Mathematics - 2024-04-23

12:13 AM

Today I reminded that the free product $G * H$ has $G$ and $H$ as normal subgroups.

Though, that raises a question... The free product $G * H$ has the direct product $G \times H$ as a factor group. What is the corresponding normal subgroup?

Jakobian

@DannyuNDos If $x, y$ are generators of free group on two elements then does $yx\in \langle x\rangle \cdot y$?

It doesn't because then $yx = x^ny$ for some $n\in\mathbb{Z}$ which leads to a contradiction

Dannyu NDos

Huh, nevermind.

Jakobian

so $G = \langle x\rangle$ and $H = \langle y\rangle$ are not normal subgroups of $G*H$ their free product

Dannyu NDos

The source of confusion was this: Given a group $G$ and its subgroup $H$, I tried to find the pushout of the identity map $\mathrm{id}: H \to H$ and the inclusion map $i: H \to G$.

I thought the result would be $(G * H) / H$.

Which is... isomorphic to $G / H$, in my second thought?

And still, $G * H$ should have $G$ and $H$ as factor groups.

Wait, $G / H$ isn't a thing; $H$ might be not normal. Dang.

Thorgott

12:36 AM

pushing out along an identity map does nothing

Dannyu NDos

12:59 AM

D:

Semiclassical

2:00 AM

trying to decide if the following sentence is grammatical: What radius circle will a particle (mass m, charge q) follow if moving at speed v in a magnetic field B?

specifically the "what radius circle" bit

leslie townes

it seems grammatical to me, but i have no idea. and, there is sometimes a gap between grammatical and good exposition.

this is something ted could have answered, because when he went to school they still taught grammar

Semiclassical

yeah, grammatical != "reads well"

(i'd hope that being grammatical is a necessary if not sufficient condition but i'm not confident of that)

leslie townes

part of the question is an implicit assertion, which is that a particle under those circumstances moves in a circle. if the point is that you don't want the student to prove/investigate/focus on that assertion, i might just state that fact - a particle doing bleh will move in a circle. now you, student, tell me the radius of that circle.

Semiclassical

yeah, they're given the formula r=mv/qB on the equation sheet, and this is a multiple choice question

robjohn

@Semiclassical it's not $\frac{mv^2}{qB}$?

leslie townes

2:06 AM

in the last physics class i took, i think we did some kind of experiment relating to this equation to compute the mass of an electron.

Semiclassical

no. mv^2/r = qvB, hence r=mv/qB

i'd say something about units but that would require me thinking about Tesla as a unit

robjohn

ah, forgot that qvB was force

Semiclassical

and life's too short for that

Xander Henderson

2:37 AM

@robjohn ur mom is force!

Dannyu NDos

4:42 AM

Which branch does K-theory belong to, algebra or topology?

leslie townes

it began in algebraic geometry, in grothendieck's formulation of the riemann roch theorem, but topological K came pretty soon after

why not both

Dannyu NDos

So it belongs to geometry then.

leslie townes

arxiv.org/abs/math/0012213

Dannyu NDos

Too sadge geometry isn't broad enough to be considered the 4th branch.

Analysis, Algebra, and Topology are the three branches, but geometry?

Lucky Chouhan

4:59 AM

@leslietownes you're pretty good at finding stuff on the internet. Are you a detective?

leslie townes

nope

Lucky Chouhan

@leslietownes do you have PhD??

leslie townes

where is all of this going. what do you really want to know?

Lucky Chouhan

@leslietownes your profession? But you're not telling me :(

leslie townes

i am a lawyer

i guess i forgot to answer that

Lucky Chouhan

5:06 AM

@leslietownes so you have done math degree then you studied law?

leslie townes

yeah, why not

Lucky Chouhan

I think, you spend more time here than in court haha no offence :)

leslie townes

i don't know where you live or how the court system works there, but in the united states, it isn't exactly a badge of honor to have one's clients appearing in court all of the time

there's a good joke in the US about a country lawyer who says "i'm the greatest lawyer, every will i have written has been upheld by the supreme court"

the underlying premise of this joke is that you are f-ing up badly as an attorney if anything you do leads to a supreme court case

Soumik Mukherjee

5:55 AM

I can't find any wiki or nlab articles on tame sets and wild sets. Can someone suggest some references where I can find the definitions and basic stuffs about those?

leslie townes

i never saw such things, what field do they arise in?

Soumik Mukherjee

3 manifolds. I don't know if they appear in other fields or not.

Jakobian

8:49 AM

@SoumikMukherjee @BalarkaSen

Sounds like parts of what you study

one potato two potato

topologically tame 3-manifold is a 3-manifold whose interior is homeomorphic to a compact 3-manifold. wild otherwise

I mean interior of a compact 3-manifold

Jakobian

10:28 AM

@onepotatotwopotato does 3-manifold mean manifold with boundary here?

one potato two potato

yes possibly nonempty boundary

Soumik Mukherjee

12:19 PM

@onepotatotwopotato Thanks

YourLordJoyBoy

1:35 PM

@onepotatotwopotato MASHED POTATOES! SWEET POTATOES! BAKED POTATO! POTATO CHIPS!

YourLordJoyBoy

2:31 PM

Um, what would be the opposite of $3y^2-2xy+6x+2$?

EE18

Hoping to get some help with the following problem:

I've found this and intend to follow the proof: math.stackexchange.com/questions/791411/…

Thus my main question for the chat here is why would my text include the hypothesis about there existing $a,b$ such that $a<b$ and $a<f(a)$ and $b<f(b)$ (it does not appear in the linked question)?

My suspicion is that the existence of these $a,b$ is already implied by the other facts and so is redundant but perhaps there in order to have me do a little less work?

If so, what I am hoping for is help in showing how the existence of such $a,b$ is indeed implied by simply the fact about $f$ being an increasing function from $\Bbb R \to \Bbb R$ (and proeprties of $\Bbb R$, naturally).

Soumik Mukherjee

3:01 PM

@EE18 Otherwise it is not true, f(x)=x+1 is an increasing function without any fixed point

Also you mean b>f(b)

EE18

I do indeed, yes

Thanks for the comment Soumik

What then am I missing from the linked answer? The hypothesis seems not to be there?

Soumik Mukherjee

@YourLordJoyBoy what do you mean by opposite?

@EE18 That each non empty subset has glb and lub

Which is not true for R

EE18

Ohhhhh

oooofffff

Soumik Mukherjee

So they added that condition

EE18

Boundedness, of course

Thanks for setting me straight, much appreciated

Soumik Mukherjee

3:03 PM

Welcome

EE18

I was wasting lots of time trying to prove that hypothesis unnecessary

Another question for folks

I've just proved Bernoulli's equality $(1+x)^n \geq 1+nx$ for $x \geq -1$ by induction but (as with most induction proofs, they feel mechanical) don't have good intuition for it

I vaguely recall Baby Rudin using it and "proving" it via the binomial theorem somehow

Something to effect of $(1+x)^n = \sum_{k = 0}^n{n \choose k}1^kx^{n-k} = \sum_{k = 0}^n{n \choose k}x^{n-k} ... \geq 1+nx$

But I can't see how one would go through the dots here

Any tips possible?

I guess I want to be able to show that $\sum_{k = 0}^{n-1}{n \choose k}x^{n-k} \geq nx$ but can't quite see how. I guess this leads to induction again so I'm in no better place

ZaWarudo

3:29 PM

Studying the local extrema in $\mathbb{R}$ of $f_{m,n}(x)=x^m(1-x)^n$ for positive integers $m$ and $n$, I noticed that $f_{m,n}(x)=f_{n,m}(1-x)$. Can I use this information to assume something and simplify the study?

Thorgott

4:16 PM

@EE18 if $x\ge0$, all terms are positive and you're just omitting all but $2$ of them and it's trivial

not sure this perspective is helpful for $-1\le x\le0$

Jakobian

$(1+x)(1+y)\geq 1+(x+y)$ for $x, y\geq -1$ is a very simple observation to make

induction seems to be the most natural

Obliv

What are some very obvious situations where every subgroup of a group is normal

I thought that the alternating subgroup was one but apparently not

in this definition of a near-ring spawned from this answer, how can a group be non-abelian under addition

addition is by definition a commutative binary operation

Alek Murt

4:38 PM

Hey guys I want to clarify something so If someone could help me would be great

For a function $F(U, x, t): \mathbb{R}^2 \times \mathbb{R} \times[0, \infty) \rightarrow \mathbb{R}^2$, the following inequality is proving that $F$ is Lipschitz with respect the variable x
$$
\|V\|_{\infty} \leq M \text { and }\|W\|_{\infty} \leq M \Rightarrow\|F(V, \cdot, t)-F(W, \cdot, t)\|_{\infty} \leq k_3\|V-W\|_{\infty}.
$$
have in mind that $U:\mathbb{R} \times[0, \infty) \rightarrow \mathbb{R}$. Can someone helpe with this please, since the $\cdot$ are confusing me or is the inequality only proving Lip…

Thorgott

@Obliv if the group is abelian

another example is the quaternion group

and IIRC you can construct every group satisfying the property from these example in some sense

@Obliv theirs isn't

Obliv

who is theirs? Why use the same word "addition" if it means something else

Thorgott

words typically don't have a single meaning

Jakobian

@Obliv commutative

Obliv

@Jakobian what do you think? Should addition have one meaning in abstract algebra*?

Jakobian

4:46 PM

@Obliv near-ring is a more general of a concept than a ring

in this case addition just means another operation on a ring, as to distinguish it from multiplication

@Obliv yes we shouldn't be possessive of our nomenclature when it doesn't make sense to be

Obliv

why call a potato a tomato? We're talking about binary operations on sets

Jakobian

Why call hydrogen oxide simply water?

you see why are you being ridiculous?

Obliv

Not really, it's taught in primary school that addition is commutative. I think it makes a hell of a lot more sense to just call the operations something else that reflects their freedom to be more than addition & mult.

Jakobian

its taught in university that singletons are closed

but thats only because they are discussing metric spaces

why are you being so possessive of this?

we can still call it the same even if concept is more general

and doesn't have all the properties we were told before

Thorgott

in primary school you are taught addition of natural numbers

Obliv

4:55 PM

well yeah, we're not going to be discussing matrices/permutations at that level

or whatever non-commutative stuff is going on in abstract algebra

Jakobian

You weren't protesting that binary operation on a group is called multiplication from what I know

Obliv

out of curiosity, what are some non-commutative addition situations for near-rings

Jakobian

thats generally non-commutative

Obliv

@Jakobian I was actually about to :D

Jakobian

so why are you objecting to near-rings but not groups?

Obliv

4:57 PM

I mentioned to thorgott a couple weeks ago that I wished we had more symbols for each operation on sets that met their respective axioms

Jakobian

Are you objecting to cross product too?

Obliv

en.wikipedia.org/wiki/… like the table here for groups, we could develop some sort of babylonian symbols lolol

Jakobian

God forbid we call it product

Obliv

mathshistory.st-andrews.ac.uk/HistTopics/Babylonian_numerals

@Jakobian yes, throw it all out

robjohn

@Jakobian Hydrogen Hydroxide or Hydroxic Acid

Obliv

4:59 PM

do it properly, make everything abstract :D

Jakobian

shrug Do whatever you want, I'll still call it product or multiplication, or addition of near-ring an addition

Thorgott

well, there are plenty of good reasons why you don't see this opinion shared among professionals

Odestheory12

Hello. Any idea how to solve the following limit? $$\displaystyle\lim_{(x,y) \to{(1,1)}}\left ( 1-\dfrac{\log^2(1+x-y)}{\sin(x-y)}\right)^{\dfrac{2}{(x-y)^3}}$$

Jakobian

Point of abstractifying things is to relate them to known concepts. So if you're trying to strip that familiarity away, you're doing it wrong

Odestheory12

I am not sure if i can do the subst z=x-y and compute the one variable limit when z goes to 0

Jakobian

5:04 PM

@Odestheory12 yeah you can

robjohn

@Odestheory12 since $x$ and $y$ are only seen in the expression $x-y$, you can definitely use $z=x-y$ and evaluate $\lim\limits_{z\to0}$

The only problem is that when $x=y\ne1$, that is you are approaching on a diagonal, you are using $z=0$, and the expression does not have a value on that diagonal, so the two variable limit won't exist on that diagonal.

Jakobian

@robjohn I'm not sure if its right to say that the limit won't exists on that diagonal, we simply don't consider those points since the function is not defined there

robjohn

5:23 PM

@Jakobian Look at the definition of a limit, the function is supposed to exist in a neighborhood of the point in question, so you need to explicitly ignore that diagonal.

Jakobian

definitely not required

leslie townes

this is that old split between continental people and calculus books

Jakobian

or see here

leslie townes

clearly, it's impossible that different conventions exist for such an everyday object

Obliv

i think u linked the same thing

Jakobian

5:33 PM

yes because the link didn't update when I clicked on it

Obliv

@leslietownes I literally just vented about this. I don't think there should be so many "conventions" for math. looks around AM I WRONG?

Maybe I am..

leslie townes

obliv: hahaha the static around limits is one of the most annoying forms of this

there are only a few that genuinely annoy me as opposed to being "ok, whatever"

Jakobian

@leslietownes Obliv is objecting to calling operation of a group multiplication, to calling cross product a product and to calling addition of a near-ring an addition

Obliv

the 2nd one of those follows from the first

Jakobian

Is what he was venting about

leslie townes

5:35 PM

i'm completely at peace with 0 maybe or maybe not being a natural number, for example, to the point that i forget which alternative i prefer

jakobian: those objections are interesting because of how novel they are. i kind of like them

Jakobian

@leslietownes I don't know, if in your definition of a limit function is supposed to exist around the point $(1, 1)$ then clearly its ill applicable to the current problem. So its not a conventional issue imo

Obliv

@Jakobian I do like this, but I think it also obfuscates what you're actually saying by making things sound similar (or in this case identical)

leslie townes

yes, you would have a strongly held opinion that your opinion isn't even an opinion :)

i kind of like the idea of not calling the operation in a group a multiplication

i don't like the other ones, but i like that one

where does it stop, though? is there a limiting principle? can you use juxtaposition to denote the operation in an abstract group

Jakobian

38 mins ago, by Jakobian

Point of abstractifying things is to relate them to known concepts. So if you're trying to strip that familiarity away, you're doing it wrong

while there is nothing wrong in not calling a group operation "multiplication", objecting to others doing so is ridiculous

Obliv

wdym by that? I'm just saying don't call $+,\cdot$ the operations for a structure if they don't mean what we mean for $\mathbb{C}$ (I use $\mathbb{C}$ since $\mathbb{R},\mathbb{N},..$ are all contained within it)

actually I shouldn't use uncountable sets

I heard just using $\mathbb{R}$ means ur talking about topology

leslie townes

5:43 PM

that's sort of what i'm concerned about. if this leads me to having to use funny notation for the group operation, i can't be on board with it

i like the background philosophical position, however

Jakobian

@Obliv not correct

and not what the post is saying

Obliv

@Jakobian How is that ridiculous? Why would I call it something else just for myself? lol

Jakobian

read

leslie townes

the question of how much structure comes along with a single letter is an interesting one, but not (as far as i can tell) one that is raised by that post

Obliv

well in this context I'm using $\mathbb{R}$ to mean a group under $+$ or $\cdot$ which the user said involves topology

Jakobian

5:46 PM

HE ISN'T SAYING THAT

Xander Henderson

@Obliv No, that isn't what the answer says.

Thorgott

we should make a list of all groups and invent a different notation for the binary-operation-that-I'm-not-allowed-to-name for every entry individually

2

Xander Henderson

The answer says that you can generate a subgroup in a particular way if you are only considering the algebraic structure (e.g. the additive group structure).

robjohn

@Jakobian In Wikipedia it says $$\lim_{\substack{x\to p\\x\in T}}f(x)=L$$ For every $\epsilon\gt0$, there exists a $\delta\gt0$ such that $\color{red}{\text{for all }x\in T}$, $0\lt|x-p|\lt\delta$ implies that $|f(x)-L|\lt\epsilon$ (I added the red for emphasis).

Xander Henderson

But if you want $\mathbb{R}$ to have the structure of a topological group, then the same object generates the entire thing, as a topological group.

robjohn

5:48 PM

So $T$ must avoid $x=y$

Jakobian

@robjohn Yes, so you want to let $T = \mathbb{R}^2\setminus \{(x, x) : x\in \mathbb{R}\}$

Xander Henderson

And even that explanation is deeply caveated.

Jakobian

I was being pedantic

Xander Henderson

@leslietownes IT ISN'T A NATUAL NUMBER!!!!!

PISTOLS AT DAWN!

Thorgott

i WILL fight you

Obliv

5:51 PM

@Thorgott u only need one notation and u just have to specify what it does on the set

Xander Henderson

@Thorgott The gloves are off, man! Let's go!

Obliv

(assuming u dont wanna go for my babylonian symbols idea)

Jakobian

I've been both considering and not considering $0$ to be a natural number in the past

Obliv

like $(G,\star)$ and $(R,\star_1,\star_2)$

Jakobian

but now I just assume both based on what's more convenient for the moment

Xander Henderson

5:53 PM

The Natural Zeroists will be the first up against the wall when the Revolution comes.

leslie townes

thank you, xander, for reminding me what the correct position is

Xander Henderson

Heh.

Glad to help.

Obliv

and only say $(\mathbb{N},+)$ when u mean addition. Use something else when u mean a near-ring or a non-abelian group

Xander Henderson

@Obliv Meh. + is a single chord on the keyboard, while \star is five keystrokes. Ain't nobody got time for that.

Obliv

actually + is two if u count shift

i dont use numpad much

Xander Henderson

6:00 PM

@Obliv It is a single chord.

Thorgott

what kinda keyboard does not have a + key??

Obliv

oh i see what u mean, i thought u mean stroke

i thought most had + be the shift+=

shift with = **

Xander Henderson

@Thorgott The + key is all the way over on the numeric keypad. If you don't want to move your finders from home, it is a chord, i.e. [shift]+[=].

Thorgott

oh damn, QWERTY is wild

Xander Henderson

(On the American QWERTY).

Obliv

6:01 PM

what do u use..

Thorgott

+ right next to enter

this is a culture shock moment for me

Obliv

WHAT... is that standard

it is for me too

Jakobian

@Thorgott I have none of those symbols

Xander Henderson

@Thorgott LEARN TO TYPE AMERICAN, JACKASS! :P

Obliv

its bizarre seeing y and z swapped

We're so different..

@Jakobian so is it not a european standard then

Thorgott

6:04 PM

@XanderHenderson no I will commit an offence

@Obliv QWERTZ is the German standard

Xander Henderson

@Thorgott We're already scheduled for pistols at dawn. I am not sure how to respond to this new offense.

Obliv

my mind is being blown at the thought of each country having their own standardized keyboard. I guess binary operations are the least of my concerns now

I'm out of my depth

same

Same

6:06 PM

why is everyone posting communist keyboards to the chat all of a sudden

Xander Henderson

@leslietownes 'Cause we're being overrun by commies, no doubt.

Jakobian

we just use the American keyboard with right alt for special letters

leslie townes

@XanderHenderson [sound of shotgun loading]

Soumik Mukherjee

We all got scammed, where are those extra 50 keys in our boards?

Jakobian

@leslietownes oh... if you want to call America a communist country shrug

Obliv

6:07 PM

imagine if germans used qwerty maybe we would have called $\mathbb{Z}$ yahlen

ok im being silly

Xander Henderson

@leslietownes WOLVERINES FOREVER!

Obliv

@SoumikMukherjee its a dlc apparently.

Jakobian

@SoumikMukherjee this is the advanced one

Soumik Mukherjee

There are 4 optional keys as well, who knows what those contain.

@Jakobian wow

leslie townes

jesus used a qwerty keyboard without any of those weird accent marks, that's good enough for me

Jakobian

6:12 PM

I've never seen the advanced one before

it looks like the standard German one but slightly more complicated

Soumik Mukherjee

Imagine having a keyboard battle against that keyboard

Obliv

Only true jedi masters know the ascii codes for all of these symbols and can bridge the worlds

Any keyboard with an alt+numpad is then equivalent just with potential for rsi/carpal tunnel

leslie townes

one of my friends did give himself a repetitive stress injury when his keyboard slowly stopped working and his workaround involved alt codes

to his defense, it is surprising how quickly you can pick it up if you absolutely need to

Soumik Mukherjee

All the other keys are so congested, yet the space key won't help them, true capitalist

Jakobian

@leslietownes keyboards are cheap though

leslie townes

6:18 PM

they are cheaper now than when he did this to himself, but, your point is well taken

Odestheory12

@robjohn hey thanks that is exactly what was making me doubt about the substitution

That*

robjohn

6:36 PM

@Jakobian Oh, no. Alt-right keyboards, now?

They are missing out on a fourth set of characters: no alt, alt, alt gr, and alt+alt gr.

Alessandro Codenotti

7:03 PM

@Thorgott on italian keyboards too (confirming this required immense mental effort since my laptop has a german keyboard)

psie

> Claim Let there be two separable metric spaces $E$ and $F$. Then $\mathcal{B}(E\times F)=\mathcal{B}(E)\otimes \mathcal{B}(F)$.

Here $\mathcal{B}$ is the Borel sigma algebra and $\otimes$ denotes the operation of product $\sigma$-algebra. In showing $\mathcal{B}(E\times F)\supset\mathcal{B}(E)\otimes \mathcal{B}(F)$, it says that it is easily verified that for a fixed open $B\subset F$, then $$\mathcal C_B=\{A\in \mathcal{B}(E): A\times B\in\mathcal{B}(E\times F)\}$$ is a $\sigma$-algebra.

I struggle with this. The empty set is in $\mathcal C_B$ since $\mathcal{B}(E)\ni\emptyset=\emptyset\times B\in \mathcal{B}(E\times F)$, but why, if $A\in\mathcal C_B$, is $A^c\in\mathcal{C}_B$?

$E\times F$ is equipped with the product topology by the way.

Jakobian

@psie $E\times B$ is Borel as its open

so if $A\times B$ is Borel, then so is $(E\times B)\setminus (A\times B) = A^c\times B$

psie

ok, thanks, I'll make a note. Is there any special property of cartesian products that comes in handy when showing that if $A_n\in\mathcal C_B$ for $n\in\mathbb N$, then $\bigcup_n A_n\in\mathcal C_B$?

I guess just distributive property

EE18

7:42 PM

@Jakobian Ya this is sort of what I did, I guess it just didn't feel like something I could connect to my grade school intuitions. Will think about what @Thorgott said here

Jakobian

7:53 PM

@psie yeah

YourLordJoyBoy

8:39 PM

@SoumikMukherjee EXTRA 50 KEYS!? :D

@SoumikMukherjee There's no hidden meaning. I mean literally the opposite.

Soumik Mukherjee

I don't know what opposite of a polynomial function means

Okay do you mean -f(x,y)?

YourLordJoyBoy

YES

Soumik Mukherjee

Just multiply by -1 then

Semiclassical

9:20 PM

opposite isn't a great word for that, at least not in english

the "negative" polynomial, maybe, but tbh i don't see a reason to use words rather than algebra for that

John Zimmerman

9:56 PM

I wish baseball season was back

oh it is back I didn't realize it had started.

Xander Henderson

10:33 PM

@JohnZimmerman college baseball is almost over.

Obliv

would it be in bad faith if for the answer to this problem

I write $\mathbb{Z}_{12}\oplus \{e\}$

I can't tell if I'd get scolded/get marks off. Personally, I think you shouldn't take marks off because the question wasn't specific..

Thorgott

10:49 PM

it's a poor question, but that answer would still be in bad faith

Obliv

Why does it say "Write $\mathbb{Z}_{12}$ as a direct sum of two of its subgroups"? Shouldn't it be "Write two subgroups of $\mathbb{Z}_{12}$ as a direct sum"

how can I write $\mathbb{Z}_n$ as a direct sum/product of anything if doing so gives me n-tuples

Thorgott

@Obliv if this is the type of text to make a big deal out of distinguishing between "internal" and "external" direct sums, that could be the reason. to a working mathematician, those two phrases mean the same thing.

Obliv

Are you saying it's okay to write $\mathbb{Z}_{12} = G_1\times G_2$?

eh I guess I should refer to another text

Thorgott

I'm saying it depends on what the text means by "direct sum"

psie

I'm a paraphrasing from Gamelin's and Greene's book:

> Let $(X_i,d_i), i=1,2,\dots,n$ be metric spaces. Let $X=\prod_{i=1}^{n}X_i$ and let $(X,d)$ be the metric space where $d$ satisfies certain nice properties, so that componentwise convergence of sequences holds. For $i=1,2,\dots,n$, let $U_i$ be an open subset of $X_i$. Prove that the subset $U_1 \times U_2 \times \dots \times U_n$ of $X$ is open and that each open subset of $X$ is a union of sets of this form.

I struggle with the second part. In the book they say it suffices to show that if $U$ is open in $X$ and if $x\in U$, then there exists open sets $U_k$ in $X_k$, $1\leq k\leq n$, such that $x\in U_1 \times U_2 \times \dots \times U_n\subseteq U$. How does this establish that $U$ is the union of open sets of the form $U_1 \times U_2 \times \dots \times U_n$?

Thorgott

11:13 PM

pick such neighborhoods for every point of $U$, take their union

psie

ah ok, I think I understand. That's a nice property open sets enjoy I guess

Obliv

is $[x,y]_n$ bad notation to denote a pair of congruence classes both from $\mathbb{Z}_n$

or is $([x]_n,[y]_n)$ better

Thorgott

tge former is wrong, the latter is correct

Obliv

also how can I write $\langle a\rangle \pmod n$ for a cyclic subgroup of $\mathbb{Z}_n$