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1:07 AM
@onepotatotwopotato but is a maximal ideal $\mathfrak{m}$ unique?
existenece is quite obvious
1:22 AM
Suppose $k$ is an algebraically closed field, $\mathfrak m$ is a maximal ideal of $k[x_1,\dots,x_n]$, and $\phi:k \to k[x_1,\dots,x_n]/\mathfrak m$ is an isomorphism. How come $x_i \equiv \phi^{-1}(x_i \mod \mathfrak m) \mod \mathfrak m$?
So, as always, whatโ€™s an example?
i'll try working over $\mathbb C$ and see what I can get
Sorta irrelephant, but sure.
ah wait, $k[x]/(x-a)\simeq k$ for any $a$ and $(x-a)\neq (x-b)$ for $a\neq b$.
uniqueness is false
oh $k$ is not $k[x]$-module...
1:46 AM
Why not?
hmm... you mean evaluation.
No, I mean letting $x$ act trivially.
that's evaluating at $1$ isn't it?
No? At $0$?
trivially you mean $x\cdot a = a$ or $ = 0$?
1:52 AM
I guess you could do yours, too.
but anyway my example doesn't work.
once I fix evaluation $0$ then $\Bbb C[x]\to \Bbb C$ by $f(x)\mapsto f(a)$ is not $\Bbb C[x]$-linear map
Because $\mathfrak{m}\neq\mathfrak{n}$ does not imply $R/\mathfrak{m}\neq R/\mathfrak{n}$, I doubt such $\mathfrak{m}$ is unique (also nobody in the internet says uniqueness). there should be a counterexample...
2:19 AM
ted: duck crash landed in the pool today while we were in it. i think we're being stalked.
i think i'm getting closer. i'm trying to study the specific case of $\phi: \mathbb C \to \mathbb C/\langle x-7,y-i\rangle$, and i'm trying to see whether, if $\phi$ is an isomorphism, it should map $x$ to $7$ and $y$ to $i$
2:32 AM
edit: flash of insight
@leslietownes And appropriately so!
2:48 AM
@shintuku Does this make sense?
no it doesn't, that statement has the elements in the wrong domain/codomain
for the specific case of an isomorphism $\phi:\mathbb C \to \mathbb C /\langle x-7, y-i\rangle$, we have $x - \phi^{-1}(x+\mathfrak m) = x - \phi^{-1}(7+\mathfrak m)$, i think if i properly understand what this implies i'll get the general case
3:12 AM
i get it: the isomorphism $\phi$ must be the composition of the two canonical maps. so $\phi:k \to k[x_1,\dots, x_n] \to k[x_1, \dots, x_n]/\mathfrak m$. now since $\phi$ is surjective, there is an element $c_1 \in k$ such that $\phi(c_1) = x_1 + \mathfrak m$. But furthermore, by our definition, we have that $\phi(c_1) = c_1 + \mathfrak m$. therefore, $\phi(c_1) = x_1 + \mathfrak m = c_1 + \mathfrak m$, so $x_1 - c_1 \in \mathfrak m$, that is, $x_1 - \phi^{-1}(x_1+\mathfrak m) \in \mathfrak m$
thus establishing the nullstellensatz ๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž๐Ÿ˜Ž
4:12 AM
i have just read a piece of bourbaki. i am now a participant in the Mathematical Canon
4:37 AM
i'm not sure that gives a surjection after some thought
it gives a surjection if you presuppose the nullstellensatz but that's what i'm trying to prove
at ted: any thoughts?
2 hours later…
6:51 AM
interesting: $R$-module category and $M_n(R)$-module category are equivalent
Is Lebesgue sigma algebra generated by any topology on $\Bbb{R}$?
Obviously such topology (if exists) would be strictly finer than the usual topology.
@shintuku some more twin prime crackpottery by me:
7:37 AM
How can I get an isomorphic $Q\otimes_{M_n(R)}P\simeq R$ as an $(R,R)$-bimodule where $Q$ is a set of row vectors and $P$ is a set of column vectors. It should be $a\otimes b\to ab$ but I can't find the inverse of this map. $R$ is a unital ring.
8:01 AM
Ah it's doable.
I asked my friend for some measure theory questions. He couldn't remember it exactly but the question has something to do with $\mu<<\lambda$ and $\lambda<<\mu$ then something something happens.
Have you seen any such questions?
it's a notion of absolute continuity
The definition of *hyperbolic* metric space is:
The geodesic metric space $X$ is hyperbolic if there exists $\delta \geqslant 0$ so that for any geodesic triangle $[x,y]\cup [x,z]\cup[z,y]$ and any $p \in[x,y]$ there exists some $q\in [x,z]\cup[z,y]$ with $d(p,q)\leqslant \delta$. Such $\delta$ is called hyperbolicity constant of $X$.

I want to prove that $\Bbb R^2$ is not hyperbolic. For this I negated the quantifiers in definition. Is this correct? I have:

$\Bbb R^2$ is not hyperbolic if for any $\delta \geq 0$ there exists geod. triangle $[x,y]\cup [x,z]\cup[z,y]$ or there exists $p\i
@onepotatotwopotato I know. But I am asking if anyone remember any questions along those lines.
I recently read something about the Hausdorff measure and its dimension. It's surprisingly interesting despite it's a measure theory.
8:07 AM
I guess we have to take $\mu,\lambda$ positive. By RNT, $d\mu=hd\lambda$ and $d\lambda=gd\mu$. So $h,g$ have to be non negative. and $d\mu=hgd\mu$. So $\int_E hg-1d\mu=0$ for all $E$.
So $hg=1$ a.e. .
But can we conclude something about $\mu,\lambda$ here?
Got it. I think he was talking about this: math.stackexchange.com/questions/3260420/…
8:21 AM
This is just a silly question emerged from ignorance, but can double sharps and double flats appear in musical isomorphisms?
@DannyuNDos What are the domains and codomains of these functions?
If X is a metrie space and A closed in X, show that a map $f:A\to S^n$ can
always be extended over a neighbourhood of A.
This means that we have to show that there exists an open set U in X containing A such that f can be extended (continuously) to U.
Is my understanding correct? Thanks.
Or, do we have to show for every neighborhood of A?
@s.harp I mean, do the symbols ๐„ช and ๐„ซ appear in any occasions?
@Koro "extended over a neighborhood of A" โ€“ that marks existence.
@DannyuNDos Morally: No, the domain of sharp differs from its codomain, so you cant apply it twice. Same for flat. Technically: It could, if you have two different sharps you might as well compose them. Practically: Never.
@s.harp Thx!
8:50 AM
@Koro See here
Dang it, ninja'd!
I knew I gotta make use of Tiezte, but got confused with the retraction part.
@DannyuNDos thanks :-).
@DannyuNDos I think you are right that only existence has been asked. I just wanted to confirm it anyways.
With existence, $U=\tilde f^{-1}(R^{n+1}-\{0\})$ works.
$\tilde f$ is a (Tietze) extension of f.
Can anyone explain why $\Bbb R^2$ is not hyperbolic?
specifically, $\tilde f$ is obtained by Tietze on every $f_j$ (jth coordinate map of f).
Yes. We can apply Titze's extension theorem for component functions
9:03 AM
I suspect that it is not necessary for f to continuously extend to all of X.
of course, it is not necessary.
9:36 AM
@SouravGhosh I am extremely thankful for this insight, Sir! :-)
9:49 AM
@Koro Tietze extension would be overkill here, and in any case, it would extend it to all of X
whereas the question is only asking about a neighbourhood
so its clear that they want you to use specific properties unique to metric spaces (not normal spaces) to furnish the extension
you can use tietze, but then youre kind of missing the point of the exercise imo
Is Lebesgue sigma algebra generated by any topology on $\Bbb{R}$?
Obviously such topology (if exists) would be strictly finer than the usual topology.
@SouravGhosh you should be able to construct such a topology, based on the fact that a lebesgue measurable set looks like $B \cup N$, where $B$ is a borel set and $N$ has lebesgue measure zero
so replace $B$ with open subsets (in the euclidean topology), clearly the result is a topology on $\mathbb{R}$
and clearly this generates the lebesgue measurable sets
but this is definitely finer than the euclidean topology
also, you can play the same game with any completions of borel sigma algebras
this is not unique to $\mathbb{R}$
as for the properties of this topology, idk, but its at least hausdorff because it contains all euclidean opens
10:11 AM
@porridgemathematics it won't, in case of S^n in codomain.
@porridgemathematics in my case, Tietze's been proven for metric spaces using the fact that two disjoint closed sets can be separated by a continuous map.
but the later statement is otherwise known as Urysohn's lemma... For metric spaces though its proof doesn't use the full generality of normal spaces.
oh okay, i missed the S^n part
yeah indeed you wont be able to then, and if tietze was only proven for metric spaces then that proof is essentially what I thought they wanted you to reproduce to show this for a neighbourhood
you could extend it to all of $X$ if you had partitions of unity, but for metric spaces you wont necessarily always have those
using basically the whitney theorem
(well the proof of the whitney theorem)
10:27 AM
@porridgemathematics this is not closed under arbitrary unions
@Jakobian less than a measure zero set then, so $U \setminus N$, where $U$ is euclidean open
this should still generate the lebesgue measurable sets but wont suffer from the previous issue
oh wait
ok maybe this doesnt work either
i thought subtracting would fix the issue, because I was erroniously just taking arbitrary unions of sets of the form $U \setminus N_{\alpha}$
where $U$ stays fixed..
but im not sure how to handle the general case
(so I just get $U \setminus \cap_{\alpha} N_{\alpha}$ and im subtracting another null set)
oh wait this could work but i retract my previous statement about it being generalizable to arbitrary completions
but it will work for all second countable topological spaces
basically we can just reduce to the case of $U$ being the same with different null set subtractions, because we only ever take countable unions when our underlying topology is second countable
so in $\bigcup_{\alpha} U_{\alpha} \setminus N_{\alpha}$, we let $O_{i}$ , $i=1,2,3,...$ be a countable basis, so that our union looks like $\bigcup_{j \in J} O_{i} \setminus N_{i}$ from the previous step, where $J \subset \mathbb{N}$ is some subset
so what I mean by that is rewrite each $U_{\alpha}$ , $\alpha \in A$, as a countable union over the $O_{i}$, then for every $i$, we are taking an arbitrary union $\bigcup_{\alpha \in K_{i}} O_{i} \setminus N_{\alpha}$, for some subset $K_{i} \subset A$
and this arbitrary union is of the form $O_{i} \setminus N_{i}$, where $N_{i} = \bigcap_{\alpha \in K_{i}} N_{\alpha}$
so this works for every second countable topological space, i.e. the completion of the borel sigma algebra for such a space is realizable as the borel sigma algebra w.r.t a finer topology
and yeah my previous effort was nonsense, because it would just generate the discrete topology..
11:10 AM
Finally I have managed to show that $\int_{0}^{\infty}\frac{\sin x}{x} dx=\frac{\pi}{2}$
11:20 AM
@porridgemathematics there is a paper by S. Scheinberg, "Topologies which generate a complete measure algebra" which discusses different topologies on $\mathbb{R}$ which generate the Lebesgue measurable sets as Borel sigma-algebra. This is one of them, in the paper it's called $T'$-topology
@Jakobian thanks for the reference, ill take a look!
@SouravGhosh how did you do it? i can think of one way via a clever contour integral trick (that i just remember and probably wouldnt be able to reproduce without memory of it), and another via a fourier transform
Contour integral.
thats using e^{iz}/z and then a semi-circular contour with a small bump at zero, right?
plus the residue theorem
ah okay, yeah i guess its not necessarily a trick, theres another way via contour integration that avoids the bump, but requires modifying the numerator to something else
11:34 AM
oh yeah, subtract one
so there is a removable sing at the origin
and then just use the residue theorem and take imaginary parts
$f(x)=\frac{\sin x}{x}$ is not Lebesgue integrable on (0, \infty) $
11:55 AM
modular representation theory gave me a headache
2 hours later…
2:10 PM
Good morning Professor @TedShifrin
2 hours later…
4:28 PM
anyone see how this inductive argument works ?
Do it by hand for $k=1$ and $k=2$
4:44 PM
@jay you can perform the swap (y_0,...,y_{k-1}) -> (y_{k-1},...,y_0) by the inductive hypothesis
then use tonellis theorem to move the \mu(dy_{k-1}) to sit next to the p_{t_{k} - t_{k-1}}(y_{k-1},dy_k), and use the k=1 case to transpose (y_{k-1},y_k) -> (y_k,y_{k-1})
you are given the k=1 case because that is the same as saying p_{t}(x,dy) \mu(dx) is symmetric in (x,y)
finally use tonelli again to move the \mu(dy_{k}) back to the right so its integrated against last
5:23 PM
Hello, Ted! What is the standard proof of boundness of the image of uniformly continuous real function on bounded set $E\subseteq R$?
I have constructed two proofs so far, first was with Heine-Borel and continuous extension, but recently after your notes on my writing I found it messy (and I didn't bother to prove continuity), so at first instead of doing it, I tried to construct a cleaner argument, so the second deals with looking at the preimage and using Heine-Borel, too. Much cleaner and shorter.
i use uniform continuous implies continuous implies bounded
trying to figure out a pithy way to say and generalize the following
shin, but continuity does not imply boundness. 1/x is continuous.
continuous over interval my bad
But we can't guess if the interval is closed.
5:27 PM
theorem is only true for closed interval, no?
Theorem I mentioned? No, it is true for every bounded set, which might be open interval (a,b).
suppose i have two vectors $a,b$ in R^3 which i project onto the xy-plane. then the scalar product of these vectors before/after projection are $a_x b_x+a_y b_y+a_z b_z$ and $a_x b_x+a_y b_y$, i.e., they differ only by the product $a_z b_z$
it's basically just splitting up $\sum_k a_k b_k$ in some way, so it's not very interesting. but i feel like i'm forgetting a name for that fact
i dunno if it has a name beyond, like, orthogonality? if a hilbert space is an orthogonal sum of two subspaces, the inner product is the sum of the inner products in the subspaces. it sometimes 'looks more interesting' if they aren't coordinate subspaces but i don't know if it has a name.
maybe your people have a name for that :)
@noballpointpen Let $E\subset \Bbb R$ bounded then cl(E) is compact. Since $f$ continuous $f(cl(E)) $ is also compact. Hence $f(cl(E)) $ is bounded. Implies $f(E) $ is bounded.
We need only continuity of $f$ .
5:46 PM
@SouravGhosh you need uniform continuity to be able to extend to cl(E), otherwise continuity will get you things like tan on (0,pi/2), which does not have bounded image
@SouravGhosh the problem is that $f$ can miss limit points of $E$, so that your $f(cl(E))$ is meaningless.
More precisely, $E$ can miss its limit points, so that $f$ is not defined on them. We cannot just use $f(cl(E))$.
@noballpointpen I dont think you need anything about continuous extensions: Consider $f(E)$, and suppose there are points $f(x_n) \in f(E)$ such that $|f(x_n) - f(x_{n-1})| \geq n$. Let $x_{n_k}$ be a subsequence converging to $x \in \text{cl}(E)$, since $f$ is UC, $f(x_{n_k})$ is cauchy , and thus convergent. Contradiction.
all I used was heine-borel
and uniform continuity
and it was basically two sentences
anyone know where i can find exercises on k-algebras
just remember UC -> cauchy subseq for image sequences
then you dont need to worry about extending anything
Why such sequence would exist?
5:54 PM
the whole point is $f(x_n)$ as I have defined doesnt exist..
if they did exist, $f(E)$ would be unbounded
since they dont, $f(E)$ is bounded.
thats why I end the proof with the phrase contradiction
Now I see.
Didn't comprehend such a definition deals with assuming unboundness.
But it still seems suspicious. You define it by a growing gap between terms.
@porridgemathematics For my proof $f:\Bbb R\to \Bbb R$
At first glance it seems to me that we could very well have a sequence in $f$ without a growing gap but which still shows our assumption about unboundness.
So it is not at all clear to me.
Am I right that you defined it basing on assumption about growing gap? Maybe I missed your idea.
Here is how I've done it in my second proof
Let $X$ be any infinite subset of $\overline{f(E)}$. Look at the preimage of $X\cap f(E)$. It is an infinite subset of $E$, which is bounded, so it has a limit point $e$ in $\overline{E}$. Applying $f$ to every Cauchy sequence in the preimage converging to $e$ we have it converging to a point in $\overline{f(E)}$. So $X$ has a limit point in $\overline{f(E)}$, so the set is compact, and since $f(E)\subseteq\overline{f(E)}$, we are done.
Overline is closure in Rudin's notation, btw.
@noballpointpen What is your question? What is the domain and codomain of $f$?
Initially I asked Ted what is a standard approach. I am looking for a cleaner and accepted way. Just to look how it could be done by expert.
Just scroll up. My question is not so far ahead.
6:07 PM
@noballpointpen what does not bounded mean?
For any given $M>0$, existence of $x$ for which $|f(x)| > M$.
Let $E\subset \Bbb R$ bounded and $f:E\to\Bbb{R}$ is uniformly continuous. Then $f(E) $ is bounded.
I am not looking for help proving it, notice.
Yes, this is the statement, Sourav.
Good, now let M vary over the integers. Fix a point z in your unbounded set, and define the following sequence: x_1 is z, x_2 is a point outside the disk of radius 2 center 0, x_3 is a point outside the disk of radius 3+ |x_2| , and so on. For the converse, assume your set had a sequence with a growing gap but was bounded. Then the growing cap sequence has a convergent subsequence, which is absurd because such a subsequence is not cauchy.
That proves unboundedness coincides with โ€œhas a sequence with a growing gapโ€
So now you should be comfortable with the growing gap sequence I defined as a standin for being unbounded
I think I got it how we could have such a sequence. Thanks for clarification.
6:13 PM
Yeah but you wanted to know about the converse
why having such a sequence implies unboundedness
if you werent then i have no idea what you mean by its not clear to you
I already told you that now I appreciated how we could construct such a sequence in assumption of unboundness.
Anyway, I am going to leave right now. Thank you, porridge and Sourav, for the talk! :)
sure but that wont be enough for the validity of my proof is all, since im actually assuming the existence of such a sequence implies unboundednessโ€ฆ
also what I gave you is a โ€œstandard approachโ€ justf fyi
@porridgemathematics Good. Then it is all done with. I am just saying you that such a sequence is tied with unboundness.
great, honestly the converse implication isnt super obvious on paper
its obvious by virtue of heine borel or visualizing a growing gap sequence
so you were right to question it
anyway night
6:44 PM
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@AlessandroCodenotti I found a cute proof that discrete space of size continuum is realcompact, you might be interested
biject it with $[0, 1]$, and using that any ultrafilter on $[0, 1]$ converges, take a local base $U_n$ of its limit $x$. If the ultrafilter has countable intersection property, then $\{x\}$ belongs to it, so it's principal.
So you just use that $[0, 1]$ is compact and first countable
@leslietownes yeah. it's just one of those features that makes orthogonality such a powerful tool
someone on main was asking for an explanation of $E[(X-EX)(Y-EY)]=E[XY]-E[X]\cdot E[Y]$. if you think of random variables as vectors with inner product $E[XY]$, then $E[XY]=E[(X-EX)(Y-EY)]+E[X]E[Y]$ is that principle in action
all first countable compact Hausdorff spaces have at most size continuum, so we can't prove anything about bigger discrete spaces this way
7:12 PM
@Jakobian The wiki page says something is realcompact if and only if its homeomorphic to some closed subset of $R^X$ for some $X$. If you have a discrete space $D$ isn't it hoemeomorphic to the "axis vectors" $\{ \delta_k \mid k \in D\}\subset R^D$?
7:43 PM
why are you saying this
it's not like I am looking for a proof of this, or using that definition
besides it's kind of not entirely true, since you need to add the zero vector
otherwise it won't be closed
8:05 PM
To porridge: actually I thought about having only a sequence which grows with a constant gap, not thinking that we could just use its subsequence to arrive at the sequence in question :) Everything else is ok and understandable.
@Jakobian cause I've never heard of this "realcompact" and looked it up
8:44 PM
@porridge, also I am not so sure why you said "wont be enough for the validity". We have a bidirectional relation: such a sequence exists iff the set is unbounded. We can either assume $f(E)$ is not bounded and have such a sequence or just assume existence of such a sequence, as you did, which is less wordy, of course; both lead to the same contradiction.
9:22 PM
Oh, it's just your wording "assuming". You could add "which is of course true", then, to make your sentence less confusing. Actually in your proof it is the implication: unboundness $\implies$ exists such sequence. Since we arrive at the condition of non-existence, this implies boundness, by contrapositive.
9:34 PM
Hello everyone. Am I right that \ddot{x} = t^3/x^3 is equivalent to the system (d/dt)(x_1, x_2) = (x_2, t^3/x_1)? My advisor says that something is wrong in this
10:12 PM
Did you forget $x_1^3$?
But otherwise this is standard. Please use dollar signs so your ChatJax will display.

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