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Obliv
Obliv
12:01 AM
mm k im supposed to just look at the degrees of the exponentials i guess
makes sense
lol
nvm im still confused
$$\frac{Nk_B(\varepsilon\beta)^2 \exp(\beta\varepsilon)}{(\exp(\beta\varepsilon)-1)^2}\approx Nk_B(\beta\varepsilon)^2\exp(-\beta\varepsilon)$$ since $\beta\varepsilon\to\infty$ as $T\to 0$ idk how this follows
 
EE18
EE18
Do you want it the math way or the physics way?
 
Obliv
Obliv
Either way :P
 
EE18
EE18
I'll let others speak to it the math way. For the physics way you simply observe that $(\exp(\beta\varepsilon)-1)^2 \to (\exp(\beta\varepsilon))^2$ in this large $\beta$ limit. Then cancel one factor of $\exp(\beta\varepsilon)$ from the top and bottom.
 
Obliv
Obliv
ohh okay
what about the other way when $\beta$ gets to be close to 0
somehow it approximates to be $3Nk_B$
 
EE18
EE18
Expand $e^x \approx 1+ x$ and see what happens :)
 
Obliv
Obliv
12:12 AM
OHH right
u can do that
bless u :P
 
EE18
EE18
Cheers
A "soft" question about math writing for folks here: how do you deal with nested brackets? For example, suppose I wanted to write an aside in brackets, and in that aside I refer to some equation, say equation (2.65). Would you choose square brackets in this case or is that bad form?
(This follows from equation (2.65)) or [This follows from equation (2.65)]
 
Thorgott
Thorgott
I think the former is perfectly fine, but when Xander arrives, he will probably be able to point you to the precise subsubsection in the AMS style guide where this is addressed
 
Obliv
Obliv
or you could say (this follows from equation 2.65)
 
EE18
EE18
Certainly I am not writing anything so important as to deserve that :) but thanks Thorgott
That's fair Obliv, but generally the brackets are there in the label too which is why i have them
 
Xander Henderson
Xander Henderson
12:33 AM
@Thorgott There isn't actually a ton there, though the inference I draw is that "This follows from (2.65))" is the preferred style (note that is just "(2.65)", not "equation (2.65)").
E.g. section 6.4
> Identifying letters ((a), (b), (c)), including their parentheses, are roman in all text. The AMS will allow italic identifying letters only if consistent throughout.
Section 13.14:
> In cross-references, equation numbers are enclosed in roman parentheses to match the original label and the parentheses are always roman.
(which only says that equation numbers should always be in parentheses, not other kinds of braces).
On the other hand, the only references that the style guide has to brackets is in the context of citations and "fences" in mathematical contexts. I do not think that AMS editors would be too happy with square braces for parenthetical content.
Personally, I would seek to rewrite the phrase.
@Obliv No. This is wrong. Again, see section 13.14 of the AMS style guide.
 
Obliv
Obliv
i don't think style can be logically deduced
 
Xander Henderson
Xander Henderson
@Obliv Okay...
 
EE18
EE18
1:01 AM
@leslietownes Just quickly coming back to this conversation -- as regards the $T$-periodic space of functions which are spanned by the trigonometric functions, I guess I gave the wrong norm right? It should be induced by an inner product looking something like $<f,g> = \lim_{T \to \infty}\frac{1}{T}\int_{-T/2}^{T/2} fg$?
 
leslie townes
leslie townes
EE18 if you are looking at periodic functions the usual norm of <f,g> would be an integral of fg over one period (which is arbitrary but usually fixed, e.g. [0,T] or [-T/2,T/2]). you might be able to write out some limit over increasingly large that ends up being the same thing because of the periodicity, but i don't know why you would want that
you can also define inner products of not necessarily periodic functions in terms of integrals over all of R, usually with some kind of "decay at infinity" hypothesis floating in the background to ensure that the thing makes sense and is finite, and sometimes (not always, and not usually by definition) expressions for those sorts of norms are written as limits like that
 
EE18
EE18
1:18 AM
@leslietownes Ah OK I think this is what's going on, that there's some equivalence to the infinite limit version because of periodicity
I'm working with that engineering text which is keeping stuff hazy, so wanted to just clarify what was going on
 
user 70432
user 70432
1:59 AM
@Obliv agreed, but preference should be given to consistency.
 
Xander Henderson
Xander Henderson
2:15 AM
@EE18 \langle and \rangle.
@leslietownes You, too. :(
 
user 70432
user 70432
:(
 
Xander Henderson
Xander Henderson
@leslietownes Though in a lot of cases, those won't really be inner products, but the dual pairing (which devolves to an inner product in spaces which are self-dual. (Not that this really matters).
 
Obliv
Obliv
2:44 AM
I personally write my inner products like this $\{\left\{\langle\left[<f,g>\right]\rangle\right\}\}$
just to be clear
more brackets means stronger seal
 
leslie townes
leslie townes
3:00 AM
xander: or maybe it is the inner product that evolves into a dual pairing. hippy music
 
user 70432
user 70432
🎢🎡🎢🎸
 
Xander Henderson
Xander Henderson
3:25 AM
@leslietownes That isn't how evolution works.
Though it may be the that the inner product is the distant descendant of some common ancestor of the dual pairing.
 
EE18
EE18
3:40 AM
@leslietownes Actually sorry for perseverating on this Leslie, but I've been searching around and not able to find a wikipage or something like that for this. What term should I be using to learn more about this equivalence between the two possible inner products on this periodic function space?
 
leslie townes
leslie townes
maybe just prove it? i could see it as a homework problem in an analysis book. if f is periodic with period T then the average value of f on [-L, L] goes to the average value of f on [0,T] as L goes to infinity. it should make intuitive sense.
 
leslie townes
leslie townes
4:07 AM
there are several duplicates of that question on the site too, all with similar looking proofs. it's maybe an expository challenge to make the argument look good. it might clean things up a little to assume T = 1 or something like that
 
John Zimmerman
John Zimmerman
5:05 AM
I realized I have no idea how to glue together closed subsets bounded by analytic boundary S^1's when I'm not in the Top Cat. Do I operate in the Top Cat and then use some deformation or rigidification process to deform the topological surface into smooth/real analytic manifold? I think that is the right direction because there are theorems guranteeing a topological manifold and smooth manifold can be deformed to a anlaytic one. I'm lost on the details though
 
John Zimmerman
John Zimmerman
5:19 AM
not sure where to start exactly but maybe in Top, and generate a topological surface S^2, and then sequentially layer on smooth then analytic structures. If the S^2 here is obtained through a gluing and one can obtain an analytic metric compatible with the base (unwrapped quotient), related through the section/exact projection correspondence allowing for a direct analytic quotient metric on the S^2. I think I should go for the low hanging fruit and that would be recovering a well defined
analytic manifold with a well defined analytic quotient metric
 
 
3 hours later…
Koro
Koro
8:09 AM
Can anyone please explain what $V\otimes_R C$ is?
 
leslie townes
leslie townes
i assume tensor product of real vector spaces, i.e. the thing in en.wikipedia.org/wiki/Tensor_product where you regard C as a vector space over R
Complexification
In mathematics, the complexification of a vector space V over the field of real numbers (a "real vector space") yields a vector space VC over the complex number field, obtained by formally extending the scaling of vectors by real numbers to include their scaling ("multiplication") by complex numbers. Any basis for V (a space over the real numbers) may also serve as a basis for VC over the complex numbers. == Formal definition == Let V {\displaystyle V} be a real vector space. The complexification of V is defined by taking the tensor product of...
 
Koro
Koro
thanks. So V and C are both considered over R.
I keep forgetting the definition of tensor product.
I remember it using its existence property diagram but forget it soon.
V, W - K vector spaces. V tensor W is a K- vector space along with a bilinear map f: V\times W into V tensor W such that every bilinear map out of V\times W into a K vector space U is composition of a unique linear map (from V tensor W into U) with f.
But I never really understood the motivation for tensor products.
What is their use? Why do they exist?
 
Mad
Mad
suppose $ R $ is a ring and $\mu: R\rightarrow [0,\infty]$ is sigma additive, how can i show that for $ A \subset \cup_i A_i: \mu(A) \leq \sum_i \mu(A_i)$ if the unity need not be in the set (ie i can not use the attribute of sigma additivity)
if it is contained, then its trivial by definition... if it is not contained?
 
Jakobian
Jakobian
9:19 AM
@Koro in this context its change of base field from R to C. That's their current use
Another use would be that they allow you to develop exterior algebra
Any time when there's an exchange from n-linear to linear you can expect tensors are involved
The usefulness of tensor products is definitely immeasurable, but I can't tell you historical reasons/from ground up motivation for their existence.
I just don't really know
 
John Zimmerman
John Zimmerman
9:54 AM
Why is the dirichlet divisor problem so hard?
see here for what the dirichlet divisor problem is all about
I don't get why it is notoriously difficult tbh
 
Koro
Koro
What does RHS mean?
pi_1: G---> GL(V_1), pi_2: H---> GL(V_2)
so pi_i g is a map from V_i into V_i
How can two maps be tensored?
 
Koro
Koro
10:20 AM
nvm, it's same as how it is done in differential geometry.
 
Jakobian
Jakobian
10:47 AM
yeah. Tensor products are, first and foremost, applied in differential geometry
"Defined as $(\pi_1\otimes \pi_2)(g, h) = \pi_1(g)\otimes \pi_2(h)$" really should be "Defined as $(\pi_1\otimes \pi_2)(g \otimes h) = \pi_1(g)\otimes \pi_2(h)$"
or in other words, the linear map coming from the bilinear map $(g, h)\mapsto \pi_1(g)\otimes \pi_2(h)$
 
Soumik Mukherjee
Soumik Mukherjee
11:39 AM
Sequence -> sequential convergence. net -> netial convergence? net wise convergence?
 
Xander Henderson
Xander Henderson
11:55 AM
@SoumikMukherjee Net convergence.
 
 
2 hours later…
user 70432
user 70432
1:58 PM
 
ZaWarudo
ZaWarudo
2:40 PM
Hi, I'm studying the uniform convergence of $f_n(x)=n\sqrt{4\pi^2 n^2 +x^2}$ on $\mathbb{R}$. I proved that $f_n(x) \to \frac{x^2}{4\pi}$ pointwise on $\mathbb{R}$. My reasoning to prove that $f_n$ is not uniformly convergent to $\frac{x^2}{4\pi}$ is the following: $\sup_{x \in \mathbb{R}} |f_n(x)-f(x)| \ge |f_n(n)-f(n)|=n^2|\frac{\sin \sqrt{4\pi^2 n^2+n^2}}{n}-\frac{1}{\pi}|$.
Hence, $\lim_{n \to +\infty} \sup_{x\in\mathbb{R}} |f_n(x)-f(x)| \ge \lim_{n \to +\infty} n^2|\frac{\sin \sqrt{4\pi^2 n^2+n^2}}{n}-\frac{1}{4\pi}|=+\infty$
This shows that $f_n$ is not uniformly convergent to $x^2/4\pi$. Is this reasoning correct?
I forgot a $4$ is the denominator of $-\frac{1}{\pi}$ in my first message, sorry.
 
Obliv
Obliv
2:56 PM
typically $\| \|$ is used for norms of vector quantities and $||$ just normal abs bars are for numbers?
 
Thorgott
Thorgott
sometimes yes, sometimes no, it doesn't really matter
 
Obliv
Obliv
Suppose $\overline{E}$ is $\sum_s E(s)P(s) = -\mu BP_{\uparrow}+\mu BP_{\downarrow} = -\mu B(P_{\uparrow}-P_{\downarrow}) = -\mu B\tanh(\beta \mu B)$. I want to set the energy levels to $0$ and $2\mu B$ instead of the $\pm \mu B$ but I'm not sure if that's the same as $$\mu B\left(\frac{1-e^{-2\mu B\beta}}{1+e^{-2\mu B\beta}}\right)$$
which is what u get if you do that. Like is what's in the parentheses a $-\tanh(\beta\mu B)$
I think it should be. I guess there isn't much of an advantage doing it this way since $\tanh$ is more recognizable
 
Soumik Mukherjee
Soumik Mukherjee
@XanderHenderson So anticlimactic
 
EE18
EE18
@leslietownes This is awesome, thanks so much :) \
So am working with my analysis text this morning which is on a last brief section about vector spaces before the fun with sequences starts next chapter
The book has a brief aside about polynomial interpolation, and proves that the existence and uniqueness of an $m$ degree polynomial to "match" a function at $m+1$ locations
They do it by explicit construction of the Lagrange interpolation polynomials without much motivation, fine
They then mention the following:
What could the last sentence possibly mean? I know things like "simple" or "explicit" can be informal notions but I fail to see why Gauss-Jordan wouldn't give us explicit descriptions of each $p_j$ and, putting them all together, the given polynomial?
 
Thorgott
Thorgott
3:25 PM
they would, it's just tedious
 
Jakobian
Jakobian
@Obliv depends on what you think of as a vector and what you think of as a number
 
Obliv
Obliv
is $\sigma_X=\sqrt{\sum_s P(s)^2X(s)^2}$? Or is it just $\sqrt{\sum_s P(s)X(s)^2}$
 
Jakobian
Jakobian
@Obliv are you talking about standard deviation?
 
Obliv
Obliv
yea, or root mean square deviation. square root of the average of the squares
so I imagine it's the latter but i'm not 100%
 
Jakobian
Jakobian
what is $P(s)$ and $X(s)$
 
Obliv
Obliv
3:31 PM
$P(s)$ is the probability of finding $X(s)$
 
Jakobian
Jakobian
the what?
 
Obliv
Obliv
hmm I'm gonna just post the relevant sections lol
I just wanted to write it in terms of an arbitrary variable $X$
oh wait it was average of the squares of the deviations
I didn't read a)
so a deviation can be given by $X_i - \overline{X}$
so $\sigma_X$ is the square root of the average of these squared
I'm so confused lmao
so I guess the "average of the squares of the deviations" can be $$\sum_s P(s)(X_s-\overline{X})^2$$
 
Jakobian
Jakobian
The standard formula is that $\overline{X} = \sum_k k\cdot P(X = k)$. But $P(X = k) = \sum_{s:X(s) = k} P(s)$ so the formula $\overline{X} = \sum_s X(s)P(s)$ seems correct
 
Obliv
Obliv
yeah I guess that makes sense
 
Jakobian
Jakobian
@Obliv here it should be neither in general, unless observations are independent
 
Obliv
Obliv
3:45 PM
yea what I wrote above was wrong
I think it could be $$\sqrt{\sum_s P(s)(X_s-\overline{X})^2}$$
 
Jakobian
Jakobian
no wait I'm spouting gibberish
I guess physics has that effect on people
 
Obliv
Obliv
bruh
 
Jakobian
Jakobian
$\sigma_X^2 = \overline{X^2}-\overline{X}^2$ where $\overline{X} = \sum_s X(s)P(s)$ as we already observed and $\overline{X^2} = \sum_k k^2P(X = k) = \sum_s X(s)^2P(s)$
so it would be $\sigma_X = \sqrt{\sum_s X(s)^2P(s) - (\sum_s X(s)P(s))^2}$
 
Obliv
Obliv
:( so it's not what I wrote
why'd he write it in english instead of explicit math ugh
 
Jakobian
Jakobian
If you assume $X$ has $0$ as an average, then its $\sigma_X = \sqrt{\sum_s X(s)^2P(s)}$ so its still somewhat valid
 
Obliv
Obliv
3:54 PM
i meant like right above
 
Jakobian
Jakobian
@Obliv this one?
 
Obliv
Obliv
yea
that one also isn't the same
"average of the squares of the five deviations"..."then compute the square root of this quantity" made it sound like what I wrote
 
Jakobian
Jakobian
well you wrote $X_s$ instead of $X(s)$ but its the same formula
as in, its also a valid formula
 
Obliv
Obliv
is it? when u multiply out dont' you get $\sqrt{\sum_s P(s)X^2(s)+\sum_s P(s)\overline{X}^2(s)}$
 
Jakobian
Jakobian
Exercise: Convince yourself both formulas are true i.e. both expressions are equal
 
Obliv
Obliv
3:57 PM
ok author
after I brb
 
Jakobian
Jakobian
@ZaWarudo $f_n(x) = n\sin\sqrt{4\pi^2n^2+x^2}$?
 
ZaWarudo
ZaWarudo
@Jakobian Yes, thanks for catching up the typo!
 
Jakobian
Jakobian
alright, then its indeed pointwise convergent to $\frac{x^2}{4\pi}$
@ZaWarudo its correct
 
ZaWarudo
ZaWarudo
@Jakobian thank you. Instead, if I consider the uniform convergence in $[0,a]$ with $a>0$, that reasoning above is not valid anymore because, even if it's still true that $\sup_{x \in \mathbb[0,a]} |f_n(x)-f(x)| \ge |f_n(n)-f(n)|$, I then must take the limit as $n \to +\infty$ and for fixed $a>0$ I have $n \notin [0,a]$ for $n$ big enough, right?
 
Jakobian
Jakobian
4:12 PM
@ZaWarudo Well... its not true that $\sup_{x\in [0, a]}|f_n(x)-f(x)|\geq |f_n(n)-f(n)|$
 
ZaWarudo
ZaWarudo
I mean, it's true for $n \in [0,a]$ but it is not true for each $n \ge N$ for some $N \in \mathbb{N}$ and this is what I need when I want to evaluate $\lim_{n \to +\infty} \sup_{x \in [0,a]} |f_n (x)-f(x)|$
Or am I missing something?
 
Jakobian
Jakobian
sure. The gist is that this argument doesn't show to disprove uniform convergence
 
EE18
EE18
@Thorgott weird that they would say no explicit solution though no?
 
ZaWarudo
ZaWarudo
Yes, I meant that. I just wanted to be sure that I understood correctly why it doesn't work. Thanks! :)
 
Thorgott
Thorgott
@EE18 they don't say that?
they just say "no simple expression"
 
EE18
EE18
4:46 PM
Got it, fair enough. Just seems weird, not like Lagrange polynomial is so β€œsimple” right? Anyway, I won’t dwell
 
Jakobian
Jakobian
4:58 PM
You're saying that you understand the word "simple" might be subjective, yet completely ignore it
 
 
1 hour later…
Obliv
Obliv
6:06 PM
uhh is $\int_{-\infty}^{\infty}\exp(-x^2)dx = \sqrt{\pi}$?
that's such a weird identity
 
EE18
EE18
As 3B1B says, if you see pi then there's always a circle lurking somewhere
In this case, the standard proof of that identity shows you where the circle lurks
I'm just learning about "arithmetic sequences of order $k$" and am hoping for some help on internalizing the definition. The $\Delta$ operator is defined as an endomorphism on $E^{\Bbb N}$ via $(\Delta f)_n = f_{n+1} - f_n$ for $f \in E^{\Bbb N}$ ($E$ is a vector space). Now I know that $(\Delta^kf)_n = \sum_{j=0}(-1)^{k-j}{k \choose j}f_{n+j}$ which appears relatively ugly, so I have very little inution for what $(\Delta^kf)_n$ constant would mean
Please feel free to ignore everything past the first sentence there, I am just sort of trying to parse that first sentence. What sort of sequence is an arithmetic sequence of order $k$?
 
Jakobian
Jakobian
6:30 PM
@Obliv yes
@EE18 polynomial
$\Delta$ is like a discrete derivative
$\Delta^k f = 0$ is similar as to say that $k$th derivative of a function $f:\mathbb{R}\to\mathbb{R}$ vanishes
 
EE18
EE18
Got it. So there's nothing about repetition of values or something like that to be said?
e.g. after $k$ elements the cycle repeats
 
Jakobian
Jakobian
cycle?
 
EE18
EE18
(1,2,3,1,2,3,1,2,3,...)
Something like that
 
Jakobian
Jakobian
what about it
 
EE18
EE18
OK I see the definition has nothing to do with that, that was my first instinct
Thanks Jakobian
 
Jakobian
Jakobian
6:56 PM
@EE18 In general those should be sequences of the form $f_n = \sum_{k=0}^N n^k v_k$ where $v_k\in E$
 
EE18
EE18
I think that's what the authors are driving at with the paragraph I gave but I'm still struggling to understand it
Will post back here in a little about it if that's ok
 
psie
psie
My basic set theory skills are a bit slow. Is there a typo in the yellow highlighted bit. Should $B$ (the second occurence) not be $Y$? This is from Folland's, but I haven't been able to find anything in his errata about this. The formula I have found about this is $(A \times B)^c=(X \times B^c) \cup (A^c \times Y)$.
Alternatively, one could write $$(A\times B)^c=(A^c\times B)\cup (A\times B^c)\cup (A^c\times B^c).$$
 
Thorgott
Thorgott
yes, it's a typo
 
psie
psie
ok, thanks, I know it's a small typo, but maybe an email wouldn't hurt, since it's in neither of his erratas, so he knows it's there
for the third edition, if it ever comes :D
 
EE18
EE18
7:11 PM
OK I am still struggling with the second sentence in the picture I gave above. I'll label my questions with (x). What I have so far: given any $p \in K_k[X]$ and choice of $x_0,h$ I know (Remark 8.19(c)) that $p = N_k[p;x_0;h]$, where the RHS is the newton interpolation polynomial for that function $p$ ((1): how do $x_0,h$ factor into the discussion here? Shouldn't any choice of $x_0,h$ give the same $N_k$ because this is a polynomial, so that if $N_k$ agrees...
with $p$ at $k+1$ spots then they're equal?)
Agh
Trying to parse it still, don't even know how to ask the question about how the formula (12.15) gets used. Will keep at it
This is it FWIW. Will keep thinking on it
 
Jakobian
Jakobian
7:30 PM
@EE18 $N_k$?
 
EE18
EE18
Hopefully is made clearer by the second picture I sent. The Netwon polynomial (of degree $k$) constructed from the given function $f$ and choice of $x_0$ and $h$
 
Jakobian
Jakobian
You certainly don't need this, its unclear to me as how author meant it, and its clear if you use induction.
so I wouldn't worry about it
 
EE18
EE18
Basically the way the author is arguing seems to be that since a polynomial of degree $k$ equals its Newton polynomial of that same degree (by Proposition 8.19(c)) in my book, it follows that $p(x_0 + nh) = N_k[p;x_0;h](x_0 + nh)$ and then I suspect I should be able to use (12.15) pictured above to calculate the RHS, and then somehow observe that the sequence so obtained is an arithmetic sequence of order $k$?
Does that roughly seem like the idea?
 
Jakobian
Jakobian
it looks to me like this argument is missing a whole leg and an arm
plus its too elaborate for no reason
 
EE18
EE18
I know, but it's the end of the section and seems to be summing up the general development, so I'm eager to follow it
Maybe I'll post a question and flesh it out in full
 
Frieren
Frieren
8:35 PM
The definition of limit for $A \subseteq \mathbb{R}$ and $f:A\to\mathbb{R}$ is $\lim_{x \to x_0} f(x)=l \in \mathbb{R}$ if $\forall \epsilon>0$ there exists $\delta>0$ such that $\forall x \in \mathbb{R}$, $x \in A \cap (x_0-\delta,x_0+\delta)\setminus \{x_0\}$ implies $|f(x)-l|<\epsilon$. But how is $x_0$ quantified?
Is it: $\forall \epsilon>0, \forall x_0 \in \mathbb{R}$ there exists $\delta>0$ such that $\forall x \in \mathbb{R}$, $x \in A \cap (x_0-\delta,x_0+\delta)\setminus \{x_0\}$ implies $|f(x)-l|<\epsilon$?
And what about $l$? Is it something like: $\exists l \in \mathbb{R},\forall \epsilon>0, \forall x_0 \in \mathbb{R}$ there exists $\delta>0$ such that $\forall x \in \mathbb{R}$, $x \in A \cap (x_0-\delta,x_0+\delta)\setminus \{x_0\}$ implies $|f(x)-l|<\epsilon$?
 
leslie townes
leslie townes
frieren: in the first statement you wrote, the [implicit] understanding is maybe that x_0 is an element of A, or at least an accumulation point of A, but there's no inherent need for quantification across a set of x_0
 
EE18
EE18
@Frieren It depends on the statement. But in general, when just talking about the definition of limit, $x_0$ is free
 
leslie townes
leslie townes
that's just a statement depending on x_0 and l that might or might not turn out to be true
if you wanted to express "lim_{x to x_0} f(x) exists", then that would be [exists l in R] [the definition of lim_{x to x_0} f(x) = l], where again this is just a statement that might or might not be true for x_0
but i don't see what you've first written above as automatically requiring that you go in that direction
the definition just gives you a language for talking about these things called limits, it's sort of up to you what you want to say in that language
 
EE18
EE18
8:58 PM
As regards fields, infinite and characteristic 0 are different. Characteristic 0 implies infinite but the converse is not true in general right?
 
leslie townes
leslie townes
EE18: correct
 
EE18
EE18
arigato leslie
 
leslie townes
leslie townes
the field of fractions of the polynomial ring k[t] will be infinite for any field k, and have the same characteristic as k
a good example to keep in mind
algebra weirdos love that field
 
EE18
EE18
interesting. so take $K$, get its polynomial ring (a domain), then take its quotient field, and that will be infinite (basically because $\Bbb N$ infinite) but $K$ need not be. Showing that it has same characteristic as $K$ isn't as obvious to me but I'll think on that
 
leslie townes
leslie townes
i'm not entirely sure what you have in mind with 'basically because N infinite.' what i had in mind was that k(t) contains k[t] as a subring, and there are pretty clearly infinitely many things in that (maybe if you had in mind something like the specific subset {t^n: n in N}, "because N is infinite" would be one way of summarizing that argument)
one reason not to summarize that argument that way is that "because N is infinite" maybe suggests the [in general incorrect] alternative suggestion that the subring of k(t) generated by 1 is infinite (which is true or not depending on the characteristic of k)
 
EE18
EE18
9:17 PM
Oh I just meant it because of how I'm familiar with polynomials being defined: as a subset of the formal ring of power series (maps from $\Bbb \to K$) with only finitely many values nonzero
 
Soumik Mukherjee
Soumik Mukherjee
Are these two equivalent? Seems like the op reversed the order
 
Lukas Heger
Lukas Heger
another example that algebra weirdos like me love is the quotient field of the ring of formal power series k[[t]], which is denoted by k((t)), it's the field of formal Laurent series
again, always infinite and same characteristic as k
 
Soumik Mukherjee
Soumik Mukherjee
Borsuk-Ulam is stronger than Brower and according to the wiki link Borsuk-Ulam is equivalent to LS
 
EE18
EE18
BTW Leslie, way back when you were mentioning some things where Enderton goes super deep on
I assume the little philosophical interlude at the end of Ch 5 is not one of them?
talking about "what is two" and stuff like that, and intentionality vs extensionality for defining a property
 
John Zimmerman
John Zimmerman
10:23 PM
Consider the bounded noncompact region $$ R=\big\lbrace t/x : t\in(1,3) \big \rbrace $$

for $x>0$. Define $\partial R=\lbrace1/x \cup 3/x\rbrace.$ Is the qotient space $R/\partial R\simeq S^1\times \Bbb R$?
 
Jakobian
Jakobian
@JohnZimmerman $\partial R = \{1/x\}\cup \{3/x\}$?
 
John Zimmerman
John Zimmerman
@Jakobian yes yes
 
Jakobian
Jakobian
@JohnZimmerman $\partial R$ is not a subset of $R$
 
John Zimmerman
John Zimmerman
@Jakobian you're right i just relaized that I wrote that open set incorrectly
should be $$R=\big\lbrace t/x : t\in[1,3] \big \rbrace$$
 
Jakobian
Jakobian
$R/\partial R\cong S^1$
 
Lukas Heger
Lukas Heger
10:30 PM
in that case $R/\partial R$ is homeomorphic to $S^1$
the $x$ and the fact that you took [1,3] instead of [0,1] doesn't change anything topologically
 
Jakobian
Jakobian
Another way to see that it can't be homeomorphic to $S^1\times\mathbb{R}$ is that the latter isn't compact, yet $R$ definitely is
 
John Zimmerman
John Zimmerman
10:48 PM
@LukasHeger that makes sense. If I take $R_k=\lbrace t/x: t \in [k,1/k]$ for all real $k \ge 1$ and take the union $\bigcup_{k \ge 1} R_k$. Gluing the boundaries like $\partial R_k=\lbrace k/x \rbrace \cup \lbrace \frac{1/k}{x} \rbrace$. Am I correct to conclude that $\bigcup_{k \ge 1} R_k \cong \Bbb B^3$?
 
Lukas Heger
Lukas Heger
I don't quite understand what and how you are gluing
 
John Zimmerman
John Zimmerman
oops I mean that $ \bigcup_{k \ge 1} R_k/\partial R_k \cong \Bbb B^3.$
 
Lukas Heger
Lukas Heger
I'm not sure if that union makes any sense. To take a union, you have to embed all the spaces you're taking the union of into a larger space first
(unless we're talking about the disjoint union, but that's another story)
 
John Zimmerman
John Zimmerman
disjoint union yes
 
Lukas Heger
Lukas Heger
well a disjoint union of at least two non-empty spaces will never be connected
but $\Bbb B^3$ is connected
 
leslie townes
leslie townes
10:55 PM
@EE18 no, although that is definitely another example of stuff that is in there without (imvho) any clear reason to be in there
 
John Zimmerman
John Zimmerman
@LukasHeger you've given me some things to think about. I was trying to define densely nested spheres that each share exactly 2 antipodal points. I was trying to take this collection and say that the union of the shells (leaves) generates the 3-ball. Do you know what "nested spheres each sharing the same 2 antipodal points" is homeomorphic to?
 
Lukas Heger
Lukas Heger
11:11 PM
no idea
 
John Zimmerman
John Zimmerman
I think it is some sort of wedge sum of spheres
 
Lukas Heger
Lukas Heger
if are you thinking about sphere packings? I'm not familiar with the terminology here
 
John Zimmerman
John Zimmerman
I will daw a picture
Hawaiian earring
In mathematics, the Hawaiian earring H {\displaystyle \mathbb {H} } is the topological space defined by the union of circles in the Euclidean plane R 2 {\displaystyle \mathbb {R} ^{2}} with center ( 1 n , 0...
 
Lukas Heger
Lukas Heger
11:26 PM
well, the Hawaiian earring is a famous pathological space. I doubt that it can be written as a wedge sum of spheres
 
John Zimmerman
John Zimmerman
 
Lukas Heger
Lukas Heger
what is your question exactly about that space?
 
John Zimmerman
John Zimmerman
It's a two point compactification of the union of a family of disjoint open intervals?
 
Thorgott
Thorgott
the Hawaiian earring is definitely not a wedge sum of spheres
 
Lukas Heger
Lukas Heger
hmm
it should be a two-point compactification yeah, if you take open intervals
 
John Zimmerman
John Zimmerman
11:35 PM
yeah gotcha - so I was trying to describe the surface version of this above
 

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