classes and matrices. I didn't understand well many things so I wish to re-learn by myself things.
don't really care for uni content, I'm rather interested in learning these
Was just looking at the math reddit that @Flowian posted and it links to an MO post that is essentially "Let's compile a list of all the full courses on various subjects that have been made publicly available"
So thanks a lot indeed, @Flowian
99% of Semester 4 focusses on Kato's Class Field Theory
for example in the local case, you can define dimension inductively: a 0-dimensional local field is a finite field and then a n+1-dimensional local field is the quotient field of a complete DVR whose residue field is a n-dimensional local field
Shinichi Mochizuki's father was Fellow of Center for International Affairs and Center for Middle Eastern Studies at Harvard University interesting stuff
Sounds nice. I applied for intro algebra TA for next semester and got accepted. I went over my past evaluations for the application (since I was asked for them) and noticed how I improved over time (I've been a TA 3 thrice)
Félicitations! ... Most people have a rocky start in teaching and those who care do improve. We've commented before on your improvement in explanations and patience in the chatroom :)
@BalarkaSen Remeber I was telling you of meeting an ISI M Math interview guy who admitted that he had no fucking idea of compactness ? Well now that results are out I have seriously no fucking idea (I remember he telling to someone near me after the interview that he couldn't even prove that if a, b are nilpotent then a+b need not to be nilpotent even after crapload of hints and time) how he got selected in M Math that too with a single digit rank...
Or more directly, is there a function $f$ such that $\lim_{\varepsilon \rightarrow 0^+} \sum_{n=0}^\infty \frac{1}{n} f(n,\varepsilon)$ exists and $\lim_{\varepsilon \rightarrow 0^+} f(n,\varepsilon) = 1$?
Such an $f$ would be a "regulator" for the harmonic series.