I like ramification sets that are diffeomorphic to $S^1$

huh those tame and wild ramification are notions in algebra

darn!

at least I learned something

Yesterday, our prof was sayin, that to check whether a subset S, of a Vector Space is a subspace , we need to ensure three things:

1. The zero vector should be in S

2. The addition in the elements of S should be closed

3. Scalar multiplication in S should be closed

But I think, the condition 1 is unnecessary.

you could replace 1 with a requirement that S be nonempty, but it isn't unnecessary. the empty set satisfies (2) and (3) but is not a vector space. (the vector space axioms separately require nonemptiness)

Your phrasing is poor. The set is closed under an operation. The operation itself is not closed.

My assertion is, " A subset S of a Vector space V is a subspace iff the subset S is non-empty and is closed under addition and scalar multiplication"

@leslietownes I am sorry missed it. He gave us 4 conditions and non-emptiness was the first one

is a subspace a vector space?

so what is "required" to be a vector space then?

@D.C.theIII the axioms of a vector space should be satisfied

dc yes. the point is that inside a 'known' vector space one does not need to check most of the axioms, because if they failed in the subset they would fail in a known vector space. hence, criteria for 'checking a subspace' are simplified.

I agree leslie. I was just trying to emphasize the need for the additive inverse (fancy way of saying $0$), but you know what I mean

roughly they are just nonemptiness, and that the 'known' scalar multiplication and addition operations also give operations on the subset

@D.C.theIII A vector space is a non-empty subset generally defined over a field F and the vector space is endowed with two operations addition and scalar multiplication that satisfies some conditions

@TedShifrin Can you comment on my assertion:

@leslietownes You may give a comment as well

ugh...I meant additive identity...........

I don’t understand what the issue is. Containing $0$ is equivalent to non-emptiness. The infinite list of axioms need not be checked because they are known to hold in the big vector space.

@TedShifrin ok

But I think we can take my proposed assertion as a valid verbatim?

Is your fuss over the use of iff as opposed to if?

You are often obscure in your questions.

No, but can u explain if say, : W is a non-empty subset of Vector space V over F, and W is closed under addition and scalar multiplication, then, 0 is in W?

I am now confused unexpectedly

That's a perfect question to practice writing a proof

Oh, if it’s nonempty, then there is some $v$ in it. Closure under scalar mult gives you $-v$, and $v+ (~v)=0$.

You never give me answers so quickly.............😢

One step here is subtle, but we discussed it a few days ago.

Maybe 2+ years ago, I did.

Nah....I was always made to struggle.....the trauma is forever present

Good!

triggers cold sweats having to think about the Lebegue number

@TedShifrin I think you mean, as scalar multiplication is closed, we have, as $v\in V$ and $0\in F$ so, $0v=0\in V$ as well, right?

Btw, I understood, it's just that we can say the same things in multiple ways.

Oh, that’s even easier. Still pay attention to the way you use “is closed.”

@sunny Sorry, I had a number of things to do offline. Here is another approach.

There are only 5 in the room currently.

wonder if that's some network thing or if it's really just us.

@Jakobian Did they prove earlier that $\alpha$'s a repeated root of $f$ iff it's a root of both $f$ and $f'$ (that is, $f(\alpha)=f'(\alpha)=0$)?

If not it's worth trying yourself, anyway

@AkivaWeinberger I was wondering the same thing.

Nope. It's after they introduce the derivative

msg, $\mathcal{M}(S_g)$.

@Jakobian That doesn't seem to have anything to do with what Akiva was asking. Was this in response to their question?

@robjohn wdym?

it does have to do with what they were saying

they introduce the derivative and jump into the proposition

$\alpha$ is repeated root iff $f(\alpha) = f'(\alpha) = 0$ should have been proved there

Of course it would have to be after they introduced the derivative because Akiva is mentioning both $f$ and $f'$

well maybe I've didn't phrase it well

So what you are saying is that it is just after they define the derivative, with nothing else said?

I should have said it's right after they introduce the derivative

Context is important

well, they introduce few standard identities for derivatives, saying they leave it to the reader

product rule, and so on

but nothing else

the proof should have been included in the proposition but someone didn't pay attention it seems

The thing that Akiva was mentioning can be proven using the product rule, but if they don't show it, then something is missing.

I mean I'm fully aware of that, it's just weird to me that they mentioned so many things (in my opinion the proof doesn't even need all that), but in the end they didn't conclude anything

I'm fully aware that you can prove it this way, I mean

I assume what happened is someone looked at it, and because it was so lengthy, they didn't realize it's incomplete, and the editorial board somehow didn't notice either

I would assume that also and move on.

I already did

coolio

can you smoothly interpolate a surface between constant sectional curvatures -1,0,1 where for all time each surface is of constant sectional curvature?

so at $t=0$ you have Euclidean plane at $t=1/2$ you have a surface of constant sectional curvature equal to $1/2$ etc.

I feel like you might need computer simulation to find the answer

what is this function on LHS?

is this cardinality?

@TedShifrin in your book you had an exercise about the regular p-gon being constructible. You gave the hint to consider what happens to $[\mathbb Q(e^{2\pi i/p}):\mathbb Q]$ if $\cos(2\pi/p)$ were constructible. Is what you had in mind just the argument that if $\cos(2\pi/p)$ were constructible, then using the field operations and $\sqrt$, you can construct $\cos(2\pi/p) + i\sin(2\pi/p)$, and thus the degree of the extension should be a power of 2, but it's not (it's power $p-1$)?

Or did you have something less handwavey in mind?

@mick Depending on which conditions?

@anak You're messing this up. The conclusion is that $p=2^t+1$; there's not a contradiction. There is an earlier exercise about constructing an angle iff both sin and cos are constructible.

@Koro from the context given I'm guessing this says that the sequence $\gamma n$ is uniformly distributed mod $1$ for irrational $\gamma$, and that's indeed cardinality

but that's only a guess... together with $\langle r\gamma\rangle$ denoting the fractional part of $r\gamma$

Since we have $1\le r\le n$, I think there's no choice than for this to be about integers, @Jakobian, as you suggested.

@Jakobian it's indeed cardinality. It became clear to me once I looked further-into the proof of the statement.

@Jakobian compact set instead of compact in your recent edit

ah well sure

@geocalc33 Yes. Work on the unit disk. You can pull back the metric on a sphere of any radius to that disk, and there is likewise a hyperbolic metric of any desired curvature on that disk. If you write it down, you can get a parameter $\beta$ in the metric (thinking of $|dz|^2/(1+\beta|z|^2)^2$), and curvature will be something like $4\beta$.

@TedShifrin Sorry, I was thinking about the contrapositive simultaneously.

Sometimes unnecessary contradictions just clutter the landscape.

Here’s the solution @robjohn

@TedShifrin But the point is that you can explicitly come up with the tower of extensions, and each has (relative) degree of a power of 2?

@anak Absolutely NOT. This is if, then, not iff. You need Galois theory to decide when the $n$-gon is actually constructible.

@TedShifrin I think you misunderstand. I am saying that if you suppose $p-1$ is a power of 2, then you can construct the explicit tower with in it $e^{2\pi i/p}$. Thus showing that the $p$-gon is constructible.

No, I do not misunderstand. You do.

This is a necessary condition, not a sufficient one.

Gauss actually figured out precisely which gons are constructible. This appears in further exercises in the Galois theory section in chapter 7.

That's if you are judging by the index alone, no? I thought if you had a tower (which for me I thought was defined as relative degrees being 1 or 2) then the statement is equivalent.

That is, $\alpha\in\mathbb C$ is constructible if and only if there are field extensions $\mathbb Q(\alpha) = \mathbb K_n \supseteq \dotsc \supseteq \mathbb K_0 = \mathbb Q$ such that $[\mathbb K_i:\mathbb K_{i-1}]$ is 1 or 2 for each $1\leq i \leq n$.

There are field extensions of degree 4 that are not obtained by consecutive quadratic extensions.

Yes.

Yes, what you just said is of course what I proved in the book. But degree alone won't do it.

I am not saying degree alone will do it.

I am saying you can explicitly make the tower.

Thus, saying $p-1=2^t$ is necessary but NOT sufficient. It sure looks like you were saying it.

You cannot.

At least, I don't remember that as being so.

I gave Gauss's result at the end of section 5.2.

I guess what you said is in fact correct, according to Gauss. But it takes work. Sorry.

If I now think about, I guess the point is that the Galois group is going to be abelian, and so the tower will follow from Galois theory.

I’m bored

I don't understand how people here will downvote a question without explain.

I've gotten tired of commenting "do not just post your homework with zero effort" ... but I vote to close, not downvote.

I only downvote questions/answers after I explicitly complain about something serious in comments and get no response for a day or so.

What specific question are you talking about?

@冥王Hades the question asks about $\angle BAO$ not $\angle NOC$, does it not?

Is overrall I see people downvote and don't explain HHAHAA.

how's everyone doing?

Not helpful, EM4.

@robjohn correct. However I proved that $\angle BAO=\angle NOC$. Can you see it or should I elaborate?

Sometimes I downvote because I just don’t like the solution

You should comment, then, on ways to improve it. Penalizing someone for being correct but having taste different from yours seems rude and inappropriate.

why do these limits go to 0 & 1 resp. as r tends to \infty?

<,> denotes fractional part here.

I am not doing well wither @TedShifrin.

Sorry to hear that, @EM4.

I am taking gap year of graduate school from personal and family problems.

Graduate school is arduous even when there are not external stresses.

@geocalc33 It occurred to me while taking my daily walk after posting earlier that we obviously need to restrict to $\beta>-1$, but if you insist on $\beta$ more negative than that, you have to shrink the disk.

@TedShifrin sometimes I do, but it takes too much effort sometimes.

Well, I don't think you should downvote something correct just because you "don't like it."

Well I do concede on one point, I am a bit of a brat

This comes as a shock to me. I would delete "a bit of."

I knew you’d say that

So uh, is trumpet getting jailed or not?

@robjohn in your answer, don't we require $x$ to be strictly less than $1$ and greater than $-1$. Suppose $x$ is $1$, then $\log (1/x)=0$, which doesn't tend to infinity. What am I missing?

Moreover, I don't see how you obtain $H_n\geq \log(n+1)$ from $\frac1{k}\geq\log(1+\frac1{k})$. From $\frac1{k}\geq\log(1+\frac1{k})$, I get $H_n\geq \sum_{k=1}^n \log(1+\frac1{k})$.

i was tempted to downvote the other day, but i ultimately didn't. like ted, i do vote to close a lot, and without comment if the reason is clear from MSE's own description of the grounds for closing.

i see voting to close as directed more at people who might answer, and not really at the asker so much.

I’m probably gonna fall asleep during one of the lectures.

i will reveal to you my ultimate secret technique, which I use when I can't risk drinking more coffee because i might not sleep until morning

get chewing gum and chew through the lecture

@algbr its complicated and not my expertise ... or simple and not my expertise but complicated for me

it would be pretty epic to interrupt a lecture by falling asleep and choking on your own gum

is it the fear of that particular event or the chewing that does the magic? who knows but it works

about a month ago my daughter came home from day care and said that you die if you swallow chewing gum. i asked if she had ever even tried chewing gum. she said "yes," but was unable to answer follow-up questions such as "when" and "what kind," which suggested to me that maybe she just wanted to be seen as worldly and knowledgeable.

now she knows the truth.

@mick But generally you are interested in number theory too, right?

The Munchkin doesn’t land far from the apple tree.

now, she likes eating chewing gum. she never went through an interim period of chewing gum while being afraid of swallowing it.

after having it the other day, she told her mom that she had turned into a ghost.

i had a doctor's appointment two weeks ago and i was chewing gum and i swallowed it because they called my name and were expecting me and i didn't know what to do

@shintuku I’d rather a lollipop

i maintain any upstanding citizen would have done the same, and must disdain sticking it under the chair like a barbarian

a quick google search revealed that it would not in fact remain in my digestive track until the decomposition of my body

@leslietownes i recall swallowing chewing gum one day just to see if I’ll die, I was happy I didn’t but surprised that it had 0 effect

@algbr yes

i believe number theory to be the foundation of math , not set theory

although that opinion might be a heritance

maybe it is more philosophy but my mentor/friends considered number theory and calculus as the real foundations ...

he gets angry from set theory

@mick Sounds interesting. Perhaps at least arithmetic is more natural than set theory.

set theory only relates to set theory and is just a matter of what axioms you accept

I mean things such as robinson or presburger airthmetic.

it has no applications

outside of itself

that is infinite set theory beyond the cardinality of the functions

Set theory has some applications in computability theory.

i guess it makes him oldfashioned

only the diagional and some fourier applications apply to him

cantors diagional i mean

The fact that some functions are not turing-decidable can be inferred from the uncountability of $\mathbb{R}$

that is cantors diagional

Yes.

other applications than the diagonal and fourier type are rare

outside set theory itself

set theory is " too arbitrary " in his mind. you can choose axioms and get the related results ... but so what ? you get what you put in

and the open problems are just " results from poor definitions "

thats very controversial i know

but if you ask a question in set theory , its like : what definitions and axioms do you use ?

But one can see lawveres fixed point theorem as the general reason behind cantors diagonal argument.

its not like that in number theory

@algbr yeah kinda i guess. like space filling curves over a surface have a higher cardinality then the space itself

but the fuss about integers vs reals is really not that shocking

@sunny $x$ is tending to $0$, why are you worrying about $x=1$?

i mean

Is there any progress in the attempt to use number theory as the foundation of mathematics?

I never heard about it.

@sunny Compute the sum. The sum is $\log(n+1)$.

the set with numbers with finite decimals is countable , a set with numbers with infinite decimals is bigger. just like infinity is bigger than finite.

the set*

@Koro $u_r$ is an integer, so $\{((1+\sqrt{5})/2)^r\} = \{-((1-\sqrt{5})/2)^r\}$

and without assuming a well-order most things are very debatable and axiom dependent

whats worse : thinking about cardinality B does not help for problems within cardinality A , if B > A

so what is the fuss ?

This is a theorem. If I take $\gamma> b$, then how does this theorem hold?

(it fails. doesn't it?)