Mathematics - 2017-03-22 (page 3 of 6)

Nope.

Tim The Enchanter

Oh great then.

Are you familiar with probability too?

A little, @Alessandro.

I know the classical stuff

@Fawad That's the relation between them, yes. There's a similar relationship between sin and sinh. (Note, though, how different they are for real values: A plot of cos(x) for real x gives a curve oscillating between -pi and pi, whereas cosh(x)=cos(ix) is symmetric but bounded below by 1 and diverging exponentially at infinity.)

1:05 PM

@Alessandro Any progress on the Hopf link?

oh, hi chat

Hi

hi

I'm quite sure the group has 2 generators, but there should also be some relations I haven't figured out

Correct.

1:06 PM

@Secret It's a different notion of 'hyperbolic', but it's fun to see videos of visualizations of objects moving at near lightspeed.

Tim The Enchanter

Here is a cute problem, consider an equilateral triangle and an arbitrary point inside it. Prove the sum of the distances from the point to the sides is invariant of the point chosen.

@Semiclassical "lightspeed" is one word now?

I also think that the complement of 2 disjoint and unlinked $S^1$ has a free fundamental group with 2 generators

feh

He guys, if we have a norm induced by an inner product, do we then always mean $\Vert v\Vert=\sqrt{\langle v,v\rangle}$, of do we just mean that the norm of $v$ can be written in terms of the inner product of $v$?

1:09 PM

I don't see the distinction. @ShaVuklia

In the first case we have a norm that is induced by an inner product in a specific way

in the second case, all we need is that it comes from an inner product

@Secret spacetimetravel.org/ueberblick/ueberblick1.html for examples of what I mean

@Semiclassical I think the second is like, you can write $||x|| = f( \langle x,x \rangle ) $ for some function $f$

@AlessandroCodenotti True. Can you prove that?

hmm.

1:10 PM

Where $f$ is not necessarily the square root

exactly

Dunno, then.

I've never seen a norm induced in another way

because my book asks me to prove that the parallelogram identity holds for any norm induced by an inner product

We always ever mean $\|v\| = \sqrt{\langle v, v \rangle}$.

1:11 PM

It may be possible, but then it would definitely be specified how

so I don't know if I can use $\Vert v\Vert=\sqrt{\langle v,v\rangle}$

Yeah, just use it.

because technically, that is a special case

okay cool

I guess one way to look at it: Suppose $\|v\|^2=g(v,v)$ for some symmetric function $g(u,v)$.

@BalarkaSen it deformation retracts to a plane with 2 fat "cylinders" running around on the surface and I can use SvK on that

1:13 PM

I contacted by teacher about this

and I just received his reply

he said that "induced by" means the square root thing

@Alessandro So you mean like a 2-plane with two cylinders attached along two parallel lines? That doesn't sound right.

For the parallelogram identity to hold, you need $2g(u,u)+2g(v,v)=g(u-v,u-v)+g(u+v,u+v)$.

No, they run in a circle

It's like a pair of tori with a plane cutting them in half

Hmm, there's probably a smarter way to frame that.

isn't it easier to come up with a counterexample @semiclassical?

I'm guessing there must be some $f(\langle v,v\rangle)$ for which it doesn't hold?

1:15 PM

That's what it looks like from the outside, the tori are actually full except for a missing $S^1$

@Alessandro I'm having a little trouble seeing that. Like, two tori, each attached to $x^2 + y^2 = 1$ and $x^2 + y^2 = 2$ on the plane by a longitude?

oh wait, but it's still a norm

i donno, I'll just stick to what my teacher said

I'll show you a drawing later, it's easier

Let me ask something easier. What does the complement of an unknotted circle deformation retract to?

A torus with the hole closed by a disk

1:17 PM

Ideally, one would like to show that $$\|v\|^2=g(\langle v,v\rangle )\implies 2g(\langle u,u\rangle)+2 g(\langle v,v\rangle)=g(\langle u-v,u-v\rangle)+g(\langle u+v,u+v\rangle)$$ only works if $g(x)=x$.

Well, the parallelogram rule holds for a norm induced (in the standard way) by an inner product

@Alessandro Ah, got it. Yes.

Anyhow, that is homotopy equivalent to $S^2 \vee S^1$

But, there's also the converse: if the parallelogram rule holds for some norm, then you can construct an inner product using this norm

(which is what I am more used to seeing)

(and that inner product will induce the norm you started out with, of course)

1:20 PM

One has for sure that $\|0\|^2=g(\langle 0,0\rangle)=g(0)=0$. So therefore when $u=v$ one has $4g(\langle u,u\rangle)=g(4\langle u,u\rangle).$

So $g(4x)=4g(x)$.

I was trying to get the knot as "flat" on a plane as possible because it looks easier to think about

But, eh, that doesn't really tell you anything you didn't know from the initial statement: $\|\lambda v\|^2=\lambda^2 \|v\|^2=g(\lambda^2\langle v,v\rangle)$

So $g(\lambda x)=\lambda g(x)$ for any $\lambda>0.$

There's probably a smart way to conclude from this that $g(x)=ax$ is the only possible example. If you then require that it agree with the usual case for some particular value, that would give $g(x)=x$.

Hi.

Hmmmm

@Alessandro Not a bad idea.

1:27 PM

So set theorists defined $\operatorname{card}(\Bbb N)=\aleph_0$

(Unless I'm missing something, $\|v\|^2=2\langle v,v\rangle$ would be a perfectly acceptable choice. So there's some additional condition required for uniqueness.)

@Mahmoud Sure, the cardinality of the natural numbers is aleph-null.

And we don't even know if $\operatorname{card}(\Bbb R)=\aleph_1$, how can we hypothesize that there exist more infinities of bigger sizes ? Such as $\aleph_2$

Well, it's not hard to create a set which must have cardinality larger than aleph-null. Just take the power set.

And then do that again, etc.

How ?

How to take the power set?

1:31 PM

Yes

Well, do you know how to construct the power set of a finite set?

I know, but I have no idea about infinite ones.

Well, it's the same idea.

You start with a set $A$, and construct a new set $2^A$ consisting of all subsets of $A$. That works even if $A$ is infinite.

In the case of $\mathbb{N}$, That will give some new set $2^\mathbb{N}$ whose elements are themselves a set of positive integers.

(also: the existence of the power set of any set, is an axiom in ZF)

And then you have to do that again with $2^\mathbb{N}$ itself. You'll get a set of sets of sets.

It's not hard to convince yourself that it works in the finite case, I think.

And if you trust the axioms of ZFC, then it should still work in the infinite case.

1:36 PM

So by taking the power set of a infinite set we can create another set of bigger size ? Always ?

Right.

Ok thanks $:)$

Mmkay.

@Mahmoud that's known as Cantor's theorem if you want a name to google

On the other hand, the question of whether $\mathbb{R}$ has the same cardinality as aleph-one...

Well, that's the continuum hypothesis.

1:38 PM

So by induction, you get a (strictly) increasing sequence of cardinalities

And that's a weird one. Godel showed (in the 30s, I think?) that the continuum hypothesis is consistent with the axioms of ZFC.

However, Cohen showed (in the 50s-60s?) that the negation of the continuum hypothesis is also consistent with ZFC!

And it's $\lnot$ is also consistent with ZFC

Secret

Axioms:
Inner products
$\langle x,y \rangle = \overline{\langle y,x \rangle}$

$\langle ax+y,z \rangle = a\langle x,z \rangle + \langle y,z \rangle$

$\langle x,x \rangle \geq 0$

Norms
$\rho (av)=|a|\rho (v)$

$\rho (u+v)\leq \rho (u) + \rho (v)$

$\rho (v)=0 \implies v=0$

Now

$\rho (\langle x,y \rangle) = \rho (\overline{\langle y,x \rangle})$

$\rho (\langle x+y,z \rangle) \leq \rho (\langle x,z \rangle) + \rho (\langle y,z \rangle)$

Which means that you can do ZFC with the continuum hypothesis being true or false; in either case, you'll get a perfectly consistent theory.

But nobody really agrees whether CH, or its negation, is "intuitively true"

1:41 PM

Right. As far as I know, there's not a principle which says whether one of these universes is the 'right' one.

(Kinda reminiscent of Euclidean vs. non-Euclidean geometry, but faaaar less intuitive.)

One proposed solution was adding the axiom of constructibility, which says that every set can be constructed. This implies CH.

There are some axioms that some set theorist consider reasonable that imply not CH though, like the proper forcing axiom

Of course, just because you can use either doesn't mean that people don't have preferences. My recollection is that Godel, for instance, very much believed CH to be true.

(in fact, it implies the generalised CH: $2^{\aleph_\alpha} = \aleph_{\alpha+1}$ )

What he would have thought of Cohen's work, though? No idea.

1:44 PM

Gödel was against CH IIRC

why?

Well, this kind of axiom gives rise to many new spaces, whose properties may be very hard to study and very "irregular"

Oh, huh. There's some history of it here: ias.edu/ideas/2011/kennedy-continuum-hypothesis

2

And evidently Cohen communicated his results to Godel directly.

I think Gödel thougth that the "right" choice was $|\Bbb R|=\aleph_2$, but I couldn't find any source last time I looked it up

Usually there's people against it, because they want the "universe" of sets to be nice, neat, controllable

The other side is in favour of it, because they want a rich universe with many exotic things in it

1:47 PM

Hmm, there's a paragraph re: aleph-two in there specifically.

Interesting, I'll read it later, I'm in class now :P

@Semiclassical thanks for sharing

set theory is complicated

"In 1972, Gödel circulated a paper called “Some considerations leading to the probable conclusion that the true power of the continuum is ℵ2.” [...] The proof was incorrect, and Gödel withdrew it, blaming his illness. In 2000, [authors] isolated three principles implicit in Gödel’s paper, which, taken together, put a bound on the size of the continuum.

And subsequently Gödel’s ℵ_2 became a candidate of choice for many set theorists, as various important new principles from conceptually quite different areas were shown to imply that the size of the continuum is ℵ_2."

@BalarkaSen set theory is beautiful though

1:48 PM

2hard4me

It's probably the field I find most interesting so I might be biased :P

hi

hi pal

Q: Norms Induced by Inner Products and the Parallelogram Law

Something I find weird, is that the axiom of constructibility implies, on the one hand, that GCH is true (so it severly limits which cardinalities can exist)

But on the other hand, it gives the negation of Suslin's hypothesis, which pretty much says that "any totally ordered set that kinda looks like $\mathbb{R}$, must be $\mathbb{R}$"

Secret

113

Let $ V $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ \lVert\cdot\rVert$. It's not hard to show that if $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ for some (real) inner product $\langle \cdot, \cdot \rangle$, then the parallelogram equality $$ 2...

1:53 PM

apparently I have to be enrolled in the class to take the exam

Secret

Semiclassical, I think by showing g is homogenous, you already proved it

2:04 PM

@MeowMix that sounds rather... normal, no?

@Akiva said he could

Well, is this about a university exam?

No, AP

I'm in 8th grade nw

Oh, okay.

Advanced placement into what grade?

2:19 PM

AP Calc BC

nice

its unfortunate

Hi

Hi

Hi

2:27 PM

so I'm sadddd

why?

I cant take it

duh

so?

ill have to be in boring math

ever heard of independent study?

2:28 PM

yeah but its just SO boring

I mean, the class

not the independent study

you can make it fun by asking lots of questions

that you know the answers to :P

no the teacher is boring too

when I show her stuff she doesn't want to see it

show her stuff that will make her think

And yet Google says you can do it without taking the class… Also, you can contact the AP people directly, but the deadline is March 1

I have -21 days left, yay!

2:32 PM

So, yeah

@Akiva She said district policy

but maybe next year?

@MeowMix Oh :(

Class is over, bye

cya

3:05 PM

@Daminark I’m not familiar with oscillation, I’m afraid, so I’m not sure if I’ll be able to show easily that a function on $\mathbb Q$ is not continuous:p What always confuses me though: a function on $\mathbb Q$, is that a function with domain $\mathbb Q$ or codomain $\mathbb Q$?

Both if it isn't specified otherwise

Vrouvrou

what do we mean by $d_1$ the distance on $\mathbb{R}^2$

?

@Secret Probably. But to be sure I'd have to think through it more, and I don't want to.

Also, pretty sure I left my lunchbag behind on the train ride this morning. Yay.

3:26 PM

ouch!

grump

and they say we developed large brains because we had to remember where we kept food :P

Yes, and we now devote those brains to math. Small wonder.

Vrouvrou

can someone help me ?

So thanks to fridges we can do math more efficiently?

3:36 PM

@Vrouvrou Without further context, I'd assume the Euclidean distance

Vrouvrou

with absulute value ?

By the Pythagorean theorem, the distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

the index $1$ bothers me

It could be that $d_p$ refers to the distance defined using the $p$-norm?

(You draw the right triangle with corners $(x_1,y_1)$, $(x_2,y_1)$, and $(x_2,y_2)$)

actually, I think $d_p(x,y) = \sum_{i=1}^n |x_i - y_i|^p$ is the correct metric

3:39 PM

@Vrouvrou Yeah, it could be what @SteamyRoot said. The $1$-norm would be $|x_1-x_2|+|y_1-y_2|$, the so-called Manhattan metric.

@SteamyRoot You either meant $d_1$ on the left or some exponents on the right

uhh

right :P

And a root @SteamyRoot

Actually, no

You can do it with a root, but also without

Huh. That would make $d_2$ the square of the Euclidean norm

@SteamyRoot Wikipedia has it with a $p$-eth root

Lp space

In mathematics, the Lp spaces are function spaces defined using a natural generalization of the p-norm for finite-dimensional vector spaces. They are sometimes called Lebesgue spaces, named after Henri Lebesgue (Dunford & Schwartz 1958, III.3), although according to the Bourbaki group (Bourbaki 1987) they were first introduced by Frigyes Riesz (Riesz 1910). Lp spaces form an important class of Banach spaces in functional analysis, and of topological vector spaces. Lebesgue spaces have applications in physics, statistics, finance, engineering, and other disciplines. == Applications == ==...

Both occur

The one with the $p$-th root is of course, the one induced by the $p$-norm

The one without is used to define $l_p$-spaces

Either way, @Vrouvrou, if $d_1$ is used somewhere without a mention of what metric it is, then it should be somewhere earlier in the lecture notes or book you use, no?

3:49 PM

@SteamyRoot Ohh, didn't see that

They even use the $d_p$ notation. So that would make $d_1$ the Manhattan metric.

Yeah, if $p = 1$ both of them coincide anyway :)

Obvious linear algebra question: Suppose I've got a nonsymmetric square matrix A whose last row is zero. This is obviously singular and the left eigenvector of 0 is just (0 0 ... 0 1).

Am I right in thinking that it also has a right eigenvector, even if I can't write it out easily?

What's a left eigenvector? An eigenvector of the transpose?

(Or, rather, the transpose of that)

If you want to view it as a column vector, yes.

I'm guessing the difference is $Av = \lambda v$ or $vA = \lambda v$

3:52 PM

But if you're doing it as a row vector (and I am---there's no ^T on said vector) then it's a left eigenvector in the obvious sense.

I guess it should have a right eigenvector as well, with eigenvalue 0

Yeah.

Won't the same one work in this case? :P

No.

Oh, wait, no

3:53 PM

It's not a symmetric matrix.

I don't know of a good way to find it, though

I think it just comes down to "n equations in n-1 unknowns."

other than just knowing that "$n$ vectors of $n-1$ dimensions each have a linear dependence"

Yeah

Hmmm... take the Jordan decomposition blah blah yada yada

Oh well.

3:54 PM

And then make the vector the coefficients of the dependence

Well, the existence of the right eigenvector is rather trivial

Since the eigenspace of the eigenvalue $0$ has dimension at least $1$.

Right.

Rank = column rank = row rank and all that.

Yeah.

The right eigenvector will be the transpose of a left eigenvector of $A^T$...

And you can turn $A$ into $A^T$ using elementary row and column operations

Silly thing is that for my purposes I probably could just say "I'll do A^T instead of A".

Heh :P

Maple SE - Area 51 Proposal

4:05 PM

Or, even easier, just take A^T to be A instead.

It's part of what seems to be a tricky spectral question.

Suppose I have a matrix exponential e^(t A) with t some parameter and A some symmetric matrix. (In my case of interest it's tridiagonal but not constant along the diagonals.)

What I thought I should be doing originally was: Let S be the upper triangular shift matrix (i.e. S is one along the upper sub-diagonal and zero everywhere else).

How is the spectrum of Se^(t A) related to A?

Or, what is the same to my purposes: What does log(S e^{tA}) look like?

If t=0, it's just log S. If t is large, it should be t A (1+O(1/t)).

Buuut what are those higher order terms?

I think now, though, that the above isn't really the right question since it always has 0 as a zero eigenvalue. Instead I should allow S to have 1 in its lower left corner, so that it serves to cycle the rows of e^(tA) rather than shift them up and drop the first one.

blah blah blah

I have proposed a Maple SE and its now in the Commitment phase. There is quite a handsome number of commitments made by new users but it worth less compare to the SE users with rep more than 200.
please please help out and commit.
area51.stackexchange.com/…
Thanks

Fawad

I have starred this message before

Btw you are trying hard

@Semiclassical Yeah, I don't think I'll be of much help there, I'm afraid

Wasn't expecting it :)

Just wanted to ramble a bit.

skull petrol

4:24 PM

Led Zeppelin - Ramble On

►

:-)

How's it going everyone?

@Sha It's a function $f:\mathbb{R}\to\mathbb{R}$ so that $f$ is discontinuous at every irrational number, and continuous at every rational number.

@Daminark Ooohhh, wow, I completely misinterpreted it then :P

okay I should be able to solve that

but I also have a short question in the meanwhile:

Why is it true that $\Vert x-y\Vert_1\leq\sqrt n\vert x-y\vert$.
We have$\begin{aligned}\Vert x-y\Vert_1=\sum\end{aligned}_{i=1}^n\vert x_i-y_i\vert$, and we also have$\begin{aligned}\vert x-y\vert=\left(\sum_{i=1}^n(x_i-y_i)^2 \right)^{\frac{1}{2}}\end{aligned}$.
To make a simpler case, it seems to me I'm trying to prove:
$\sqrt{a^2}+\sqrt{b^2}\leq\sqrt 2\cdot\sqrt{a^2+b^2}=\sqrt{2a^2+2b^2}$.
I still don't see where this $\sqrt n$ comes from... could someone help with that one?

4:41 PM

I've seen this before, by way of Cauchy-Schwartz?

The point of $\sqrt{n}$ is that it comes from the dimension

Try to picture the unit balls for both norms

So I'm just gonna let $z = x-y$ here

$\|z\|_1 = \sum_{i=1}^n 1*|z^i| \le \sqrt{\sum_{i=1}^n 1^2}\sqrt{\sum_{i=1}^n |z^i|^2} = \sqrt{n}\|z\|_2$

Hey @Ted!

@Daminark Um ... I guess I missed something. Hi, Demonark.

@ShaVuklia regarding the proof in the n=2 case you state, note that $2a^2+2b^2=(a+b)^2+(a-b)^2$ which is never smaller than $(a+b)^2$.

@Ted What'd you miss?

I linked to your comment to Sha. I don't want to say too much, but I assume you meant to be misleading.

Wait the problem is to prove its nonexistence

4:51 PM

G'night, @MikeM.

Right. OK, just wanted to make sure the non was clear.

Mike Miller

hi

Haha, yeah

I put that on my topology final for my two students last spring.

Nice

We had it as a bonus problem

I was only able to do that, and the one about an isometry in $\mathbb{R}^n$ which fixes the origin being linear

If you look at the problem sets I sent you, there's a cool one about an example of a global inverse function theorem on $\Bbb R^n$.

Yes, isometries being linear is a nice exercise.

4:54 PM

There were some others which were pretty cool but kind of impenetrable

The bonus ones were challenge problems in Sally's book

And Soug just found ones he thought were cool and sent them over to us

I remember one problem was to prove that a continuous function $f:\mathbb{S}^{n}\to\mathbb{R}^n$ had the property that $f(x) = f(-x)$, I think

I wrote my students a baby robotics problem as an application of manifolds, but I was surprised they weren't enthralled. ...

Um, no, you didn't state that last one right ...

Oh wait

That's better! But yeah none of us figured it out and someone found it proven in Peter May's book

Hi @Ted

Oh wait that's still wrong

Demonark ... for some $x$ ... This is Borsuk-Ulam, essentially.

Hi @Alessandro :)

4:58 PM

Right, that's it

Hey @Alessandro!

But yeah, there was one about a subset of the plane having at most one point such that it was isometric to itself minus the point, and one about compact spaces being homeomorphic iff the spaces of functions mapping each to $\mathbb{R}$ were isometric.

Hi @Dami

There were prob more but I forget offhand

user193319

I am having some trouble understanding the solution to problem 4 in the following link: math.wvu.edu/~hjlai/Teaching/Math541-641/I-2.pdf

Specifically, why must f(A) be in {C^2, D, CD, −D, −CD}?

f(A) doesn't need to be in that set, one of f(A),f(B) and f(C) needs to (pidgeonhole)

@user193319: He said one of em. $f(A)$, say. ... Yeah, pigeonhole principle. Just count.

5:05 PM

they assume it's f(A) without loss of generality since you can do the same exact proof if you assume it's f(B) or f(C) instead

user193319

Oh, I see. Thanks everyone.

I realized today when doing an algebraic topology exercise that there are no injective morphisms from a nonabelian group to an abelian group, which is very obvious in hindsight but I had never noticed it explicitely

Right, @Alessandro.

G'day @Balarka.

Hi, @TedShifrin @Daminark

5:10 PM

It's tough to have a nonabelian subgroup of an abelian group, @Alessandro. :P

Yo @Balarka

@Daminark Oh I see, I did not recognise Cauchy-Schwartz immediately :P

That's fair, I only know because I've seen it before

@TedShifrin oh, derp, I thought about picking $2$ non commuting elements $a,b$ and then doing something like $\phi(ab)=\phi(a)\phi(b)=\phi(b)\phi(a)=\phi(ba)$ so it's not injective, which is even longer than needed

Oh so the oscillation definition of continuity is basically this, let's say your function is $\mathbb{R}\to\mathbb{R}$

5:13 PM

@Alessandro: Yup, image of injection is an isomorphic subgroup :)

In fact any morphism from a nonabelian group to an abelian one must kill the commutator subgroup.

Its oscillation at a point $a$ is $\lim_{\delta \to 0} \sup_{|x-a|<\delta} f(x) - \inf_{|x-a|<\delta} f(x)$

And the projection from a group to itself with the commutator subgroup modded out, is called the abelianisation (sometimes the image of that projection is called that too)

@Semiclassical But this way we get $2a^2+2b^2\geq (a+b)^2$, or $|a+b|\leq\sqrt{2a^2+2b^2}$, but we have that $|a+b|\leq |a|+|b|$, so is that really helping me here?

@BalarkaSen right, it's a consequence of the first isomorphism theorem plus the fact that the commutator's subgroup is the smallest subgroup such that quotienting by it produces an abelian group

5:16 PM

@Daminark oh, let me fine-read that:P

Now, a function is continuous if and only if its oscillation is $0$, use that plus Baire Category and you can prove that a function can't be continuous just on $\mathbb{Q}$

@Alessandro Or just that $f([a, b]) = [f(a),f(b)]$, which is zero.

first isom says such a map $G\to H$ gives a map $G/[G, G] \to H$

(passing to cosets)

sure. I might need to quotient by a bigger subgroup to get an isomorphism though

true.

@BalarkaSen oh, of course. I have a tendency of using much stronger than needed theorems while thinking about abstract algebra apparently :P

5:21 PM

"By the classification theorem of finite simple groups..."

hi folk

The Question

Hi everyone!

Hey @Eric and @TheQuestion

@Daminark OK, I finally understand what your question is even about:P Now I'm wondering about the proof: am I going to show that this function can't be continuous on $\mathbb Q$, or am I going to show that this function will also have to be continuous on at least some irrational element?

because as of now, I'm not sure which direction to take

Also, I understand the oscillation definition

and somehow I have to use that $\mathbb Q$ is the union of a sequence of nowhere dense sets

I hope you don't mind I'm going to google this:P because as of now, I'm lostXD

Well in that case I'll give a hint

The set of points where a function is continuous must be of a certain type, which you can prove by the oscillation.

And $\mathbb{Q}$ couldn't be that type of set, since that would violate the Baire Category theorem

5:31 PM

can't you play a Banach Mazur game to show this

@Eric I'm not sure, thing is you couldn't necessarily swap out $\mathbb{Q}$ with any first category set, I don't think

$\mathbf{Q}$ is countable, so whichever player is trying to avoid it can do so in the limit.

Is the complement of a set which is dense and first category also dense?

Hi what up

Hey @Akiva!

5:34 PM

@Daminark the complement of a first category set is dense

@ShaVuklia I don't understand the question. You posited that $|a|+|b|\leq \sqrt{2a^2+2b^2}$, and I pointed out why that's true.

Right I'm being stupid

Whether that helps you with your larger problem, I don't know. But that's not what you asked.

@Semiclassical true, I get the general problem, so that's good enough i guess:d

@Eric I don't think that works in general, not all comeagre sets are $G_\delta$

5:37 PM

@Daminark I hope you don't mind I come back at this another time. I only learned about dense sets, nowhere dense sets, Baire theorem, etc. yesterday, so the definitions are all shaky in my head:P In any case, it gives me too much of a headache atm and I need to study for multivariate calculus nowXD

I might have misunderstood what you want to prove with a Banach-Mazur game actually

Lol that's entirely fair @Sha

where exactly do I make a claim that comeagre sets are $G_{\delta}$? @AlessandroCodenotti

$G_\delta$ sets are the sets of points of continuity of functions between metric spaces, while comeagre sets are those characterized by Banach-Mazur, that's what I meant with I don't think it works in general

The Question

I don't want to be rude but see that I should not ask to ask here and just ask... So here is my question: Is it correct to say that "for every positive real number (r) there exists an integer k such that |k|r > 5" is true?

I would say that it is true by the completeness axiom, I see the statement as "There exists an integer (k) that is a least upper bound to the set of all real numbers 5/r" which is a subset of the real numbers.

is this reasoning correct?

5:43 PM

Oh sure I wasn't saying it was, just that it works for $\mathbf{Q}$.

Then we agree :)

Brody

How is it proven that $a_n\to a$ and $b_n \to b$ imply $a_n b_n \to ab$? I've seen a proof in class using $|a_n - a|<\epsilon/(|b|+1)$ and $|b_n - b|<\epsilon/(2m)$ where $|a_n|\le m$.

The proof seems correct. Can boundedness be used instead for both epsilons?

I guess what made me worried earlier is that first category sets can be $G_{\delta}$, so that it'd be a problem if the B-M argument made for $\mathbb{Q}$ could translate over.

But it might not actually translate

btw @Daminark my grades came in, so yours might be up too

looks like I got my first A-

so sad

Algebra? Darn

Yeah mine are up as well

Brody

5:53 PM

Take this... $\forall\epsilon > 0,\exists n_1,n_2 \in \Bbb N: |a_n-a|<\frac{\epsilon}{2m_2}$ and $|b_n-b|<\frac{\epsilon}{2m_1}$ for all $n\ge\textrm{max}\{n_1,n_2\}$, where $|a_n|\le m_1$ and $|b_n|\le m_2$ for all $n\in\Bbb N$.

Now, surely $|b|\le m_2$?

Yeah @Daminark, got As in all my other math classes

Nice

Next quarter is actually gonna be the first one in which I'll be doubling on in math

oh lol ive like only either doubled or tripled

Does anyone know how to write a norm nicely in latex? The dot is not in the right position this way: $\Vert.\Vert$

I didn't think you doubled with 207

And hmm, testing $\|.\|$

Weird

5:57 PM

you're not allowed to, but I have every single quarter since

I do $\| \cdot \|$

Nice

the problem is

if I use \dot

I get this

$\Vert\dot\Vert$

ok that makes sense actually

what about $\cdot$ though