Mathematics - 2018-04-05

Leaky Nun

00:55

Does anyone see a constructive proof of $\operatorname{Hom}(G, \Bbb Q/\Bbb Z) = 0 \implies G = 0$?

where $\operatorname{Hom}$ is the abelian group of group homomorphisms from the abelian group $G$ to the abelian group $\Bbb Q/\Bbb Z$

@AkivaWeinberger

Akiva Weinberger

Why am I being summoned

OK so there's no map from $G$ to the rational circle and that means $G$ is trivial…

In other words, you need to show that every nontrivial abelian group has a nontrivial map to the rational circle.

Leaky Nun

please kindly re-read the fifth word of my question :)

Daminark

"Constructive"

Okay nevermind

Leaky Nun

:)

Akiva Weinberger

@LeakyNun I don't see what it changes

In any case, I think I see a way to do it if we can use choice…

Leaky Nun

01:02

Constructivism (mathematics)

In the philosophy of mathematics, constructivism asserts that it is necessary to find (or "construct") a mathematical object to prove that it exists. In standard mathematics, one can prove the existence of a mathematical object without "finding" that object explicitly, by assuming its non-existence and then deriving a contradiction from that assumption. This proof by contradiction is not constructively valid. The constructive viewpoint involves a verificational interpretation of the existential quantifier, which is at odds with its classical interpretation. There are many forms of constructivism...

Akiva Weinberger

@LeakyNun Oh, I thought you meant "abelian"

Semiclassical

01:20

@EricSilva Math prof: A machine for turning coffee into theorems. Grad school: A machine for turning intelligence into mental illness.

(That is an overly bleak/pessimistic rendering, to be sure)

Zee

01:36

fact : on the undergrad level there is 0 correlation between creativity and grades

Crazy fact : creativity is NEGATIVELY correlated with performance during graduate school

Does anybody know a good way to learn programming for a mathematician? Am probably gonna get kicked out and I need a skill

Actually I wouldn’t call myself a mathematician but you get the jest

Leaky Nun

maybe learn to not be anti-tech first

Zee

Am not anti tech anymore , am not anti anything, I just wanna make money and die in 30 years

Leaky Nun

anti latex?

Zee

01:51

God bless you , god bless you all, I hope you have a wonderful day

Leaky Nun

maybe define matrix formally

oh wait, it can't be done

Zee

02:02

I heard Gromov say it in a video , idk if it’s true , am just parroting what he said

Probably said something completely different and I fuccked it up

:)

Do you have anymore stuff ?

God is within us all ; I hope not , otherwise god is dead

Semiclassical

"God is dead. And, as a matter of fact, I also do not feel too well…"

Leaky Nun

@Zee high level Grothendieck point of view?

Zee

That’s just me overcompensating for the fact am failing my courses :)

Leaky Nun

never met anyone who knows what a matrix is?

Zee

again just acting like Gromov without knowing what am saying

Is this the best you can do?

Am disappointed

Leaky Nun

02:15

ok

Zee

I literally got the worst grade on my real analysis midterm “by a quantum leap” is what my professor said

Xander Henderson

Well, quanta are small, so the next lowest grade must have been just a hair higher than yours, right?

Semiclassical

I approve of this analogy.

Zee

he meant a huge leap

Even if it don’t make any sense

0celo7

saying you got the worst grade by far doesn't make sense given your usual attitude

AI confirmed

Leaky Nun

02:19

@0celo7 you don't know what AI is

Zee

I wish I was an AI

0celo7

more than you @LeakyNun

Leaky Nun

AI is smarter than you think

0celo7

not this one

Zee

Ya, am like an AI from the 50s

I believe am fully mechanical with no electronics , except this phone am typing on

I been talking smack to you guys for weeks . Why are you being such wimps ?

0celo7

02:24

@LeakyNun See. This is not a real person.

Leaky Nun

|> leaves

0celo7

02:34

Some people are just dead-set on being annoying

2

David Zhang

hey guys, quick question

say you consider the lie algebra gl(n, C) as a real lie algebra of dimension 2n^2. does anybody know offhandedly what its complexification is?

is it just the direct sum of two copies of gl(n, C)?

Secret

03:37

When Ted left this room forever, it will be 2020

2020 is the year where the best and the worst things globally will happen as polarisation achieve its saturation point

10 hours ago, by Ted Shifrin

I got chewed out for not being nice to a gentle flower on main yesterday (despite hours of having helped over the months) and now this room is nuts. I think I need a long sabbatical.

Recall Ted is a glue that stablise the math chat room with his unique session after some old members have already left and most math mods and RO are inactive

Leaky Nun

@DavidZhang yes, just like any complexification, according to wiki

David Zhang

@LeakyNun awesome, which wiki article are you looking at?

Leaky Nun

@DavidZhang complexification

David Zhang

@LeakyNun afraid I'm not seeing it, can you point me to where on the page?

Leaky Nun

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 be a real vector space. The complexification of V is defined by taking the tensor product of V with the complex numbers (thought of as a two-dimensional vector space...

basic properties

alternatively, you can use the universal mapping property of the tensor product to construct a map V^C -> V+V

and use the standard maps into the tensor product to construct a map V+V -> V^C

and show that the maps are inverses of each other

Forever Mozart

03:52

I need to invent Ifification

Secret

what will it does. Generates if statements?

David Zhang

04:14

Oh @LeakyNun I know this is true for vector spaces. What I'm asking about is the complexification of gl(n, C) as a Lie algebra, i.e., when you complexify the real Lie bracket on gl(n, C), is the resulting operation isomorphic to the direct sum Lie bracket on gl(n, C) + gl(n, C)

but if this isn't immediate to anybody here I'll give it some more thought

Leaky Nun

no idea

Fawad

05:20

$\log i =\dfrac{\pi}{2}$

Silent

05:54

I am not quite sure why $\lim\limits_{h\to0}\dfrac{f(x-h)-f(x)}{h}=-f'(x)$, assuming $f'(x)$ exists. Anyone?

Leaky Nun

@Silent substitute h -> -h

Silent

06:10

@LeakyNun like this: $\lim\limits_{-h\to0}\dfrac{f(x+h)-f(x)}{-h}=-\left(\lim\limits_{-h\to0}\dfrac{f‌(x+h)-f(x)}{h}\right)$? But still, how does $\lim\limits_{-h\to0}\dfrac{f(x+h)-f(x)}{h}=f'(x)$?

Leaky Nun

I mean, $h\to 0$ and $-h \to 0$ are kinda the same thing

$h \to 0$ doesn't mean $h$ approaches $0$ from the positive side

it's both sides

if you want rigorous proof, just use epsilon-delta

Silent

ok.

Leaky Nun

reuse the same delta

Silent

@LeakyNun where?

Leaky Nun

@Silent in your epsilon-delta proof

Silent

06:30

@LeakyNun Since $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}=f'(x)$, for every $\varepsilon>0$, there is a $\delta>0$ such that $0<|h|<\delta$ implies $\left|\dfrac{f(x+h)-f(x)}{h}-f'(x)\right|<\varepsilon$. Now if $0<|-h|<\delta$, then, since $|-h|=|h|$, we see $\left|\dfrac{f(x+h)-f(x)}{h}-f'(x)\right|<\varepsilon$. Is this correct?

Also, does $\lim\limits_{h\to0}\dfrac{f(x-h)-f(x)}{h}=-f'(x)$ hold for one-sided derivative or not?

Leaky Nun

check your answer for typo

@Silent no idea

Silent

@LeakyNun is there a typo?

Leaky Nun

don't you want to prove things about f(x-h)

Silent

Yes, but then substitution meant that i had to check about $\lim\limits_{-h\to0}\dfrac{f(x+h)-f(x)}{h}=f'(x)$

25 mins ago, by Silent

@LeakyNun like this: $\lim\limits_{-h\to0}\dfrac{f(x+h)-f(x)}{-h}=-\left(\lim\limits_{-h\to0}\dfrac{f‌(x+h)-f(x)}{h}\right)$? But still, how does $\lim\limits_{-h\to0}\dfrac{f(x+h)-f(x)}{h}=f'(x)$?

Leaky Nun

well $\lim_{-h\to0}$ is abuse of notation, so you can't prove that

Silent

06:35

oh!

@LeakyNun how do we know that its abuse of notation?

Leaky Nun

because you never defined it formally

we only define things when it is one variable

not an expression

Silent

ok

ÍgjøgnumMeg

09:32

How does one go about showing that a hypercube is a convex body?

Leaky Nun

@ÍgjøgnumMeg by applying the definition of "hypercube" and "convex"

ÍgjøgnumMeg

@LeakyNun The "convexity" bit is easy but the book says that a convex body should be a closed, bounded, convex subset of $\Bbb R^n$ and I dunno how to show that the hypercube is closed and bounded lol

Leaky Nun

how do you define those two terms?

ÍgjøgnumMeg

a subset $S$ of $\Bbb R^n$ is closed if every limit point of $S$ is in $S$ means closed right?

Okay the boundedness is obvious lol, just the closed part that I'm confused about

Leaky Nun

and how do you define limit point

Tobias Kildetoft

09:39

@ÍgjøgnumMeg It is given by non-strict inequalities, hence it is closed

(that can actually be turned into a formal argument)

ÍgjøgnumMeg

I guess.. $\alpha \in \Bbb R^n$ is a limit point of $S$ if every neighbourhood of $\alpha$ contains a $\beta \in S$

and $\beta \neq \alpha$

not really sure how to apply these definitions though lol

ÍgjøgnumMeg

09:55

If $\alpha \in \Bbb R^n$ and $r \in \Bbb R^+$ a neighbourhood of $\alpha$ is defined as the set $\lbrace \beta \in \Bbb R^n : \lvert \lvert \alpha - \beta \rvert \rvert < r\rbrace$

is this $r$ just some arbitrary fixed real number?

user21820

10:38

Calculators are making us dumb. — Mohammad Sakib Arifin Feb 28 '17 at 6:48

@LeakyNun and most important of all, how do you define "define"? =D

Leaky Nun

11:11

@user21820 right

Anderson Felipe Viveiros

12:12

May I post questions links here?

Well, here it goes: math.stackexchange.com/questions/2722666/…

user21820

@AndersonFelipeViveiros I suppose you just proved that you may.

=P

Anderson Felipe Viveiros

Haha It is "possible" but maybe not "allowed" or "well received" :D

If someone has an understanding of varifolds, please, give me some help with that (above) question.

user21820

@AndersonFelipeViveiros I don't so I guess you'd have to wait for someone else. This chat-room's description says "Associated with Math.SE; for both general discussion & math questions alike. Just ask; don't ask to ask." so I think you're safe. =)

Leaky Nun

12:50

@user21820 do you like horseshoes

user21820

@LeakyNun I'm not sure what that question means, but no I don't have a particular fancy of horseshoes.

Leaky Nun

13:13

Horseshoe lemma

In homological algebra, the horseshoe lemma, also called the simultaneous resolution theorem, is a statement relating resolutions of two objects A ′ {\displaystyle A'} and A ″ {\displaystyle A''} to resolutions of extensions of A ′ {\displaystyle A'} by A ″ {\displaystyle A''} . It says that if an object ...

user21820

@LeakyNun Aha so you were trying to bait me eh? =P

Leaky Nun

@user21820 in some sense

user21820

You should change your username to Sneaky Bun.

Leaky Nun

you should change your username to loser21820

user21820

Hahaha...

What did I lose, by the way?

Leaky Nun

13:22

no idea

user21820

I lost no idea?

Wow.

Leaky Nun

har har

haar

user21820

Then where did my last idea go? Come back!

Leaky Nun

measure

user21820

English.

Such fun.

Oh yes you made be remember... "Incredible" originally means "not credible".

So...

I was amused when I first saw your link to...

in Logic, 14 hours ago, by Leaky Nun

http://incredible.pm/

Leaky Nun

13:25

try to do the NAND calculus exercises lol

working with a completely new symbol with intro and elim rules

user21820

@LeakyNun Oh I didn't notice that let me see.

Leaky Nun

well I don't know if it's new to you

user21820

It's new. Which one do you think is the hardest?

Leaky Nun

heh, I haven't finished all of them

just go through each one lol

seriously, I do those things with no motivation

I just try each block until they succeed

I have no intuition for NAND calculus

user21820

I am lost.

So you were right after all.

Whoever invented NAND calculus had too much free time and no computer.

@LeakyNun: I'm not going to waste more time on it. =P

Leaky Nun

13:40

wow, the grandmaster of logic vs incredible pm

0 - 1

user21820

@LeakyNun Very funny; I just can't be bothered to figure out what it's doing without actually converting it into normal propositional logic.

Once you do that, I bet it becomes easy, because it's just a matter of using a given set of propositional theorems to prove tautologies.

Maybe not super-easy, but still, only 3 rules, can't be hard, right?

Leaky Nun

@user21820 yeah right

it's not like minimalism is always hard to use

user21820

@LeakyNun I shan't give in to temptation.

user193319

14:30

0

Q: Vitali Covering Lemma Proof

Why may we assume that each interval in $\mathcal{F}$ is contained in $\mathcal{O}$? What warrants this reduction? Why is statement (4) true? If $x \in E - \bigcup_{k=1}^n I_k$, then $x \in E$ and $x \notin I_k$ for every $k=1,...,n$. Given some $\epsilon > 0$, there exists $I \in \mathcal{...

1

Q: Corollary 5 in Royden-Fitzpatrick's Real Analysis: Convergence in Measure

Corollary 5: Let $\{f_n\}$ be a sequence of nonnegative integrable functions on $E$. Then $$\lim_{n \to \infty} \int_E f_n = 0 ~~~~~~(5)$$ if and only if $$f_n \to 0 \mbox{ in measure on } E \mbox{ and } \{f_n\} \mbox{ is uniformly integrable and tight over } E ~~~~~(6)$$ H...

real-analysis measure-theory proof-explanation

Secret

[Random]

Let $J$ be an unordered index set

Let $S$ be a set with the same cardinality as $J$. Then for all elements $s \in S$ they can be indiced as follows:

$s_j$

Now we can compute $\mathscr{P}(J)$ to create a new index set

We will use this to index $\mathscr{P}(S)$

In particular, by using $\mathscr{P}(J)$ we can unorderedly index all elements of $\mathscr{P}(S)$ and thus allowing us to plot a dense set of $\mathscr{P}(S)$

If this is not clear enough, we can redraw this as follows:

Here's a more tidied up "animation":

Abhas Kumar Sinha

15:15

hi

BAYMAX

16:02

Hi

in Differentiable Manifolds, 2 mins ago, by BAYMAX

Suppose $\sigma = (yz + x^2z^2 + 3xy^2z)dx dy dz$ then how do we find a 2 - form $w$ such that $dw = \sigma$

Secret

Therefore if axiom of choice is discarded, Baire Category theorem fails in a complete pseudometric space in that you cannot form a sequence, thus you cannot show it converge to some limit point and hence the resulting set can be not dense

The above diagram exploit the property of disconnected non hausedoff topological spaces in order to place all elements of a powerset "side by side" so that arbitrary open sets can then be visualised

Silent

16:34

Suppose $f$ is a twice differentiable function from $(a,\infty)$ to $\Bbb R$. For $h>0$, how do we get (using Taylor theorem) $f'(x)=\dfrac1{2h}[f(x+2h)-f(x)]-hf''(\gamma)$ for some $\gamma\in(x,x+2h)$?

@Secret Where did you get this? Did you make it yourself?

Secret

yeah, and possibly incorrect or misleading

Silent

looks beautiful here.

I got my answer to post above.

Anderson Felipe Viveiros

16:59

2

Q: The $\mathbf{F}$-metric induces the weak topology on the set of bounded varifolds

Some preliminary definitions and notation: (1) Given a vector space $\mathbb{V}$, we denote by $G_k(\mathbb{V})$ the $k$-grassmannian of $\mathbb{V}$, i.e. the set of all $k$-dimensional vector subspaces of $\mathbb{V}$; (2) Given a differential (or riemannian) manifold $M$, we denot...

general-topology measure-theory geometric-measure-theory metrizability

Sha Vuklia

17:16

@Leaky do you know any graph theory?

Leaky Nun

@ShaVuklia very little

Sha Vuklia

hm, wanna try? or not? I have a question

Leaky Nun

go ahead

Sha Vuklia

Leaky Nun

lol nope

thanks

Sha Vuklia

17:18

hahahahah

alrighty

Rick

19:06

if a,b and c are chosen at random from 0 to 10, what's the probability none of' them are greater than 4?

I tried by graphing x+y+z=1 .. all points on the surface in the +x +y +z quadrant are possible

now how do I apply the constraints? how will it look like?

Akiva Weinberger

4^3/11^3, no? 4^3 ways of succeeding, 11^3 total possibilities

Alessandro Codenotti

@Rick What's the probability that one of them is greater than 4?

Akiva Weinberger

11 'cause there are 11 numbers from 0 to 10

Rick

btw a b c can be any real numbers

Akiva Weinberger

Ohh

4^3/10^3 then

Rick

19:16

oh sorry

I forgot to mention the most important part of the question

a+b+c=10

forget that " from 0 to 10" in my first message

a, b and c are real numbers chosen at random such that $a+b+c=10$ and $a \geq 0 $, $b \geq 0 $, $c \geq 0$.

What is the probability that no number is greater than 4

It's like there's a 10 m long thread cut at 2 random points giving those 3 numbers

user193319

Let $X$ be some locally compact space, and for every $n$ let $f_n : X \to \Bbb{C}$ be a function that vanishes at infinity. I was able to show that for every $\epsilon > 0$, there is an $N \in \Bbb{N}$ such that $|f_m(x) - f_n(x)| < \epsilon$ for every $m,n \ge N$. This means $\{f_n\}$ is a uniformly Cauchy sequence, right? Does this mean there is a function $f : X \to \Bbb{C}$ such that $f_n \to f$ pointwise on $X$?

Rick

@AkivaWeinberger how will we do it then?

Anonymous

@Rick Uh, use the technique of coefficients ?: $(x^{0}+x^{1}+x^{2}+x^{3}+x^{4})^3$

Anonymous

Find the coefficient of $x^{10}$ from there.

Rick

a b c can be any real number

Anonymous

19:27

Ah, I see

user193319

I think it does imply that such a function exists, but I still have to prove that $f$ vanishes at infinity, too.

Anonymous

Then I'd probably go for the geometric approach. Consider the $x+y+z=10$ plane

Anonymous

All your solutions for $a,b,c$ would lie on that plane

Rick

@Blue yup

how do I apply the constraints from there?

I'm not able to visualize it

Anonymous

Then find the area of that plane (in the first quadrant), for which $x\leq 4, y\leq 4,z\leq 4$

Anonymous

19:29

That plane should be easy to draw

Anonymous

The intercepts are just 10 on each axis

user193319

Actually, I don't see how to show that $f$ vanishes at infinity, unless $f_n$ converges uniformly to $f$.

And I know that $C_{\infty}(X)$, the set of all continuous functions from $X$ to $\Bbb{C}$ vanishing at infinity, forms a $C^*$-algebra, so $f$ better be in $C_{\infty}(X)$.

Rick

@Blue I drew the plane

I'm not able to understand how these new constraints will cut the area

Anonymous

OK. Now just try find the area of that plane for which $0\leq x\leq 4$.

Anonymous

Remember that we're only working in the first "octant"

Rick

19:42

hmm.. I assumed a new shifted coordinate system where $X + 4= x $ ; $Y + 4 = y$ ; $Z + 4 = z$

Anonymous

I don't see why you need to do that.

Rick

@Blue how will you do that? the x+y+z = 10 is tilted and the x=4 plane is straight

Rick

20:01

@Blue well, I did that because the numbers should not exceed 4, so if I have a number $t$, $4-t$ should be positive, so calling $X = 4 - x$, putting in the original eqaution, $ X + Y + Z = 12 - 10 = 2$, the points on this plane are the only ones allowed

I took the coordinates wrong intially

Anonymous

@Rick How I would find the area? I'd just find $A=\int_{0}^{4}\int_{0}^{10-x}\frac{dydx}{\cos(\theta)}$, where $\cos(\theta)$ is $(1/10,1/10,1/10).(0,0,1)$. The answer would be $3A/A_t$, where $A_t=\int_{0}^{10}\int_{0}^{10-x}\frac{dydx}{\cos(\theta)}$.

Anonymous

I think that should be right unless I've overlooked something. I need to leave now, though. We can discuss tomorrow.

Anonymous

Uh, I think there could be some overlaps between those 3 areas. So we might have to remove those extraneous areas.

Anonymous

Anyway, gotta go now

Kasmir Khaan

20:19

@MatheinBoulomenos hey mathein :D

@MatheinBoulomenos check your email when you have time =p this is very unlucky situation we got in -.-' anyway i need a conformation :D

Rick

ok

Alessandro Codenotti

What does the dual of $\mathcal{C}^0_b(\Bbb R)$ with the $||\cdot||_\infty$ norm look like?

user525966

Why isn't (F -> T) false?

Alessandro Codenotti

"If it rains I'll take an umbrella" is true, even if it's sunny

skullpatrol

Vacuously, yes.

PabloZ392

20:34

Hello :) For which values of "a" : $x(t)=1/a+ce^{-at}$ will be bounded ?

Alessandro Codenotti

Hi @Dami

user525966

@AlessandroCodenotti let's say P = "you mow my lawn" and Q = "I give you money"

If you mow my lawn and I indeed give you money, that's a true implication

If you don't mow my lawn and I give you money, that's still a true implication?

Alessandro Codenotti

"If I mow your lawn you'll give me money" is a true statement, regardless of whether I mow it or not

user525966

And, if you don't mow my lawn and I don't give you money, also true?

Alessandro Codenotti

we're assigning a truth value to the statement as a whole, not the single possible outcomes

user525966

20:44

but once you mow it then it's modus ponens and you can infer that i gave you money

right?

Alessandro Codenotti

right, but in this case you know that "if I mow your lawn you'll give me money" and "I mowed your lawn" are true

so you have more information

@0celo7 pinging you because you probably know the answer (or where to find it) to my question a few messages ago (it's in not Brezis or maybe I just couldn't find it)

user525966

p implies q = (not p or q)

p and (p implies q) = p and (not p or q) = (p and not p) or (p and q) = false or (p and q) = (p and q) right?

how does this then "equal" q?

Alessandro Codenotti

well it's not really equal, from p and (p implies q) you can deduce q, but I don't know what you mean with = here

user525966

i dont know the symbol you're supposed to use

just trying to "simplify" the boolean algebra

from one thing rearranged to the other

Alessandro Codenotti

The symbol is $\vdash$, but you can just say that "q can be deduced from the other two"

Anyway p and (p implies q) is not "the same" as q, because from the first two you can deduce q, but from q you can't deduce the first two

While p and (p implies q) and p and (not p or q) are "the same" because from the first you can deduce the second and viceversa

(I'm being informal with "the same" here, the technical term is that they are logically equivalent, one holds iff it the other holds)

user525966

20:52

what's the difference between $\vdash$ and $\implies$ then?

those two seem pretty similar

Sha Vuklia

Anyone here familiar with Galvin’s theorem, which states that $\chi_l’(K_{n,n})=n$?

Alessandro Codenotti

That's somewhat subtle and would deserve a more in-depth explanation of what I can give here. But $\implies$ is part of the language, while $\vdash$ is an external consideration saying that from some sentences another can be deduced

@Sha Do more logic and less graph theory! :P

Daminark

Hey @Alessandro!

@AlessandroCodenotti >:( graph theory is great

Alessandro Codenotti

Anyway this is not something you should worry about now, just forget about the existence of $\vdash$ for the moment, it's probably the best choice pedagogically if you're just starting to learn logic

user525966

I know it's considered more meta..logic? metalanguage?

but I don't really know what that means

Alessandro Codenotti

20:56

You can also ask in the logic chatroom, there's @user21820 who is both more knowledgable and a better teacher than me

@Daminark Do you happen to know something about that question?

skullpatrol

Conditional sentences in English grammar | Condicionales en Inglés

►

Sha Vuklia

@AlessandroCodenotti lol, I wish:P

Alessandro Codenotti

@ShaVuklia You should check out the De Bruijn–Erdős theorem if you want to see a cool result in graph theory which can be proved with logic/model theory techniques

Sorry for all the pings, I can't English tonight...

Sha Vuklia

lol, infinite graphs:p

I will look at it one day, but now I'm too much a noob, because I haven't really studied logic yet, and I haven't come across infinite graphs yet:p

but the theorem itself is pretty self explanatory and intuitive!

Alessandro Codenotti

Infinite graphs are the same as finite ones, they have a lot of vertexes :P

Sha Vuklia

21:01

so that's cool

@AlessandroCodenotti hahahahaha:p fair enough

Alessandro Codenotti

I can explain you the model theory proof after you've seen the compactness theorem if you're interested

Sha Vuklia

I have a lot of time pressure these days... ehhmm, let me come back at it when I have more time, and when I've done at least some logic:p I'm nót uninterested!

Alessandro Codenotti

Sure, there's not hurry

Sha Vuklia

alright then:)

Daminark

@AlessandroCodenotti what does the b stand for?

Alessandro Codenotti

21:07

bounded

continuous and bounded functions $\Bbb R\to\Bbb R$ (otherwise the sup norm is problematic)

Daminark

Hmm, so it's one of those things where I'd want to say that if we could extend it to a continuous function on the circle and then use Riesz we'd be good

We can't because it's not necessarily "nice at infinity"

But I feel in any event that it'll be some space of measures, with the heuristic being that it's "almost" the space of continuous functions on a compact space

Alessandro Codenotti

Yeah, I think it's some space of measures, but I need some more precise description of it. Does it contain the probability measures on $\Bbb R$? If so are they dense in this space?

(this all business is motivated by the fact that I'm trying to understand if convergence in distribution of random variables is actually weak convergence in some Banach space in disguise)

Because $X_n\to X$ in distribution (or weakly) $\iff$ $\Bbb E[f(X_n)]\to\Bbb E[f(X)]$ for all $f:\Bbb R\to \Bbb R$ bounded and continuous

Which, if $X_n$ is a random variable on $(\Omega_n,\mathcal{F}_n,\Bbb P_n)$ and $X$ is a random variable on $(\Omega,\mathcal{F},\Bbb P)$, is the same as saying $$\int_{\Omega_n}f(X_n(\omega))\mathrm{d}\Bbb P_n(\omega)\to\int_\Omega f(X(\omega))\mathrm{d}\Bbb P(\omega)$$ for all $f:\Bbb R\to\Bbb R$ bounded and continuous

Anyway I should go to sleep instead, I'll think about this tomorrow! Bye

Daminark

21:59

Catch you around! Looked it up and it turns out you need something called the Stone-Cech compactification

user193319

$Mathematics$ Modern Abstract Analysis

For functional analysis, measure theory, and related areas. M...

0celo7

22:35

@AlessandroCodenotti I think it's a subspace of the dual of $C^0(S^1)$.

Ah, just bounded functions. Well then a subspace of the dual of $L^\infty(S^1)$.

2

Q: Dual space of continuous functions

Let $C_b(\Omega,V )=$ { $ f:\Omega\rightarrow V $ } is the Banach space of all bounded continuous functions in Banach space $V$ with a norm $\|\cdot\|$ defined as $\|f\|_\infty=\sup _{x\in\Omega}\|f(x)\|$. Let $C_b(\Omega)=C_b(\Omega,\mathbb R)$. For a normal topological space $\Omega$ ( $T_4$-sp...

fa.functional-analysis measure-theory real-analysis

@AlessandroCodenotti For a textbook treatment, see Conway (A course in FA), V.6.

That's the other FA textbook I like.

Unfortunately there's no explicit description of $\beta\Bbb R$.

It's the spectrum of $C_b(\Bbb R)$ ;)

8

Q: Stone–Čech compactification of real line

I know that $[0,1]$ and a unit circle $\mathbb{S}^1$ are one-point compactifications of $\mathbb{R}$ under some suitable homeomorphism. But how does one construct the Stone–Čech compactification?

general-topology compactness

Joe Shmo

23:28

quick and dirty algebra question y'all

Fargle

Sounds fun.

Joe Shmo

let phi: (R, +) -> (C, *) be defined by phi(x) = exp(2 * pi * i * x). Then phi(R) = S^1, and phi is a homomorphism

also phi(x) = 1 <=> x in Z. I.e.(?) ker(phi) = Z?

oh yeah, and in S^1, 0=1

so ker phi = Z, yeah

nevermind. thanks for the help

Leaky Nun

@JoeShmo but you didn't ask anything

Joe Shmo

you got the funny part!

i did sprinkle a couple question marks in there, so yeah.

alan2here

hey

A genuinely (and generalisably) "hard" problem

involves "search" right

else it'll becomes "easy"

Well, after a while, and tend towards (however) reaching "trivial".

Is this not so?

$Mathematics$ Modern Abstract Analysis

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