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12:00 AM
@TedShifrin So i was about to ask about this
 
So what does that mean? He wasn't sure it converged?
Notice I didn't get anywhere close to division, @EE18.
 
Zorich doesn't comment much on it but after we've here constructed a bijection between the reals (again, axiomatically defined but presumably arrived at somehow) and this restricted subset of decimal sequences, if one wanted to say these decimal sequences WERE the reals then we'd have to demonstrate a field homomorphism between the two sets?
 
multiplication with cuts is hilarious too, rudin himself gives up with it and says some of the cases are the same as the others
 
forgive me I don't know if that's the correct term
 
@TedShifrin yes
 
12:02 AM
@EE18 field isomorphism ... probably ordered-isomorphism.
 
ah true, though we've already got the bijection bit but that's well taken
OK, will note that down
 
I respectfully suggest adopting something like the LUB axiom for the reals and moving on to actual analysis. If you want to come back to a construction of the reals later, do it later.
The rabbit holes are not particularly instructive.
 
@leslietownes don't make fun of my Dedekind cuts
 
my interest here is actually somewhat piqued (even given my starred comment in the sidebar) since my background is roughly related to computers and these representations are important therein
in fact i've got a little digital arithmetic book going on the side too right now and it seems that much of what it's taking for granted is spelled out here
nice to see the dots connect
 
I wish the Supreme Court of the US were as easily ignored as my suggestions.
 
12:06 AM
@TedShifrin i forget which exercise it related to, but it was something like, we had sum a_n where a_n was given by a recursive formula which guaranteed it fast enough decay for the thing to converge. they weren't sure if it converged, and inequalities did not convince them, and they tried to discuss it in terms of decimal digits (which "surprisingly" weren't easy to analyze) and got lost
 
Giving up on Halmos was surely listening to your suggestions!
 
I assume this was a postdoc, leslie?
 
ted: yes
 
Not that I, as a tenured professor, didn't occasionally get lost and mess up ... I'm sure. :)
 
i thought about joking and saying it was charles pugh
no it was someone who was clearly talented at whatever they did, but whatever they did had nothing to do with analysis, and they simply should not have been asked to teach analysis
probably a staffing shortage or people just not caring
a last minute replacement when the 'real' professor decided to stay in the riviera
 
12:08 AM
I remember when I was in the middle of teaching my multivariable math course as the cancer was getting worse and worse (ultimately, I got so anemic I couldn't stand up and had to spend a week in the hospital ... pre-surgery) ... and my brain just could not work, and I did get confused trying to do an $LDL^\top$ decomposition of a symmetric matrix at the board. The kids were truly very upset seeing me ...
I actually never taught the actual real analysis course during all my years as postdoc and professor. The Spivak course and the multivariable math course were a hell of a lot more fun to teach.
Of course, one of my grumpy colleagues did complain when I taught the graduate complex analysis course several times that I was not qualified to do so (he was an operator algebraist, of course). Operator algebras are allowed, but not complex geometry, apparently.
 
i would have taught rudin chapter 0 as a semester length course
week 10: case 6 of the associative law of multiplication of cuts
 
shoots leslie on 5th Avenue and no one cares
I suppose I shouldn't joke like that. :(
 
@TedShifrin this is kind of funny, at a lot of places grad complex analysis is a kind of 'stepchild' class that nobody wants
odd to think of someone being possessive about it
 
He wasn't all that possessive. He just didn't think I could do a competent job. This was a guy who — whatever he taught — copied the textbook onto his paper and then transcribed the paper onto the board. Award-winning teaching.
 
he was one of my professors too
well... he reminds me of one anyway
 
12:14 AM
m. riesz
or did he have his assistant do the copying, and was it his own textbook
 
That is far from a uniqueness result.
 
i forget the story
a tiny bit of early 20th century central european methods in 21st century USA
 
No, 20th century.
 
12:26 AM
Did any of you ever use or read Hoffman and Kunze?
Just finished Chapter 1 (again, had to stop a while back). One thing which confuses me is the constant reference to what seems vaguely defined as a system of equations
there's a constant underlying isomorphism between said systems and augmented matrices, but the systems are just vaguely defined enough that i can't seem to make this all precise. what is a system of equations?
 
linear equations?
 
ya exactly
 
if a book does this right, they are referring to a list of linear equations, and a specific way of writing them ('variables' specified, including an order of them, and 'constants' on one side)
 
will be a hard question if haven't read it as it's very much a book-related question i imagine
 
books that take this approach often aren't as interested in "sounding formal," but they often are quite formal
 
12:28 AM
you got to specify you mean linear equations, not all systems are linear
 
you just have to skim past the kind of musty 19th century language
 
Here for example is the definition (and Jakobian you're absolutely right re: linear, as you can see from the picture)
 
$$\begin{cases} a_{1, 1}x_1+...+a_{1, n}x_n = b_1 \\ \cdots \\ a_{m, 1}x_1 + ... + a_{m, n}x_n = b_m\end{cases}$$
 
2 mins ago, by leslie townes
if a book does this right, they are referring to a list of linear equations, and a specific way of writing them ('variables' specified, including an order of them, and 'constants' on one side)
 
does this count as proper then? :)
 
12:30 AM
there's nothing 'vaguely defined' about any of that
 
@EE18 they defined it right there...
 
the focus on systems of equations is kind of old school, but it's not vague
 
the elements $a_{i, j}$ are in a field and so are $b_i$
and you are looking for solutions in $x_1, ..., x_n$
 
But I'm not sure I understand what a "system of linear equations" is. Is it a function, a relation, or what sort of structure?
 
@leslietownes linear
 
12:32 AM
one sec, will quickly look for the tacit isomorphism i mentioned
 
EE18: under that definition, a system of linear equations is a list of linear equations of that form
 
@EE18 a list of propositions with free variables
 
different lists being different systems [potentially having the same solution set or whatever, but different systems], systems aren't point sets or functions or anything like that
in subsequent pages there's probably all sorts of stuff about point sets and functions that you can associate with a 'system of equations,' but the 'system of equations' itself is just a list of equations
 
those propositions being equations (of course)
 
again, i'm not saying this isn't a clunky and old school thing to put in the foreground on page one, or whatever, but it's not vague
good 19th century vagueness would be something like, "a linear system of equations is a system that can be expressed in the form [blah]," which maybe leaves a whole lot of wiggle room about what might not be in that form, but could be "expressed" in that form
 
12:35 AM
I wouldn't say its old-school
its what we think of when writing matrix equation of the form $Ax = y$
 
good old vagueness is, like: a function is called continuous if a minor change in its argument results in a minor change of its value
 
its very fundamental to our thinking
I'd say its eternal if anything
 
@Thorgott even better :)
 
but this is really weird to me
@EE18 don't lie that you don't know what systems of linear equations are, you did them in high school
 
Of course I know, but what I mean is that I know in the same way that in high shcool I knew that a function took an input to an output, not that a function is a particular type of relation which in turn is a specific type of set
So I'm still struggling to find an example of what I mean, but hopefully this theorem is OK as a proxy
Systems were defined as equivalent if one could obtain the other (and vice versa) by forming linear combinations of the former
 
12:39 AM
whats the question
 
This theorem then claims that if two matrices are row equivalent then the corresponding homogeneous systems that they represent are equivalent
Obviously, there is some tacit isomorphism between the systems defined above and these matrices, and that isomorphism "preserves" linear combinations when in the "system view" and row operations when in the "matrix view"
so all of this is basically me trying to make the above precise. how can i make this notion of isomorphism precise when "system" is sort of vague to me as a mathematical structure. With matrix there is no problem: it's a function from $\{1,...,m\} \times \{1,...,n\}$ to $\mathbb{F}$
 
this is too much even for me
surely you can use "transport of structure"
why would anyone formalize this is beyond me
you won't call this isomorphism unless there is structure to preserve
 
EE18: a system in the above definition (list of equations) is, in terms of the 'data' being specified, "just" (i.e., optionally thought of as just) an ordered list of coefficients and right hand sides
which could, for example, be rows of an augmented matrix, or not
 
I guess it does though, right? If $AX = 0$ represents the homogeneous equation (as in my first screenshot) and $M(A)$ is the matrix corresponding (via said isomorphism, but I'll call it a bijection for now) to that system, then if I modify $AX = 0$ to an equivalent system $A'X = 0$ obtained from, say, multiplying one equation by $c$, then $M(A') = e(M(A))$ where $e$ is the elementary row operation of multiplying the corresponding row by $c$
surely that's structure-preserving?
Got it. OK, I will digest that and, to Ted's point, not worry too much more. But this was all sort of messing with me as I read Chapter 1 so I figured I'd ask. Appreciate everyone's time as always!
 
i'm not 100% sure of the desire that is failing to be met here. would you insist upon something like "x_1 + 2 x_2 + 3 x_3 = 4" to have a function that goes with it? if we encode this as "(1,2,3,4)" is this less "vague," and if so, why?
you're comfortable swapping or scaling rows of a matrix, but not swapping or scaling individual equations in a list?
 
12:51 AM
I guess the desire is that everything else in the book is defined as some thing from math, whereas HK call a system of equations a "problem"
 
@Jakobian Wow!
 
you might think, what is your level of comfort with "a matrix is a rectangular box of scalars" as a definition, is it supposed to be some set theoretic thing and if so what
 
it's a function from $\{1,...,m\} \times \{1,...,n\}$ to $\mathbb{F}$ :)
 
is there a difference between the box of numbers [[1,2,3],[4,5,6]] representing the matrix "A" in Ax = b, or representing both A and b in an "augmented matrix"
if you insist on a box of numbers only having one role (e.g. apparently as a function) i guess i understand why this feels confusing
 
@leslietownes I suppose no, insofar as given one i can always obtain the other
Removed
(assuming a solution exists)
To summarize then, I have here that if $[A b]$ is row equivalent to $[C d]$, then $AX = b$ and $CX=d$ (where this latter statement is considered as an equality in $\mathbb{F}^n$).
 
2:01 AM
then Ax = b and Cx = d are [equivalent? feels like a word is missing there]? you're missing the "statement" that you're considering as an equality in F^n
if the number box [A | b] is row equivalent to the number box [C d] (where A, C mxn and b, d mx1 column vectors), then {x in F^n: Ax = b} = {x in F^n: Cx = d}, where the = in the middle of all of that is literal equality of subsets of F^n
notice that if you were given just a 2x3 box of numbers, without being told it was that augmented matrix that you cared about, {x in F^3: [first two columns of the box] x = [third column of the box]} is not the only subset of some F^k that you could associate with the box
 
2:16 AM
oof yes i missed equivalent solution sets
thank you for filling that in for me
got too excited to start my aside in parentheses
@leslietownes Possible to elaborate on this? I don't follow what you mean
 
i had a typo up above but the idea is that a 2x3 box of numbers can represent an augmented matrix for a system Ax = b with a given 2x2 matrix A and a given 2x1 b, or it could just be a 2x3 matrix "A" to be considered in its own right as a function, with its last column not representing a "right hand side"
which isn't contrived, its pretty common even in systems land to consider solvability of Ax = b for multiple values of b and to analyze this in terms of studying just the box of numbers A with no right hand side glued onto it
its just not that unusual for there to be multiple things associated with the same box of numbers, arising not from 'vagueness' in any setup but from the usefulness of representing things with boxes of numbers
and maybe the more familiar uses just seem more familiar than the less familiar ones
when textbooks choose to introduce matrices in a very literal way as boxes of numbers (and not functions in the first instance, but rather things that can have various functions and/or point sets associated with them in various context) they are just trying to avoid insisting on centering one use as 'the' use of a box of numbers, with all other uses being second-class
one of the joys of teaching linear algebra is springing systems of equations, matrices, and five or six different subsets of R^m or R^n associated with any given matrix on everybody, at essentially the same time, and fending off weeks of ill formed questions like "help what is the row space of 3x = 5" and "how do i solve [[1,2,3],[4,5,6]]"
 
3:09 AM
2
Q: Bijective Continuous Map

user284331Does there exist a map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$ such that $f$ is bijective and continuous but not open? Intuition tells me that this would contradict the invariance of domain. However, I stumble in arguing so.

this must be a duplicate, right?
 
3:29 AM
(removed)
 
3:50 AM
"it's basically the same as plus, only there's 100 in it" - my daughter, on integer multiplication (perhaps speaking with reference to tables that she has used to compute both operations)
 
100? Huh?
now it’s time for $2\times 3=3\times 2$, finally.
 
4:10 AM
@leslietownes I'm wondering what she means by the same
 
I felt very confident in my algebra exam like so that means either I got a 50/100 or 100/100
i finished in like half the allotted time (spent too much time on a 1st isomorphism thm problem) then spent the remainder sketching 3 terrible incoherent proofs for a bonus question
it was like for $n\geq 2$, show that $x \in \mathbb{Z}_n$ is nilpotent or a unit iff $n$ is a power of a prime
i did the converse I think, but the other direction I forgot how to logic
 
@Obliv if you were truly confident then you wouldn't think you'd get a 50/100
 
well idk his standards. My answers made sense to me, but that hardly matters
 
If not a prime power, there are zero divisors other than nilpotents.
 
I've done so many assignments/exams before where I've felt confident only to get a mediocre grade so yeah, not gonna get my hopes up.
@TedShifrin I couldn't think of how to break up the cases of what an element could be. I know you can have zero divisors, units, neither, and now nilpotents
 
4:16 AM
Based on what you do here, you cannot reliably tell when you’ve written nonsense or a good proof.
 
I knew that if it was a power of a prime, then the units have no shared prime factors and if they do share prime factors, then they're nilpotent
 
oK, and if not a prime power, there might be nilpotents but there must be non-nilpotent zero divisors.
They also forgot to say nonzero $x$.
 
like suppose $n = p_n^k$ where $k \geq 2$ then $x \in \mathbb{Z}_n$ has either only $p_n^j$ for some $j<k$ where you can show that it's nilpotent, or some prime not in $n$'s factorization so we can't have it be a unit
 
@TedShifrin whats the exercise?
 
$x$ is a zero divisor in $\Bbb{Z}_n$ when $x$ shares prime factors with $n$ right @jakobian
hence why $\Bbb{Z}_p$ has all units for prime $p$
I was writing with a pen because I forgot a pencil and I didn't want to waste any ink by crossing stuff out so I wrote like 3 separate incomplete proofs with no obvious distinction b/t them
I felt so embarrassed I told my prof not to read it and I even wrote "sorry for this nonsense" ughh
I do enjoy math though, I gotta get over the mind games I get in my head while taking exams so i can just lock in
 
4:27 AM
I feel like you have a lot of misconceptions about this. But also probably because you're younger
 
Hi
 
Every nonzero $x\in \Bbb Z/n$ is a unit or nilpotent iff $n$ is a prime power.
@Jakobian
 
I see. I never did this one, let me think
 
Nor I, but it’s a fine question.
Obliv, note I’m doing the foward direction by contrapositive.
 
@TedShifrin isn't zero nilpotent?
 
4:32 AM
so if not a unit nor nilpotent, then it can't be a prime power
 
Oh, who knows, Jakobian. I guess it must be. I added the nonzero.
Wrong logic, Obliv.
 
ran into the following kind of question which i can't really remember seeing before
 
If not a orime power, then ….
 
I did that direction normally I think, probably incorrectly though.
 
I can certainly easily prove it if I knew what the question was beforehand
 
4:35 AM
I thought you said you did the backwards direction, Obliv.
 
would be more interesting if it were "Find all $n$ for which $\mathbb{Z}/(n)$ consists only of units and nilpotent elements"
 
Suppose i have a function $f(t)$ which is only nonzero for $-T<t<T$ for some $T>0$. Does there exist a function $g(t)$ such that $g(t)=0$ for $t\leq 0$ and $f(t)=a g(t)-g(t+1)$ for all $t$?
 
I think I can just make cases like $(x,n) = 1$, then its a unit. if $(x,n)= x$ then we're finished, or $(x,n) < x$ meaning it shares the prime with $n$ but also has another factor which would mean it has another zero divisor in $\mathbb{Z}_n$ that isn't itself. @TedShifrin
 
Seems wrong and unnatural, Semiclassic. No clue.
 
i mean the ideas are there I'd just have to formally write the proof
 
4:37 AM
it's kinda wacky yeah
 
@TedShifrin I thought backwards direction meant if $n$ is a prime power then either $x$ is a unit or nilpotent
 
@Semiclassical and $a$ is?
 
yes, I thought that was what you said you’d proved.
 
woops. some real number.
 
yes, so isnt that contrapositive redundant? like if not a prime power, then not a unit nor nilpotent
 
4:39 AM
i think it should be the case that $0<a<1$ (based on the context)
(i'm confident in it being real, but not the in the domain)
 
Then $f(-1) = 0$ but what if $-1\in (-T, T)$?
 
scratch that, i am confident: $0<a<1$ for certain
 
Obliv. Seriously, review your logic.
 
$A \iff B$ I did $B \implies A$ and you're saying $\neg B \implies \neg A$ is equivalent to $A \implies B$? oh wait..
 
Contrapositive is equivalent to the original implication. The converse is very different.
 
4:41 AM
you're right lol
 
Duh.
 
It just looks funny because we're doing $B \to A$ both times
 
No.
 
@Jakobian oof, i see what you mean (both in the math sense and for the underlying physics)
take $T=1/2$ then
(my test case has been $f(t)$ as the indicator function on the interval $[-1/2,1/2]$, so i didn't see that before)
 
Obliv, Assuming B and assuming not B are entirely different.
 
4:44 AM
@Semiclassical whats the relevant physics for this function question
 
trying to solve for sending a 1D sound pulse towards a barrier of finite width
when the pulse hits the surface of said barrier, it will both reflect and transmit
the transmitted part will then hit the outgoing surface and then, again, reflect or transmit
which means that you'll keep having a wave inside the barrier bouncing back and forth forever, but losing more and more intensity each time
 
$g(t) = 0$ for $t\leq 1/2$
 
how are you deducing that? i buy it but don't see the logic immediately
 
$f(t) = -g(t+1)$ for $-1/2 < t < 1/2$
 
@Jakobian my guess: the experience of manually looking something up in a printed table is more or less the same, regardless of what you're looking up
 
4:50 AM
@jakobian I read your profile, I think you updated it? Happy belated birthday
 
@leslie What was Munchkin’s reference to 100?
 
@Jakobian okay, yeah, i see it
what this should lead to, i think, is something like $g(t)=f(t)+a f(t-1)+a^2 f(t-2)+\cdots$
 
and $ag(t) = g(t+1)$ for $t \geq 1/2$
 
@TedShifrin she probably was referring to the successor function in that multiplication & addition are both fundamentally tied to it and that the only difference b/t them is that mult. tables go to the 100's since $10\cdot 10$ is in fact $100$ , $11\cdot 10$ is $110$, etc.
obviously has a high level understanding of hyper operations
 
@Obliv thank you. I don't remember my birthday being there though
 
4:56 AM
i googled pisces its b/t feb 19-mar 20 I think
I'm a virgo :0
 
ah
 
@TedShifrin what are you
 
I’m Aquarius
 
the smarter way to do what i'm trying to do is probably just to figure out how to simulate the acoustic wave equation in mathematica
 
earlier in Feb
 
4:57 AM
well happy belated bday to you too. I'm not a nut but just going off of the descriptions for the signs always seem to match my image of the person
 
leslie claims I’m consistent with double or triple Aquarius
 
i am in fact a fish in real life
 
LOL
 
a sturgeon?
 
I wonder what @leslietownes is
 
5:00 AM
@TedShifrin is it because I'm late matured, or a bottom feeder?
 
I'm not a fisherman but I believe sturgeon are very hard to handle/catch or something idk
 
I just find the name amusing. Usually smoked, too.
 
I think of like 1 real fishing trip I've ever done, my first catch was a rainbow fish which swallowed my hook and my uncle had to basically rip it open to get it back
safe to say I was traumatized by the experience, and I think we threw it on the rocks in an attempt to throw it back into the lake.
 
Poor fish.
 
that's funny
 
5:41 AM
Waves are funny. Why do they propagate ad infinitum? i guess they dont if ur a physicist? shrug
*engineer maybe more like it
 
6:05 AM
@TedShifrin double or triple? What is that?
 
6:16 AM
@TedShifrin the times table, unlike the addition table, does have 100 on it. must have been wild to see such a big number on a table
 
I can now prove the twin prime conjecture topologically and though it's quite different from Furstenberg's infinitude of primes, the last step is analogous.
It's not my "proof" from last night
that had major flaws but I've overcome them
@mick I have proven it. If you'd like to see the proof, ping me here, but prereqs are understanding how Furstenberg's proof works
 
7:02 AM
@robjohn Moon sign and rising.
@leslietownes Oh, duh, the “whole” table.
 
Oh so its actual furniture?
 
they're platonists at her kindergarten, they took her on a field trip to see the number 100
 
That's not plantonist, those are botanists leslie
 
 
4 hours later…
11:28 AM
helloooooo
 
 
4 hours later…
3:54 PM
4 hours later…
 
Lets wait next four too
 
@Balarka are you familiar with the Puppe proof of Blakers-Massey?
 
@Thorgott I only know the proof in Hatcher
 
4:10 PM
ah ok
 
5:07 PM
My book (Amann Escher) somewhat in passing says the following: "To simplify our discussion we introduce the following concept: A set M is called an infinite system, if there is an injective function f : M → M such that f(M) ⊂ M. Clearly the natural numbers, if they exist, form an infinite system. The significance of such systems is seen in the following theorem proved by R. Dedekind: Any infinite system contains a model (N, 0, ν) for the natural numbers."
I recall from Halmos that we proved that the natural numbers (as constructed set theoretically via what I think Jakobian you called the von Neumann model) has a bijection onto some subset of any infinite set.
My question is, is this the statement alluded to above by Amann Escher? Or is a "model" stronger than a bijection?
 
Hi @Jakobian. Pinging you as a daily habit.
 
Sahaj, you know that annoys him. Do not do it to be funny.
@EE18 A model is a technical term from mathematical logic. It means you have a version of the natural numbers with all its structure. Nothing to do with a mapping or a bijection.
 
5:27 PM
the Puppe proof of Blakers-Massey is really cool, for the record, but ive been in great agony cause the proof only appears to work for open rather than excisive triads, which is enough for p much all applications, but violates my sense of aesthetics
 
can i read Aluffi chapter 0 if im just a beginner
 
you're going to great lengths to avoid Hatcher somehow @Thorgott :P
 
@TedShifrin I wanted to congratulate him on hitting 10k reputation. He didn’t come online though.
 
@Thorgott I assume that an "open triad" is some kind of polyamorous relationship which admits the occasional guest?
While an "excisive triad" is more rancorous, with two members wishing to expel the third?
What kind of nonsense is this?!
 
@TedShifrin it doesn't
 
5:38 PM
@Xander You're far too influenced by our former president and his felonious 91 character
Oh, OK, @Jakobian. :)
 
@Sahaj hi daily habit
 
@TedShifrin :(
 
Your cold getting better, I take it?
 
@nickbros123 yes
 
@TedShifrin Yeah, I am mostly back to fighting form.
Though this week has been exhausting. So much lecture to record, after missing last week.
 
5:42 PM
@Thorgott do you have any idea on how to obtain an open continuous map from a continuous map?
 
But I am caught up on all of my lectures, and caught up on the grading in two of three classes.
I deal with the third class tomorrow.
@Jakobian You can try to invite in a third, but a lot of maps just don't swing that way.
 
Xander, do you have any relationship issues lately?
 
@Jakobian No, just attempting humour.
 
Congratulations on hitting 10k rep on M.SE Jakobian.
 
@Sahaj thanks
 
5:44 PM
All mathematics vocabulary sounds dirty if you say it right.
 
@XanderHenderson Humo(u)r in the generalized weak sense.
 
Hey, baby. Wanna see my Hausdorff dimension?
@TedShifrin weak-*, even.
 
Xander is very lively today
 
@Xander I am reluctant to award you any stars, for fear that that will encourage you even more.
@Jakobian Yes, sadly for us, he is back to "normal."
 
@TedShifrin :D
@TedShifrin :'(
 
5:47 PM
What branch of math is popular in ocean?
 
I'm actually getting a little punchy. I've been in the office since 4 this morning, and am waiting for one more meeting to start before I can go home and eat soup.
 
Ans: Sea Star Algebra
 
It is definitely lunchtime if breakfast was before 4 AM.
@SoumikMukherjee That's as bad as the abelian grape.
 
@TedShifrin Breakfast? I had some coffee...
 
Very unhealthful, that, @Xander.
 
5:48 PM
@TedShifrin Meh... I've never eaten breakfast. It is too early to eat.
 
I literally cannot function without breakfast.
 
Makes me nauseous to eat before noonish.
 
@TedShifrin What is abelian grape?
 
@SoumikMukherjee What's purple and commutes?
 
Abelian grape
 
5:53 PM
@BalarkaSen that proof is unreadable to me. the Puppe proof is very accessible. you should check it out.
 
I think the mistake I made is that I would need to not have the map $f^{-1}:\text{RO}(Y)\to \text{RO}(X)$ when $f$ is continuous and open, but some of the form $\text{int }\overline{f^{-1}(U)}$ perhaps
 
@XanderHenderson it's complicated
@Jakobian that seems more reasonable
 
6:08 PM
@BalarkaSen as one should
 
6:43 PM
the congruence classes of $F[x]$ mod $x-a$ for some field element $a \in F$ are polynomials degree 2 or greater in the form $(x-a)(bx^n+cx^{n-1}+..)$?
makes sense right
 
@Obliv no
What you listed is elements of equivalence class corresponding to $0$ element
equivalence classes are represented by elements of the field $F$
each polynomial $f(x)$ is congruent to $f(a)$
 
wait before u get technical, I just want to describe them like $[f(x)]$ so
 
I'm not waiting
 
so the cong. classes of mod $x-a$ are $\{f(x)+h(x)(x-a)\mid f(x),h(x) \in F[x]\}$ i adapted the set described in my book for cong. classes of $2x+1$ for $x^2+1$ in $\mathbb{R}[x]$
where it would be $\{2x+1+k(x)(x^2+1)\mid k(x) \in \mathbb{R}[x]\}$
 
@Obliv Any $f(x)$ can be written as $g(x)(x-a)+$constant. Where the constant is $f(a)$
 
6:55 PM
@Jakobian you were saying yesterday that I should consider the map $f\mapsto [f]$ between the function spaces $C_c(\Omega)$ and $L^p(\Omega)$. What confuses me is that the output of that function is a set, i.e. the equivalence class $[f]$. Is there a metric that measures the distance between sets? The reason I'm asking is because I'm trying to construct an isometry between these spaces.
 
@psie "What confuses me is that the output of that function is a set,"
you shouldn't think like that
well if you're trying to find a point to be confused about, then sure this could be one of them
but this isn't just a set
 
should I think about $[f]$ as a representative function of the equivalence class?
 
for example, yes
 
@SoumikMukherjee why must it be a constant, $f(a)$? can't it be an irreducible polynomial
 
in any case you shouldn't think of it as a set
 
6:59 PM
@psie You are overthinking it.
 
this is completely wrong
sets have nothing to do with this
its not really about set theory even though we define equivalence relations using sets
 
Yes, the elements of $L^p$ are equivalence classes of functions. But you can just think of them as functions. Distance in $L^P$ is measured using the $L^p$ norm: $d(f,g) = \|f-g\|_p = \left( \int |f-g|^p \right)^{1/p}$.
 
if you're confused about that then you have way more to be confused about that you were just plainly ignoring apparently, e.g. functions
 
@Obliv This is just division algorithm, you are dividing by a degree $1$ polynomial.
 
functions are defined using sets so clearly you should be confused about them too
 
7:02 PM
oh.. @SoumikMukherjee so the congruence classes are just constants? ;O
 
you aren't - why?
because you're not thinking in terms of sets
why do you need to think of equivalence classes as sets?
 
@Obliv Yes
 
when you take an element of $L^p$ what do you think of? You can think of it as a function
you can think of it as an equivalence class, too
 
thinking is overrated
 
but its a specific instance on equivalence class and it has its own meaning
 
7:05 PM
Honestly, I am more worried about the metric in $C_c(\Omega)$. Isn't that just the space of compactly supported continuous functions? There isn't an "obvious" natural metric there. The uniform metric?
 
you can't just forget about the meaning of what those equivalence classes are supposed to represent
@XanderHenderson $L^p$
 
Though compactly supported continuous functions are $p$-integrable, so perhaps the idea is to think of $C_c(\Omega)$ as a "subspace" of $L^p(\Omega)$?
Of course, I've come into this conversation late...
 
on $C_c(\Omega)$ its standard to equip the metric of compact convergence iirc
this works by first taking an exhaustion by compact sets of $\Omega$
its then a Frechet space
 
@Jakobian Sure, you can do that, too.
 
yeah... I think this is the approach in distribution theory
 
7:11 PM
@Obliv Here you are looking at things modulo $x-a$, which means $x-a$ is acting as the zero element. So $F[x] /x-a$ is just $F[a]$, which in turn is just $F$ as $a$ is already in $F$. In the other example, $x^2+1=0$ implies $x=i$, so $R[x] / x^2+1$ is isomorphic to$ R[i]$
 
@Jakobian It's been too long, but most of distribution theory seems to run through the $L^2$ norm. I'm not sure why you would bother with another metric. But, again, it's been a decade.
 
@SoumikMukherjee very cool, so what's the homomorphism such that the kernel is $x^2+1$?
 
Ok, thanks for the inputs, I will have to distill this further.
 
it maps $R[x] \to R[i]$
 
@Obliv Evaluation at $i$
 
7:20 PM
@XanderHenderson its what you do to define standard topology on the space of distributions
don't ask me why such topology is chosen - I don't know
this sounds like a question for @leslietownes
oh wait I think I recall now... its about some equivalence between continuity of maps
 
Oh duh, lol. I'm not super comfortable with this stuff yet thank you for your patience @Soumik
 
$\mathcal{D}$ is the space of smooth functions with compact support
 
Why don't people write iff instead of if and only if?
@Obliv np
 
people write that all the time
mostly in less formal contexts
 
Is it okay to write $(f(x),g(x))=h(x)$ to mean gcd for polynomials? or even $(f,g)=h$ for example
 
7:29 PM
topology on $\mathcal{D}$ is not that of compact convergence
its weak topology from subspaces $\mathcal{D}_K\subseteq \mathcal{C}^\infty$ of smooth functions with support in $K$
so I think its uniform convergence on $\mathcal{D}_K$ actually
($\mathcal{C}^\infty$ has compact convergence)
"its weak topology from..." not precisely...
not sure if it coincides with it
 
nvm got it
 
7:57 PM
@TedShifrin Nor has mine changed, over at least ten years!
 
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