Are there parts of mathematics that are not useful
It's inaccessibly annoying to see unanswered and questions that are deliberately ignored from answering in this chat, that it will be better to never have those questions asked in the first place
I hate RHV, but I hate seeing dangling questions a lot more
If I see anymore of these, I am putting the source out of view
> If you knew you are being ignored, STOP ASKING, you are only annoying us with loads and loads of question that reminds the existence of being ignored
I don't think there are many graph theory experts here. I have seen many graph theory questions got unanswered, at times users saying they knew nothing about it
For analysis it depends, usually when Daminark is on, there is some spike in analysis
Tbh, unlike most users, I can ignore huge reams of messages because I am the most prolific reams of message poster. However:
If the message contains a reminder of someone ignoring someone or the concept of hell ban, it is so annoying that should I have expert hacking skills, I will hack the chat out of existence
I am that kind of person that will start WWIII just to make absolutely certain no human beings can ignore each other anymore (and also the inverse concept, trolling)
My resentment on ignore and disengagement is not something a Berkeley cardinal can capture
In set theory, Berkeley cardinals are certain large cardinals suggested by Hugh Woodin in a seminar in Berkeley in about 1992.
A Berkeley cardinal is a cardinal κ in a model of ZF with the property that for every transitive set M that includes κ, there is a nontrivial elementary embedding of M into M with critical point below κ. Berkeley cardinals are a strictly stronger cardinal axiom than Reinhardt cardinals, implying that they are not compatible with the axiom of choice.
A weakening of being a Berkeley cardinal is that for every binary relation R on Vκ, there is a nontrivial elementary embedding...
@AkivaWeinberger ah no wonder why polynomials are so hard to factorise when it gets so large. Also besides encryption, I suspect this also have implication on number theory due to polynomials play a huge role in finding factors, sum of squares etc. of numbers
So pretty much most of the large polynomials are irreducible
@MatheinBoulomenos I understand why we want integral and separated thinking about the classical definition of variety, but I have no intuition for the finite type, why do we require it?
So integral separated schemes of finite type over $k$ are equivalent (as a category) to varieties over $k$ (variety=separated prevariety). If I drop the "separated" I get prevarieties over $k$. What kind of bad things happen by dropping integral or finite type?
Separated should be free since every morphism of affine schemes is separated
And if $A$ is a domain $\mathrm{Spec} A$ should be integral since all of its localizations are also domains so the sheaf is a sheaf of integral domains on a basis
The way to enable developer settings on Android really feels like it is "secret", even though it is perfectly well documented. It just has the feel of an easter egg.
That is something I and Holo were discussing in another room. We conclude a model for that will be very weird. What we knew however is the analoguous case where arbitrary union of finite set is finite can be modeled by the set of all absolutely convergent sequence since they are closed under pointwise multiplication and pointwise addition
Isn't a model just an algebraic structure that evaluates to true for a given set of propositions? Then it seems the ring of absolutely convergent series with their limits identified with finite sets and addition identified with set union can evaluate the statement "arbitrary union of finite set is finite" to true?
actually wait...
the concatenation of two series that converge to different limits will produce a divergent series
o nvm then
There is a trivial model though if the empty set is the only set that exists in the universe. Clearly, an arbitrary union of the empty set is empty and the empty set is finite
We can identify the empty set with each element in the ring of all convergent series that limits to zero. Then any concatenation, pointwise addition and pointwise multiplication will always give a series that sum to zero
and we can identify these binary operations with set union
Well concatenation, summing or multiplying any series in that ring will only produce a series that sum to zero and we identify this with the empty set. That seemed to satisfy the desired axiom that "arbitrary union of the empty set is empty", no?
I want to build a set theory with that axiom, and try to retain as much ZF as possible. For that we are not sure how much of ZF axioms we can retain if we insert this weird axiom
Yeah that's what I mean with unary unions, for every $A$ you can take $\bigcup A$
But there is no binary union in the axioms $A\cup B$ needs to be defined as $\bigcup X$ where $X=\{A,B\}$ (and pairing is needed to show that $X$ exists)
I don't disagree with you, we're really saying the same thing, you do have unions over arbitrary sets, but they are all finite
My point was that $\bigcup_{n<\omega}A_n$ is really $\bigcup X$ where $X$ is a countable set containing each $A_n$, but no such thing exists in $V_\omega$
Yeah, that's why I am initially mistaken because I forgot that union $\bigcup_{n<\omega} A_n$ is a countable set hence is not an element in $V_{\omega}$
Anyway, the more interesting axiom I want to include is this paradoxical axiom "finite union of finite set is infinite". One version of phrasing this is the same as asking the question "Is there exists a model in ZF where $\aleph_0$ is singular". The discussion in the logic room seemed to suggest the question is still open. Tobias have pointed out ZF does not allow that, but what remains open is what axioms we have to discard to make $\aleph_0$ singular
I mean, we do have models of ZF where $\aleph_1$ is singular, so it seems weird we cannot extend that to $\aleph_0$
(clarification: Analogous to how singular cardinals are defined, $\aleph_0$ will be singular if its cofinality is finite)
So I am a big fan of Hypergeometric terms and Series, I found the Wikipedia basically just vaguebooked a lot, yet also said very little, anyone else enjoy these things maybe wanna throw me a reference and or pay for the a book on the subject for me?
i guess id solicit to ted "smacking" me if it ends up in a peer reviewed publication on the subject mentioned above being sent my way, i mean its still a dirty that no shower will ever be able to wash away but it would feel better knowing it wasn't for money or crack, but a reliable academic resource
Why is it the case that there is no morphism from $\mathrm{Spec}\Bbb Z$ to $X$ if $X$ is a $0$-dimensional scheme? I see that this would give us that the identity on $\mathrm{Spec}\Bbb Z$ factors through a 0-dim scheme and this should be a contradiction, but I don't see why exactly
any ring hom $f : A \to \Bbb Z$ would need to be surjective, and $\ker f$ is a prime ideal of $A$, hence maximal ideal, so $A/\ker f \cong \Bbb Z$ would be a field, contradiction...
A set $T$ is subfinite if $T \subset A$ where $A$ is finite, $|T|=n$ for some $n\in \Bbb{N}$ and $|\mathcal{P}(T)|<|\mathcal{P}(n)|<|\mathcal{P}(A)|$
Rationale: Powerset axiom only said it will produce all subsets of a given set. It does not really restrict us as to how many subsets are there in a given set, thus leaving open a possible loophole to axiomise a set that has fewer subset than the finite set it bijects to
one possible consequence of this is that for these sets, many partitions will become empty
So I could have anormalous set with 5 elements {a,b,c,d,e} such that it's power set is only {{a,b},{b,c,d,e}}
The Star Wars Room is currently investigating just how many normal rules are screwed by the existence of these sets
In other news, we build new systems by first defining the objects we want to include, derive all consequences from them and then decide what axioms is needed at the last step
I want to share an exercise that I liked, of my real analysis course: Let $f$ be a continuous function on $[0,1]$. Show that there is a point $x$ such that $f(x)=x$. This is interesting because it's easy to convince yourself with the geometric meaning.
Is there a way to express it like in French? With just saying the set. I speculate that it's common on working with a sole set. There has to be a way to simply state it.
the definition of solvable is that the composition factors of any composition series of $\mathbb{Z}$ are abelian, but $\mathbb{Z}$ does not have any composition series
@mathsresearcher Because it might not be due to the way you defined it. You get $d(a,u) = \epsilon_1 - \epsilon_2$ which is not necessarily smaller than $\epsilon_2$.
@Eran Sure they do. They have exactly one composition factor, namely themselves.
I need some help finding the primitive element of ℚ(¹³√7, ⁵³√7) over ℚ. Usually the linear combination works out, but proving this one without an insane number of computations seems to be impossible
I'm reading an MSE post and am confused about one point. In this post, they define the following equivalence relation on $\Bbb{R} \times \{0,1\}$: $(x,0) \sim (x,1)$ if and only if $x \neq 0$. What's confusing me is, how do I, for example, determine whether $(x,0)$ and $(y,1)$ are equivalent under this relation? I'm use to equivalence relations being defined for arbitrary points in the space, not specific points like $(x,0)$ and $(x,1)$.
@GodotMisogi it really isn't my field of expertise excuse the pun but the wiki page says that the linear combination is the generator of the multiplicative group of non zero elements of the field, I must be missing something here definition wise because I thought that would make it the primitive element
Consider the following equivalence relation on $\Bbb{R}$: $a \sim b$ if and only if $a - b \in \Bbb{Q}$. Is the projection map from $\Bbb{R}$ to $\Bbb{R}/\sim$ an open map?
I'm writing up a problem and I'm realizing (yet again) I don't know my definitions properly
What I thought I knew: A $d$-dimensional affine subspace of $\mathbb{R}^n$ is defined as $\Gamma=\{x\in \mathbb{R}^n: Ax=b\}$ for some choice of $A\in\mathbb{R}^{d\times n}$ and $b\in \mathbb{R}^n$
But that doesn't seem to be the wikipedia definition
I've made a start on answering an old, unanswered MSE question. I could do with some help, please. My next post directly below is a link to a recent post of mine in the group theory chatroom about it.
I'm forgetting set-builder notation. Suppose I want to write the set of $y\in A$ such that $y=f(x)$ for some $x\in B$ subject to some condition on $x$.
Most obviously, I could just define a subset $B'\subseteq B$ on which said condition is satisfied.
And then said set should be something like: $\{y\in A| \exists x\in B(y=f(x))\}$
But I can't remember if that's the right notation.
Is my reasoning right? I have $f_n(x) = \sqrt{x^2 + \frac{1}{n^2}}$ for $x \in \mathbb{R}$, so I conclude that it's pointwise convergent $f_n \to |x|$, and moreover it's uniformly convergent to $|x|$, because $\left | \sqrt{x^2 + \frac{1}{n^2}}- |x| \right | = \frac{1}{n^2(\sqrt{x^2 + \frac{1}{n^2}} + |x|)} \leq \frac{1}{n^2} \to 0$
and that's because $\sqrt{x^2 + \frac{1}{n^2}} \to |x|$ and it's decreasing for any $x \in \mathbb{R}$, therefore I can make denominator smaller by $\sqrt{x^2 + \frac{1}{n^2}} \leq |x|$, so I'd get in denominator $n^2 2|x|$ and if $|x| > \frac{1}{2}$ then $\frac{1}{n^22|x|} \leq \frac{1}{n^2}$, otherwise if $|x| \leq \frac{1}{2}$ then $\frac{1}{n^22|x|} \leq \frac{2|x|}{n^22|x|} = \frac{1}{n^2}$
Let $p,q$ be prime, I need to prove that (1) any group of order $p^a q^b$ is solvable for $a,b\in \mathbb{N}$ if and only if (2) there are no simple groups of order $p^a q^b$ for $a,b \ge 1$. Already proved the first direction, struggling with proving (2)--->(1)
The question I'm trying to answer has only been viewed on MSE 59 times in two years and nine months.
Thus I think it's safe to assume that perhaps it was just overlooked and/or forgotten about, so no detailed answer was provided. The old comments are helpful though. Maybe the OP got an answer elsewhere but didn't share it here.