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12:00 AM
I don't follow the statement of the prop
did you mean to write $\overline{k}$ everywhere?
my bad, I missed an overline
$M$ is a maximal ideal of $\overline{k}[x,y]$ good catch
yeah there you go
and no
I think Lemma $1$ does not actually have an overline
but for sure, you need the $\overline{k}$ for the actual prop. bc of what you said
ok, so you're just asking why non-zero prime ideals of $\overline{k}[x]$ are of the form $(x-a)$, yes?
I guess also why is $P$ prime, but yes
the preimage of a prime ideal under a ring homomorphism is a prime ideal
assuming rings with unity, I think this is true yeah. de ja vu lol
12:04 AM
this also works without units
for the other part, remind yourself that polynomial rings over fields are PIDs and what it means to be algebraically closed
hmm I think it does... I have to write out the proof but I am thinking the kernel of a given ring hom with the quotient map (for quotienting the target ring by its prime ideal) will have kernel a prime ideal in the source ring because both of these quotient rings will be isomorphic as integral domains
or something like that
not isomorphic, but one will embed into the other, which is enough
this also follows from a one-line computation by definition, tho
oh ... hmm well the source ring modded by the kernel of the composition maps will at least be isomorphic to a subring of the target ring modded by its given prime ideal
and I think a subring of an integral domain is an integral domain
and lol i overcomplicated things, but it's ok
and $\overline{k}[x]$ is a PID for sure so its ideals are $(x-a)$ and are prime surely and so they're all like that because, since $\overline{k}$ is algebraically closed all irreducibles are just of that form... kinda obvious in hingsight
thanks for giving me the right hints @Thorgott
12:39 AM
$u$ is computationally universal (in the sense of the UTM theorem) iff
\forall f : \mathrm{RE} \qquad
\exists g_1, g_2 : \mathrm{R} \qquad
f = g_1 \circ u \circ g_2
where RE is the set of recursively enumerable functions and R is the set of recursive functions. Is this definition correct? Or does R need to be replaced with another set like PR, the set of primitive recursive functions?
what is the idea for the 2 recursive functions?
one is an index I imagine and the other?
Recursive in the sense of computability theory.
yeah I know
maybe just explain in words though
I get that $g_i$ are total computable functions, so basically indices
but idk why you post and precompose
The idea is that $g_2$ formats the desired input into $u$'s "programming language" and $g_1$ formats the output back to the desired format.
so I was asking for intuition since I had never seen this
12:44 AM
Because $u$ might be universal but have a restricted input/output format.
I thought $u$ was supposed to be universal
Oops. I meant $u$.
e.g. $u$ might deal with binary strings $\{0,1\}^*$ as input and output, whereas $f$ deals with natural numbers $\mathbb{N}$.
Just as a rough example. You can imagine more elaborate versions of this.
you generally fix an encoding for this sort of thing, so I am guessing $g_2$ is going to be superfluous and $g_1$ will necessarily be in $R$
Right, the idea is that $g_2$ handles the encoding part.
changing encodings I do think is just another total function
12:46 AM
and $g_1$ handles the decoding.
That's a good way to put it (encoding and decoding).
yeah that's basically what you're doing
good stuff
I suggest you look at page 24 by Cutland in his book "computability- an intro to recursive function theory"
looks like what you are saying kinda
Thanks. I just haven't seen the universality property stated in this form, which seems very clean.
idk if it's an iff type of thing
You mean the $\rightarrow$ or $\leftarrow$ direction?
lmao backwards
there we go
it is not that clear to me that it should be true
12:54 AM
Hmm. What would be the definition of "universal" then?
I see the page in Cutland's book. Thanks for the reference. It seems to be describing (a fixed instance of) computability in terms of computability on natural numbers. But the ideas are very similar (with an extra quantifier).
the universal function for $n$-ary computable functions is the $(n+1)$-ary function $\psi^{(n)}_U(e,x_1,...,x_n) = \phi^{(n)}_e(x_1,...x_n)$ where equality is really equality of partial functions and the $U$ subscript recalls the unary case
basically it simulates p.c. functions of the same arity (-1) when given the index for that particular p.c. function
maybe look at Kleene's $T$ predicate
Thanks. I wonder if R needs to be restricted to something like PR.
and yeah I don't really know if what you're talking about should refer to universality at all. it kind of could be suggested that the $g_2$ could play the role of an index, but idk, I'd have to see a proof
idk if that would hurt you rather than help you, but it still seems nice to think about
I guess I could define $u : A \rightarrow B$ as universal iff there exist total-computable $f : B \rightarrow \mathbb{N}, g : \mathbb{N} \rightarrow A$ and a natural-universal $h : \mathbb{N} \rightarrow \mathbb{N}$ such that $h = f \circ u \circ g$?
Basically you can emulate $h$ through $u$.
I mean, idk what natural-universal is, and this is all very much focused on the maps more than the indices, so I am not sure it would be the same
1:08 AM
By "natural-universal" I mean any function $\mathbb{N} \rightarrow \mathbb{N}$ which is universal in the sense of the UTM theorem.
with some $\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ bijective pairing, like Cantor's pairing function.
Ok, this is a bit more elaborate than I thought.
$h(e,x) = f(u(g(\langle e,x \rangle)))$ or something like that.
where $h$ takes the place of $u$ in en.wikipedia.org/wiki/UTM_theorem#Theorem.
well just from the get go I am sort of confused because you say $u$ is universal so it should be defined $\Bbb N \times \Bbb N \rightarrow \Bbb N$
so if I try to make sense of this you might basically be doing a change of encoding of a utm
Yeah, I guess I'm generalizing universality from the two-argument, natural-number case $\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ to the more general case $A \rightarrow B$.
yeah I guess you could pair it off that way. it's kind of nice to think of it as having at least 2 inputs pedagogically, but yeah you could do that
Right, the input can be unpaired into a "code" and "input", where the code selects an index in an enumeration of partial computable functions.
wait also idk what $A$ and $B$ are. should probably be related to $\Bbb N$ somehow
1:19 AM
I guess I can reword it as: $u : A \rightarrow B$ is a universal function if and only if
\forall C \quad
\forall D \quad
\forall f : (C \rightarrow D) \cap \mathrm{RE} \quad
\exists g : (C \rightarrow A) \cap \mathrm{R} \quad
\exists h : (B \rightarrow D) \cap \mathrm{R} \quad
f = h \circ u \circ g
where $C$ and $D$ range over some appropriate domain of discourse.
and $A$ and $B$ I imagine?
is $g:(0,\infty]\rightarrow \Re \mid g(x)=\sqrt{x}$ a surgective function?
right I see you're trying to generalize the sources and targets, but still idk if this whole thing should be true. also it is a $\Pi^2_2$ sentence so it's vastly more complicated than utm's usually are defined to be
Hmm... you mean $A,B,C,D$ should be "finitary" in some sense, right?
@Bohemianrelativist Plot the graph of the function.
e.g. $\mathbb{N}$, $A^*$ for any finite set $A$, etc.
I guess one should consider what sort of domain/context R and RE "reside in".
well it's still kind of tricky because you're quantifying over functions rather than elements. and also idk if restricting to finite sets gives you what you want at all.
tbh idk if this description of universality fits the bill like I haven't seen a proof really
and yeah, usually all this stuff is done over the natural numbers so that's a start
(also I should have written $\Pi^1_2$, but yeah)
@Bohemianrelativist What's the definition of "surjective"?
I guess one usually first fixes a "universe of objects", like $\mathbb{N}, \{0,1\}^*$, before one starts even talking about RE.
I wonder what definition encompasses objects like $\mathbb{N}$ and $\{0,1\}^*$ in a comprehensive and unified way. I suspect something type-theoretic (algebraic data types?).
For now we could restrict ourselves to something concrete like $\mathbb{N} \cup \{0,1\}^*$ and ask whether the biconditional holds in that context.
second thing doesn't really fit the bill, but this comes to mind for the first thing
also $\Bbb N \cup \{ 0 , 1 \}^*$ is pretty much just $\{ 0,1 \}^*$, Cantor Space. you could do computability there probably but I don't know loads about that
depends on your model of naturals but yeah
@user76284 for every y in the codomain of g, there is at one x in the domain of g such that g(x)=y. So I think $g(x)=\sqrt{x}$ is surgective. However, the Wikipedia says "A function has a right inverse if and only if it is surjective", but I doubt $g(x)=\sqrt{x}$ has a right inverse.
@Bohemianrelativist What's the codomain of your $g$?
at least*
1:44 AM
and I mean since it's trivially compact Hausdorff so it comes from a codensity monad so you can take cofiltered limits on there, so there has to be a sense in which the natural numbers are sitting in there
@user76284 $\Re$ real numbers
but uh, yeah, that's literally nonsense that helps bat cobwebs away
@Bohemianrelativist And what's the range/image of $g$ you're seeing in your graph?
@user76284 positive real number.
That graph is from the Wikipedia.
@Bohemianrelativist Ok. There's your answer.
@BigSocks Thanks for your help by the way. I'll take a closer look at the book you mentioned, in case this is discussed there.
1:52 AM
glad to help- let me know if you figure something out. would be cool to see a proof too
@user76284 what's the answer? is $f(x)=x^2$ a right inverse of g? does $g(f(y))=y$ always hold?
is there a difference between left/right inverse in this context?
hmm, i guess there is
I wonder why in the graph $y=\sqrt{x}$ is always positive.
because \sqrt is defined as the positive square root
at least for real x
it has a good attitude.
2:04 AM
@Semiclassical really? but that the Wikipedia writes $g: [0,\infty)\rightarrow\Re \mid g(x)=\sqrt{x}$ makes me consider $\sqrt$ means both positive and negative square roots.
a function has to associate to each element in the domain a unique element of the codomain
because $\Re$ should mean all real numbers, positive or negitive.
so it can't mean both positive and negative square roots, you have to settle for one (and usually you settle for the positive ones)
@Thorgott but the Partial Inverses Section of the Wikipedia says one can also consider $g=\pm\sqrt{x}$ as a multivalued function.
then it doesn't have codomain the real numbers
2:16 AM
@Thorgott if what's the codoman of g if it is treated as a multivalued function?
depends on how you want to formalize multivalued functions
I advise not thinking about this, it's pretty useless
"at least for real x" is the big qualifier here
if you want to make sense of sqrts of complex numbers, then the question is harder
my purpose of checking $g(x)=\sqrt{x}$ is just to see if the statement "A function f has a right inverse if and only if it is surjective" is really true.
It is true. A multi-valued function $f: X \to Y$ is not what is called a function $f: X \to Y$.
it certainly is true
2:34 AM
@Semiclassical For sure.
Howdy @MikeM. Happy inauguration!
happy no more [redacted] day
Thanks. I'm watching John Carpenter movies.
I know him not.
2:36 AM
@Thorgott so $g(x)=\sqrt{x}$ is surgective and has a right inverse $f(x)=x^2$?
The Thing, Halloween, The Fly(?)
The Fly is Cronenberg
They Live is John Carpenter
2:36 AM
Big trouble in little china right now
oh thats a good one
Escape from new york earlier
I still probably plead ignorance. But I probably know more French movies than some of you. :)
@Bohemianrelativist you're not telling me everything there is to know about $g$
John Carpenter isn't exactly high art
but meh
2:38 AM
@Thorgott the domain of g is [0,\infty) and its codomain is $\Re$ (confined to be the positive branch since you say this is necessary to be a function)
Then $g$ is not surjective with that codomain
3:01 AM
@MikeMiller why not? for any positive real number y, you can always find a x in $[0, \infty)$ so that $\sqrt{x}=y$ - it's just $y^2$.
I didn't read the parenthetical at first --- I would have just written the codomain is $[0,\infty)$.
In which case yes, $g$ is surjective (in fact, bijective) and has right inverse $f(x) = x^2$ (with the same domain and codomain, and it is also a left inverse).
I would have written the codomain is $[0,\infinity)$ and not have fixed it, but that is a different story.
@MikeMiller Remember what ol' Jack Burton says...
@Bohemianrelativist The codomain is $\Re$. Is $-1$ in $\Re$?
Say I have a nice top space and two subgroups of it's fundamental group. Say I know the covering spaces corresponding to these subgroups. Is there some topological construction that obtains the covering space corresponding to the intersection from these two?
3:19 AM
@Thorgott covers are X' and X'', if you take X' x_X X'' you get a covering space of X which is connected because X' and X'' are. Now a loop in X lifts to a loop in this iff it lifts to a loop in both factors (right? i didn't check) so I think that does it.
@robjohn give me your best shot, pal...
@MikeMiller It's all in the reflexes.
Oh yeah, that makes sense. The scenario already looks like a pullback diagram, so should've thought of that.
@user76284 I think so. I think the reason the Wikipedia takes the codomain of $g(x)=\sqrt{x}$ to be $\Re$ rather than $[0,\infty)$ is because it considers g to be the right inverse of $f(x)=x^2: \Re\rightarrow[0,\infty)$.
The claim about how loops lift is just the uNiVerSaL pRoPeRTy of the pullback
Now what if I ask the same question for the generated subgroup instead of the intersection? Result should be some mutual quotient, but it doesn't seem as intrinsic
3:38 AM
Seems like... quotient of X' x_X X'' by the relation generated by the two pulled back relations?
@Bohemianrelativist No.
You are confusing the range/image of a function with its codomain.
The codomain is extra information that comes along with the function. It can by anything you want, as long as it contains the image.
I can declare the codomain of $g$ to be $[0, \infty)$, or $(-\infty, \infty)$ or $(-4, -3) \cup [0, \infty)$, or something else.
The image or range of $g$ under the domain $[0, \infty)$ is always $[0, \infty)$.
Forget about inverses. $\sqrt{x}$ is defined to be the unique, non-negative $y$ such that $y^2 = x$.
@robjohn Just got there. Forgot that was a punchline.
Im only trying to picture this in my head, so this is prone to be dead wrong, but the covers corresponding to one free summand in the wedge if two circles are real lines with a circle attached at each integer. It doesn't look to me as if quotienting their fiber product will give the cayley graph of F_2?
Yeah you are right
Seems like you need a more intricate geometric construction
If there is one at all
3:55 AM
On one hand, it seems dubious from the categorical pov, cause the maps "go the wrong way". Otoh, you can describe what corresponds to a generation explicitly in the galois correspondence for galois extensions and that corresponds to covering theory in special cases, so there might be something to say, still.
@user76284 $\Re$ includes $(0,\infty)$, so to define to the right inverse f to g, should we apply $\Re$ or $(0,\infty)$ to f?
4:16 AM
Hello guys.
Pls have a look link in this example-2 I don't understand the solution of equation 3.11.
Pls need your help
I infer the following

f: R →[0, ∞) ∣ f(x)=x² is not injective, so has no left inverse; however, it is surjetive, so has a right inverse g: [0, ∞) → R ∣ g(x)=√x such that f(g(x))=x.

g: [0, ∞) → [0, ∞)∣g(x)=√x is injective, so has a left inverse f: [0, ∞)→ [0, ∞) ∣ f(x)=x², and it is surjective, so has a right inverse f: [0, ∞)→ [0, ∞) ∣ f(x)=x² such that g(f(y))=y.
4:59 AM
Guys I am using frixion pens for my notes
With these pens , you can erase what you have written
but on YouTube and internet I found that if I keep my notes on the car dash or somewhere out where it is hot . My notes will be erased
If you know about these pens , then pls tell me if I should use these pens or not
I am preparing for jee
1 hour later…
6:05 AM
@Physicsismylife i answered a question for you yesterday and then you deleted it!
Oh yes.It happened by mistake. I did save your answer with a screenshot. @copper.hat Thank you for your answer.I will reopen that question.I have some doubts in that I will ask you later. I will first build my concepts more strong and then see if I really have a doubt in that or not.
2 hours later…
8:12 AM
Hello! Is anythone familiar with Bayes decision?
I have a question:
Q: Apply Bayes decision problem to coins-problem

pinguWe choose arbitrarily one of the 12 apparent similar coins. However the coins, in reality, belong to two groups, the 9 are "fair" while the other ones have the probability to appear heads equal to $\frac{6}{9}$. We are making a throw and after having seen the result, we are given the following ch...

must calm the inner troll.. must calm the inner troll..
@robjohn Hi! Do you maybe have an idea to my question stated?
9:13 AM
If I set $f(x) = 0, f \in C[a,b]$ and let $g(x) = x - h(x)f(x)$. How do I define $h(x)$ so that g(x) is lipschitz continuous on [a,b], given that I don't know the behavior of $f$?
well I do know how $f$ behaves to the level it's continuous
9:29 AM
wait a minute I meant contraction, but probably is the same method
Would it work if I say $\vert f \vert \leq M$, and define $h(x) = \frac{x}{2M}$? It seems to work
4 hours later…
On driving laws and the complex numbers
Q: Metamathematics of buts

arsmathSomething I learned (probably in middle school) that always bothered me is that the truth value of "and" and "but" are basically the same. If you were going to assign a truth-functional interpretation of "but" in first-order logic, it would be the same as "and". There's been a explosion of logic...

I had to read the title twice
2:05 PM
read noah schwebers comment
yeah I saw it, epic
But, the sky being cloudy notwithstanding, it wasn't raining :3
But, even though I wrote "The sky was cloudy, but it wasn't raining", "But, the sky being cloudy notwithstanding, it wasn't raining" written by Jim earned more points in the exam.
He's an algebraist and he's a sensible person
But, even though I said "I wrote "The sky was cloudy, but it wasn't raining" but Jim wrote "But, the sky being cloudy notwithstanding, it wasn't raining" and earned more points., "But, even though I wrote "The sky was cloudy, but it wasn't raining", "But, the sky being cloudy notwithstanding, it wasn't raining" written by Jim earned more points in the exam." as written by Jim earned more points in the exam.
@Astyx driving in the UK? just take the complex conjugate of driving in france
2:17 PM
Let $M$ is an orientable surface and $\alpha\in \pi_1(M)$ is non-trivial. Let $f:\Bbb S^1\to M$ be a loop representating $\alpha$.
Let $M_\alpha$ be the covering corresponding to the subgroup $\langle \alpha\rangle$ of $\pi_1(M)$, and $f_\alpha:\Bbb S^1\to M_\alpha$ be the lift of $f$ w.r.t. the covering $p_\alpha:M_\alpha\to M$.
Now, w.r.t. the universal covering $p:\widetilde M\to M$ consider the lift $l:\Bbb R\to \widetilde M$ of the map $\Bbb R\xrightarrow{\exp}\Bbb S^1\xrightarrow{f_\alpha}M_\alpha$. My question is why $l$ is a proper map i.e. inverse image of a compact set is compact?
@User873110 $\widetilde{M} = \widetilde{M_\alpha}$ is the universal cover of $M_\alpha$ as well, which has fundamental group infinite cyclic generated by $\Bbb Z$. This $\Bbb Z$ acts on $\widetilde{M}$ by shifting the preimage of a basepoint on $f_\alpha$ around; this preimage is exactly $\widetilde{\ell}(\Bbb Z)$. Since the deck transformation action is properly discontinuous, this set can't accumulate anywhere.
This actually tells you $\tilde{\ell}$ is proper.
I mean OK I wrote it in a more convoluted way than I should have. The preimage is an infinite discrete set in virtue of being a fiber of a covering map. The point is $\widetilde{\ell}$ is a curve joining all of these points in the fiber, and is $\Bbb Z$-equivariant (deck action) as well.
so you are talking about integer translations in the universal cover
Let $V$ be an $n$ -dimensional vector space over $K$ and $T \in L(V)$ such that $(T-\lambda I)^{n}=0,(\lambda \in K)$ and $(T-\lambda I)^{n-1} \neq 0 .$ Prove that there is a basis $\mathcal{B}$ of $V$ such that
\lambda & 1 & & 0 \\
& \ddots & \ddots & \\
& & \ddots & 1 \\
0 & & & \lambda
I know How to tackle such kind of problems ?
@User873110 Yes, so for $\widetilde{\ell} : \Bbb R \to \widetilde{M_\alpha}$ there's an action of $\Bbb Z$ on the domain (integer translations) and on the range (deck transformation action of $\Bbb Z = \langle \alpha \rangle = \pi_1(M_\alpha)$). The map $\widetilde{\ell}$ is equivariant under these actions.
i.e., $\widetilde{\ell}(n \cdot x) = n \cdot \widetilde{\ell}(x)$.
*want to
2:31 PM
That immediately implies properness, do you see that?
Or should I explain why
wait for a few moment, I will try first.
OK cool
I guess the inverse image of a compact set will be a union of finitely many translates of some compact set of $\Bbb R$, but why I do not know.
@BalarkaSen I guess the inverse image of a compact set will be a union of finitely many translates of some compact set of $\Bbb R$, but why I do not know.
can you explain now?
2:46 PM
For example, you can argue that the compact set will contain only finitely many points of $\widetilde{\ell}(\Bbb Z)$, because that is an infinite discrete set.
Can you generalize this to show that $\widetilde{\ell}$ is in fact proper? Remember properness means $x_n \to \infty$ implies $\widetilde{\ell}(x_n)\to \infty$
This is one of the equivalent ways to state properness
@mathsstudent Write out what this matrix form implies about how $T$ acts on such a basis. These equations will be explicit enough so that you can guess a way of finding such a basis.
@Balarka can polynomial growth with arbitrary exponent be realized as growth rate of balls in some riemannian manifold?
3:02 PM
@Thorgott Let $v \in \mathbb{V}$ be such that $(T-\lambda I)^{n-1} v \neq 0 .$ Then
\left\{w_{1}=(\T-\lambda I)^{n-1} v, w_{2}=(T-\lambda I)^{n-2} v, \ldots, w_{n}=v\right\}
is basis of $V$ since it nilpotent operator and nilpotent operator have basis $v,Tv,T^{2}v,....,T^{n-1}v$
yup, that does the job
I don't understand how I can get matrix representation @Thorgott ? Could you explain me that for atleast one column
What goes wrong if you just plug this basis into the definition?
I got first column as T(v) = lambda v then We can write it as first column lambda and all other entry 0 but after that I got confused ?
@Thorgott I don't understand the question. $\Bbb R^n$ has volume growth rate $n$.
3:15 PM
@mathsstudent IF you don't tell me explicitly what you have and where you're stuck, I can't help you
@Balarka the growth function on $\mathbb{R}^n$ is $r\mapsto Cr^n$, polynomial of exponent $n$
@Thorgott So I have basis as I told ; now I apply individual basis on $T$
@Thorgott yes, so growth rate is $n$
what is the question?
can you realize arbitrary exponents?
non-integer, that is
So I apply T on basis ser and trying to write its coordinate vector
3:20 PM
what? groups have integer exponents
what is a group with growth rate $\pi$ man
So we have T(v) ..
doesn't exist by Gromov
but I don't see how the case of a general manifold would reduce to groups
Fractal groups?
Idk if that's a thing
e.g. if our Riemannian manifold is the universal cover of some compact Riemannian manifold, Schwarz-Milnor implies it's quasiisometric to the fundamental group of the covered manifold, hence the growth rate has integer exponent if it is polynomial at all
T(v) = a_{0} v +a_{1}(T-lambdI)v +......+((T-lambdaI)^n-1v
3:22 PM
what is $v$?
@Thorgott yeah this is what i mean
i didnt realize you were asking for growth rate of riemannian metrics on the Kirby manifold
yeah, but what about Riemannian manifolds that aren't universal covers of compact guys? is there some more general trick?
Basis I have define there
my answer to that is dont know, dont care
$\left\{w_{1}=(T-\lambda I)^{n-1} v, w_{2}=(T-\lambda I)^{n-2} v, \ldots, w_{n}=v\right\}$
3:24 PM
ok, so apply $T$ to an element of this basis and tell me what you get
@Thorgott Can I take T(v)=lambda v here
you're supposed to do an explicit calculation
I never knew that there is only one portrait of Legendre, which is hilarious: en.wikipedia.org/wiki/Adrien-Marie_Legendre
This isn't even his final form
3:42 PM
Can any mathematician here (i.e. has completed at least a bachelor's in pure math) check the accurateness of this answer that I gave? Note that I hold a bachelor's in CS and a master's in AI, that's why I'm asking for someone that knows more than me in terms of pure mathematical formalisms. Preferably, someone that holds a bachelor's or master's in pure math should be able to spot possible improvements or inaccuracies in my answer.
In any case, any feedback is welcome. It's possible that even someone that doesn't hold any degree is able to spot some mistakes or possible improvements, so feel free to provide any feedback.
Is there an obvious argument for why sets a positive distance apart in a metric space can be separated by uniformly continuous real valued functions?
@nbro It seems mostly correct, but probably using an example throughout your explanation (such as the coin toss that is confusing the author of the question) would enhance the clarity.
Can you take Tietze for granted?
@MikeMiller sure
$f_A(x) = \text{dist}(x, A)$ and $f_B(x) = \text{dist}(x, B)$ ought to be uniformly continuous, right?
Now consider a function $g_A(x)$ which agrees with $f_A$ on $B$ but is uniformly positive
And consider $h_A(x) = f_A(x)/g_A(x)$
That seems to vanish on $A$ while being 1 on B
Something like this ought to work, maybe?
3:55 PM
what do you mean with uniformly positive?
Bounded below by a strictly positive constant
ah ok
Seems reasonable, I'll think about the details
I think something like this should work yeah
4:11 PM
@Krijn My answer is already quite long and I tried to give a general answer that addresses most of the misconceptions and misunderstandings. I really would love to see other answers that specifically address the examples in the OP's post. I will certainly upvote them if I think they are correct, so, please, feel free to write an answer there :)
I could actually give another answer that addresses more specifically the questions, but I would really like to see another person's perspective
That's fair
4:43 PM
@Thorgott please help I am not able to solve it yet
write down the expression you have to calculate
Just tell me is $T(T-\lambda I)(v)=T(T(v)-\lambda v)=T^{2}(v)-\lambda T(v)$ true
@mathsstudent true
Ok so suppose I restrict $n=2$ then $(T-\lambda I)^{2}(v)=0$
Now we have basis ; $\{v,(T-\lambda I) v\}$
Now $T(v)=a_{0} v+a_{1}(T-\lambda I)_{v}=a_{0} v+a_{1} T(v)-a_{1} \lambda v$
So $a_{0}=\lambda$ and $a_1=1$
Is it correct ?
4:54 PM
assuming $(T-I\lambda)v\ne0$
Ya it is given it is non-zero
that's indeed how it works, the general case works analogously
Ok actually I am doing some silly mistake while calculation so I got second coordinate incoorect ;
Thanks @Thorgott @robjohn
Q: How do I learn the value function for a POMDP with a single-step horizon (bandit)?

jdizzleConsider a single-step POMDP, where an agent receives a belief state $b(s)$ over some fixed set of contexts $s$, selects a single action, and receives a single reward, before the episode ends. I understand that the belief state value function, $V(b)$, is piecewise linear and convex, with a single...

Maybe someone here is interested in this stuff
5:21 PM
Hi guys. How $\frac{dg}{dt} = q\cdot\nabla f(p+tq)$ where $f\in \Bbb R^n\to \Bbb R$ and $g(t)= f(p+tq)$?
chain rule?
I know it. but is it really dot product appear here?
my try is this: $\nabla f(p+tq) \circ q$ and ...
If you explicit what the dot product is, you get precisely the chain rule
ie $\sum q_i {\partial/\partial x_i} f(p+tq)$
I know, but how it is related to composition operator?
So say you have a field $F = \Pi_{p \in P} \mathbb{F}_p /U$ where $U$ is a nonprincipal ultrafilter over $P$, $P$ being the set of prime numbers. $F$ is of characteristic $0$. Does it have any definable (without parameters) proper subfields?
I feel like this might be obvious but I am missing it
furthermore, is it obvious that $x^n - m = 0$ has no solutions in $\Bbb F_p$ for infinitely many $p$, for all $n,m \in \Bbb N$?
5:32 PM
Small applause for myself for just finding a nice mistake I made and fixing it
$n,m \geq 2$ I guess
@BigSocks I have no idea what you are doing, but doesn't every field of char 0 have Q as a proper subfield
right well maybe I should add other than $\Bbb Q$
I see no reason why $\Bbb Q$ should be definable
What does that first product notation mean?
5:36 PM
($F$ is isomorphic to $\Bbb C$ btw)
(although it would also be interesting to know how $\Bbb Q$ would be definable there)
What does definable mean?
@AlessandroCodenotti I've heard $\mathbb{C}$ has quite some subfields
@AlessandroCodenotti How do I see this? I thought you needed to take the product over the closures of finite fields... none of them are closed so idk how it is $\Bbb C$
Ah right sorry, I hallucinated a closure
@BigSocks because of this $F$ is not algebraically closed in fact
5:39 PM
@Astyx a set is definable if you have a relation on the domain that is satisfied for precisely the elements in the set
so they satisfy some first order formula
you haven't specified a language but I assume it's the usual $+,\cdot,0,1$
@AlessandroCodenotti right well that makes sense
@AlessandroCodenotti yea
@Astyx It's an ultraproduct
oh yeah forgot to answer that
$F$ isn't just $\Bbb Q$ is it? that would be really weird right?
like intuitively it feels like it shouldn't be true
but if it has no proper subfields and is of char $0$, maybe? I feel like I am missing something obvious
no, $F$ is uncountable
(it has cardinality continuum to be precise)
depending on the $U$ you pick you can get nonisomorphic ultraproducts here, so you there's no hope for $F$ to be isomorphic to something nice
5:46 PM
aha! remind me why this is true though
right, I should have said "elementarily equivalent to"
Even up to elementary equivalence $F$ can have different properties depending on $U$
@BigSocks That's a general fact, any ultraproduct of finite but arbitrarily big structures over a countable index set has cardinality continuum
right like it isn't, but that should have been my question anyway
finite but arbitrarily big I think is the catch...
thanks for the help. had to review facts
The point is that you can find continuum many functions $\omega\to\omega$ that are all bounded by some exponential $n\mapsto 2^n$ and such that any two such functions agree on finitely many points
Then you use them to build an uncountable family of sequences that are not equivalent mod $U$
hmm sounds like a standard proof- where can I find this?:
to see that the elementary equivalence class of $F$ depends on $U$ just look at some polynomial that has a root modulo finitely many primes and doesn't modulo infinitely many others (like $x^2+1$ that has roots or not depending on whether $p$ is $1$ or $3$ mod $4$)
@BigSocks If I remember correctly I read about it in "Models and Ultraproducts" by Bell and Slomson, but it's probably in many books
5:54 PM
lmao that book is right on my desk. guess it's time to open it up huh
math.stackexchange.com/questions/1417688/… ah nice, someone wrote it in stackexchange as well if you want
thanks, @AlessandroCodenotti
@AlessandroCodenotti Are there polynomials for which this does not hold?
What counts as trivial?

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