Mathematics - 2016-03-31 (page 2 of 2)

manshu

13:02

hello...i am trying to find the name of the topic in which this happens: p->(q v r)
here the v is not alphabet..
it is symbol..

user147690

@manshu Propositional logic?

manshu

@AlexClark seems like it...thanks...

Mats Granvik

13:21

-1-1=-2

skill patrol

13:52

@MatsGranvik so?

is that surprizing?

Mats Granvik

@skillpatrol no, I guess not.

Mike Miller

14:11

morning

Balarka Sen

Morning, @MikeMiller.

user147690

Evening

robjohn

@Evinda so, depending on $\lim\frac y{x^3}$ the original limit can be anything.

Balarka Sen

By the way, I went through the proof you gave yesterday that $\text{dim} T_x X \geq \text{dim} X$. Note that an essential step of the proof that the collection of smooth points, namely, points such that $\text{dim} T_x X = \text{dim} X$, is open. I don't think this is trivial, because it's not clear why singularities are going to be the same as manifold singularities (at least in $\Bbb C$. and if you try to do it by rank the proof goes circular).

I have a fix for this though: just pick a hypersurface birational to $X$, for which it's clear. Then you can show that the set of smooth points in $X$ contain an open set. Now you do the same proof: look at all the points for which the Jacobian has rank $\leq d$ (that's closed). So you have a closed set which contains an open set, giving you a reduction of $X$ breaking irreducibility.

But probably you already knew this. In that case nevermind.

Mike Miller

I didn't really need them to be open... just dense

I didn't know how to prove they're dense but I firmly believed it, I just asked you if it was true so I could use it :)

Balarka Sen

14:20

ah, ok. I proved it's dense (namely, by proving that it contains an open set) above :)

Mike Miller

well, if you're happy

Mats Granvik

Would it be worth the effort to construct a matrix such that the row sums of it sum to the reciprocal of a Riemann zeta zero, trivial or non-trivial, when feeding in the same zeta zero? In other words, zeta zero as input, and zeta zero as output.

Now on second thought I don't know if it is possible.

ramsay

for a function to be defined: limit should exist and function must be defined at that point , right?

Balarka Sen

14:35

For a function $f$ to be defined at say $a$ you just need $f(a)$ to make sense, that is, that value needs to exist. No limit condition is required.

When $f$ is defined at $a$ and $\lim_{x \to a} f(x)$ exist and is $ = f(a)$, then $f$ is said to be continuous at $a$. To be continuous at $a$ you need to be defined at $a$, but not the other way around.

DeNiSkA

hi,
limit condition is required, take greatest integer function and limit doesn't exist at integral points (take $n$)but $f(n)$ still exists, and you know GIF is not continous (*at integral points*), so limit condition is neccesary.
make sense?

Balarka Sen

@DeNiSkA Read the message ramsay posted.

It said "for a function to be defined", not "for a function to be continuous".

DeNiSkA

sorry, i thought continous because usually those conditions are taken for continuity, sorry mate ;)

Balarka Sen

No problem.

DeNiSkA

haha!!

ramsay

14:47

What does this symbol "∏" mean?

Mats Granvik

Product

ramsay

product?

Mats Granvik

$$\prod _{n=1}^{5} n = 120$$

The matrix I have in mind is this:
$$\rho _1=\lim_{s\to 1} \, \frac{\zeta (s) \zeta \left(s \cdot \rho _1\right)}{\zeta '(s \cdot \rho _1)}$$

daOnlyBG

Good morning everyone

(well, hello)

quick question: does anyone know how to align everything to the left in MathJax?

I tried the whole \begin{align} and \end{align} but that aligns everything to the right

Mats Granvik

Remove the outer $ signs.

$\begin{align} and \end{align}$

daOnlyBG

14:55

I did remove them

I found a workaround. Thanks!

Balarka Sen

15:14

Suppose $X$ is an algebraic variety. Then any rational function $f$ on $X$ regular at a smooth point $p$ as a Taylor series - more precisely, there is an injective map of local rings $\mathcal{O}_{X, p} \to k[[t_1, \cdots, t_k]]$ defined as follows: $f - f(p)$ vanishes at $p$, hence is in $m_p$.

Thus choose a set of local parameters (generators of the cotangent space) at $p$, say $t_1, \cdots, t_k$, and choose $a_i$'s such that $f - f(p) - \sum_{i = 1}^k a_i t_i \in m_p^2$. Similarly choose $b_i$ so that $f - f(p) - \sum_{i = 1}^k a_i t_i - \sum_{i, j = 1}^k b_i t_i t_j \in m_p^3$. The map is defined by $f \mapsto f(p) + \sum_{i = 1}^k a_i t_i + \sum_{i, j = 1}^k b_i t_i t_j + \cdots$.

Now, if $X$ is a complex variety, there is a concrete characterization of the embedded copy of $\mathcal{O}_{X, p}$ in $\Bbb C[[t_1, \cdots, t_k]]$ - it's the ring of convergent Taylor series ring of rational functions regular at $p$.

But if my base field is arbitrary, there is no such notion of convergence. Can I still characterize $\mathcal{O}_{X, p}$ out of all the formal power series?

If so, how?

Mike Miller

15:59

lotta words

Balarka Sen

how about "how can I characterize taylor series of rational function regular at smooth pt $p$ of $X$ from all the formal power series in the local coordinates at $p$? for varieties over $\Bbb C$ we have a notion of convergence which helps. what about arbitrary fields?"

Mike Miller

fewer words for sure

Danu

16:17

@MikeMiller I don't know anything about other cohomology theories

I also don't know Mayer-Vietoris :\

Mike Miller

I don't know when I brought that up

Danu

You can follow the arrow back to the message I replied to.

Mike Miller

That's not true, because I primarily use SE on my phone.

Danu

Ah, okay. You mentioned that in response to my blabla about analogies between Leray coverings and SVK

Mike Miller

Ok... SVK doesn't work when the intersections are disconnected. MV is the homological version of SVK which makes no connectivity assumptions

Balarka Sen

16:20

@Danu To answer this in a few words: Mayer-Vietoris is an analogue of van Kampen for homology.

Danu

Oh, that's why I don't know MV---but that seems to justify what I was saying

SVK $\simeq$ MV $\simeq$ Leray stuff?

Mike Miller

It's the assertion there's an exact sequwnce $H_*(U \cap V) \to H_*(U) \oplus H_*(V) \to H_*(M)$

Danu

Ah, okay.

Mike Miller

I'm pretty sure that's not what I said but sure...

Danu

that's the thing that is the same in all the above?

I was replying to Balarka, Mike

Balarka Sen

16:23

I don't know which message was a reply to what I wrote, except possibly the "that's why I don't know MV" one.

Danu

The one after it, as well

user17629

Guys, a question.. if we have the product of linear operators, equal to another one,that implies invertible of each one?

Balarka Sen

I don't know Leray stuff. But it all depends on what you mean by the equivalence sign.

Danu

@BalarkaSen $\simeq$ means (there) "there exists some vague analogy"

possibly made precise by Mike's comment about an exact sequence

Balarka Sen

ok, sure.

user17629

16:26

I meant. Let $\tau, \sigma = \iota$, where each one belongs to $\mathcal{L}(V)$, then $\tau$ and $\sigma$ are invertibles?

Balarka Sen

I should also probably note that by that exact sequence there Mike really means the long exact sequence which repeats that bit again and again, in case it's confusing (thus the $"*"$ in the indices). If you knew this, nevermind.

user17629

$\tau \sigma = \iota$

Mike Miller

@Danu I think you're being too broad. There is indeed a common theme in all of these, but it's not exceptionally deep or whatever. It's "Cut the space into smaller, easier pieces and work with those"

In the Leray covering you're cutting it into the simplest possible pieces. Now note that Mayer-Vietoris, as stated, doesn't involve more than two open sets.

Danu

But specifically this thing with "cut it into pieces which are themselves trivial and then look at how the intersection fits in"

Which was in practice how most applications of SVK that I've seen went

Balarka Sen

in svk the pieces need not be trivial.

or at least in the most general version of svk.

Danu

16:30

Definitely not, but many applications use this

It's an important special case

Balarka Sen

It's just widely applicable. The general version is more useful because it's even more widely applicable :)

Danu

Anyways, I got the message by now---what I was on about was not particularly interesting :)

GPhys

17:05

@AkivaWeinberger I think I proved AC with ZF+IST; I'm going to discuss it with my advisor and I'll get back to you

Peter Sheldrick

17:22

is it harder to find the link to this chat on the math.se main site than it used to be?

Danu

@PeterSheldrick Should be top-left (though that doesn't work for me, for some reason)

Peter Sheldrick

i have a vague memory that it used to be on the right side (as well?)

and the link was right there, it wasnt inside some drop down menu

Danu

OK. Must've been before my time.

r9m

17:47

@DanielFischer if $f$ is a holomorphic (and vanish as $\Re(z) \to + \infty$) can we say $\displaystyle \int_0^{\infty} \frac{f(x) - f(\lambda x)}{x} = f(0) \operatorname{Log} (\lambda)$ where, $\Re (\lambda) > 0$? (something like a frullani for complex parameter)

need some sleep .. headache .. bbl

rorty

18:34

how do you number equations in mathjax (like in a question post)

r9m

18:44

@rorty with \tag{1} .. yes ... that's what I usually do .. refer it with the equation number tagged

rorty

If you refer to it later, do you say $(1)$ or is there a more elegant way to refer?

GPhys

19:00

@AkivaWeinberger Assuming the statement that given a collection of sets $X=\{A_i\}$, one can construct a set $B$ with the property that $\lvert B\cap A_i\rvert =1$ for every $A_i$ is equivalent to AC

Akiva Weinberger

@GPhys Disjoint $A_i$, otherwise $X=\{\{0\},\{1\},\{0,1\}\}$ invalidates it

GPhys

Where each $A_i$ is pairwise disjoint nonempty

Yeah

Balarka Sen

Anyone noticed we got a new game ("Unikong") for April Fools?

GPhys

@AkivaWeinberger Basically, what I tried to do originally DOES work, I just didn't try hard enough to standardize it

I'm on my phone right now; I'll write it more formally when I get home in a little bit

Akiva Weinberger

It's not April Fools' Day yet here @BalarkaSen

Random Variable

19:16

@r9m It holds for $\lambda >0$. And the function on the right side is analytic on the half-plane $\text{Re}(\lambda) >0$. So I guess we would need to argue that the integral is analytic on the same half-plane.

Balarka Sen

I think they're april fooling us on the high scores. how the hell can one score 40,000 on that game??

also the string of 0's after each of the high scores clearly shows that this should be some sort of a bad joke.

GPhys

@AkivaWeinberger Let $X'$ be the finite set that has for standard elements the standard elements of $X$. Because it's finite, there exists a $B'$ such that $\lvert B'\cap A_i\rvert =1$ for all standard $A_i$. Define $B={^\mathsf{s}\{}x\in\bigcup A_i :x\in B' \}$

r9m

@RandomVariable yes .. that'd be my argument too .. (choosing a trapezoidal contour similar to the one I mentioned in the link in p/m .. )

GPhys

$\square$

r9m

@BalarkaSen I saw that .. what the hell was that a unicorn dropping fiery potty? :P

Balarka Sen

19:25

It was a genetically modified unicorn with an internal unicorn biogun, obviously.

But, yes, in short - a unicorn who drops fiery potty.

GPhys

@AkivaWeinberger It's important enough that it deserves my closer examination later; and maybe a separate attempt to prove other statements of AC

Agawa001

sup @r9m

r9m

@BalarkaSen LOL

@Agawa001 sup :)

Michael

Hey guys, given an n dimensional euclidean vector, how can one find all the combinations of perpendicular lines?

r9m

ah! .. unicorn on top, reputations, badges, and the potty drops (downvotes) makes sense .. :P LOL

Balarka Sen

19:38

oh there's a level 4. that's new.

r9m

whaa? .. how do you get past the first level? :o

Mary Star

The center of a group G is a normal subgroup of the group G, right?

Balarka Sen

@r9m don't try to kill the unicorn.

just try to get past it by getting all the shiny diamond thingys

I think the level 4 thing is a glitch. I shot it down with a gun after level 3, and it got me to level 4.

r9m

@BalarkaSen I had all the shiny diamonds (I think) but I won't level up

Balarka Sen

level 4, 5, 6 are just the same as 1, 2, 3. It's just a glitch like Mario has.

I did the same thing with level 6, and it got me to 7. :P

Now I know how I can get 40000

@r9m Weird. Did you get the topmost diamond? The one the unicorn guards?

r9m

19:43

@BalarkaSen sure ..

Tobias Kildetoft

@MaryStar Yes

Balarka Sen

@r9m Try it again.

Ah. There's actually a real difference between the levels mod 3. I think the speed the potty drops gets quickened by a factor of 3 too.

r9m

@BalarkaSen hah .. got it .. till level 5 after that I decided to shake hand with the trolls :P (apparently you can hold back the unicorn with enough gun fire :P)

Balarka Sen

Yup.

And holding back the unicorn with gunfire in level 3 is only how you'd get to level 4.

Or at least, I didn't do it before, and it didn't take me to level 4.

Darn. I only made it to 25640.

Well, I am not going to play it anymore. Got to get work done.

DeNiSkA

hello,
writing $\sqrt{-2}=\sqrt{2}i$ is a valid step, *right*?
then why can't we write:
$$\frac{1}{i}=\frac{1}{\sqrt{-1}}=\sqrt{\frac{-1}{1}}=\sqrt{-1}=i$$
and this is completely wrong, wikipedia article on square roots confused me :(

Balarka Sen

19:57

Your second step is wrong. $1/\sqrt{-1} = (1 \cdot \sqrt{-1})/(\sqrt{-1}\cdot \sqrt{-1}) = \sqrt{-1}/(-1) = -\sqrt{-1}$.

DeNiSkA

nope, i am not multiplying and dividing by $\sqrt{-1}$
rather i am writing $1$ as $\sqrt{1}$ and then i am writing it as $\sqrt{\frac{1}{-1}}$

Mary Star

@TobiasKildetoft Ah ok... Does it hold then that $gZ(G)=hZ(G)$ iff $g^{-1}h\in Z(G)$ ?

Tobias Kildetoft

@MaryStar Yes, that holds for any subgroup

Balarka Sen

Oh, I see. Then no, $\sqrt{ab} = \sqrt{a}\sqrt{b}$ is not correct (at least not modulo signs). Easier example: e.g, take $a = -1, b = -1$.

DeNiSkA

is there any rule for this fallacy (i want to know why this is incorrect)

Mary Star

20:04

Why? From which of these properties [en.wikipedia.org/wiki/Normal_subgroup#Definitions ] does this follow? @TobiasKildetoft

Tobias Kildetoft

@MaryStar I said any subgroup, not just normal ones. You should have seen this when cosets were introduced

Balarka Sen

The reason this happens is because $x^2 - ab = 0$ has two roots: $+\sqrt{ab}$ and $-\sqrt{ab}$.

Akiva Weinberger

@DeNiSkA $\dfrac1{\sqrt{-1}}\ne\sqrt{\frac{-1}1}$

Balarka Sen

$(\sqrt{a}\sqrt{b})^2 - ab = 0$, so $\sqrt{a}\sqrt{b}$ is also a root of $x^2 - ab = 0$.

Akiva Weinberger

$\sqrt{a/b}=\sqrt a/\sqrt b$ is only guaranteed to be true for $a,b>0$

Balarka Sen

20:05

But how do you know which one it is, $+\sqrt{ab}$ or $-\sqrt{ab}$?

DeNiSkA

@BalarkaSen hmm, making some sense!

then why $\sqrt{-2}$ is written as $\sqrt{2}i$
even $a,b$ are not greater than $0$

Mary Star

@TobiasKildetoft Ah ok... Thanks!! :-)

Michael

Given 3 integers $a$, $b$, and $c$, find values of a b and c such that $a+2b-c=0$

Balarka Sen

As Akiva said, for $a, b> 0$, $\sqrt{a}\sqrt{b}$ has to be the root $+\sqrt{ab}$ because square root of positive number is positive, and thus it cannot be a negative number $-\sqrt{ab}$.

DeNiSkA

@BalarkaSen No! i didn't get this (sorry)

Balarka Sen

20:10

What do you not get about it?

Tobias Kildetoft

@DeNiSkA It is not really correct to write $\sqrt{-2} = \sqrt{2}i$ just as it is in fact not really correct to write $\sqrt{-1} = i$.

Balarka Sen

$(+\sqrt{a})(+\sqrt{b})$ is a positive number, as $a, b > 0$. Also we know $(+\sqrt{a})(+\sqrt{b})$ can be either $+\sqrt{ab}$ or $-\sqrt{ab}$. But the latter of these is negative, and a positive number cannot be equal to something negative.

So the only possibility is $(+\sqrt{a})(+\sqrt{b}) = +\sqrt{ab}$.

DeNiSkA

@TobiasKildetoft hey but we represent $\sqrt{-1}$ as $i$ then why it is incorrect

Tobias Kildetoft

@DeNiSkA $\sqrt{-1}$ is not a well-defined number

DeNiSkA

@BalarkaSen oh! got it!

Michael

20:15

What exactly is a clopen set?

DeNiSkA

if $\sqrt{-1}\ne i$ it means we have to represent $\sqrt{-24}=j$
$\sqrt{-34}=k$
......and so on . it means there is no point of using $i$

Tobias Kildetoft

@DeNiSkA No, $i$ is one of the square roots of $-1$. $-i$ is the other. Neither is a better choice than the other.

Semiclassical

hiya

Balarka Sen

Tobias is making the point here that $\sqrt{-1}$ is not a well-defined number. It can be either $-\sqrt{-1}$ or $+\sqrt{-1}$, using my notation.

DeNiSkA

it means $i=\sqrt{-1}$ or $i=-\sqrt{-1}$

it can be any of either, am i right?

Michael

20:20

is topology without tears a decent book?

Balarka Sen

@DeNiSkA Note that whenever you say $\sqrt{\cdot}$ and $-\sqrt{\cdot}$, you are inherently making a choice. Why is this choice well-defined? How do you define $+\sqrt{\cdot}$ and $-\sqrt{\cdot}$ when your arguments are negative numbers? This is the point Tobias is making. This is not a well-defined concept, so you make a choice as you move along.

OK, I gotta get back to work.

DeNiSkA

got it! (*not completely*)
and bye

Balarka Sen

20:35

@DeNiSkA The point I am trying to make is this. If $a$ is a positive number, $x^2 - a = 0$ has two solutions. One is negative, one is positive. The positive one you call $\sqrt{a}$, the negative one $-\sqrt{a}$. Fine, that's well defined and everyone is happy.

But the minute $a$ is a negative (e.g., $-1$), the two solutions of $x^2 - a = 0$ are indistinguishable. I can choose one of the solutions to be $i$, and call it $i$ and work with it. But I can never choose is canonically. There is no unique choice (the "fancy math" reason is that $\Bbb R$ has an order on it, $\Bbb C$ doesn't).

E.g, if $a$ and $b$ are solutions of $x^2 + 1 = 0$, we know we have $a = -b$. So you say we can choose $a$ to be $i$? Wrong. We also know that $b = -a$, why can't I choose $b$ to be $i$? That seems equally sensible.

So it's like having two completely identical objects one after another, and then flipping the order the two objects are in and asking which order is "nicer". This has no canonical answer- any order works, because the objects are identical.

There are is a "mathy" ways to word this ("Galois groups acts on roots transitively"), but this is what it comes down to.

Vrouvrou

Hello, is this space is a Hilert space: $L^q_h=\{u: \Omega\rightarrow \mathbb{R}^N, measurable,~\int_{\Omega} h(x) |u|^q<\infty\}$

where h\in L^{\infty}

strictly positif on $\Omega$

bounded set from R^N

Evinda

Hey @DanielFischer
Suppose that we are looking for a limit $\lim_{(x,y) \to (0,0)} g(x,y)$.

If we find the limit along the line x=0 and the limit along the line y=0, and they are equal does this mean that the orginal limit is also equal to these ones?

PVAL

A Hilbert space is a vector space with an inner product satisfying a list of properties. What is your proposed inner product? If I understand you right, that thing is not even a real vector space. It'd be a good idea to have some familiarity with vector spaces before trying to learn about Hilbert spaces.

Vrouvrou

@PVAL a norm is given ||u||=\int h(x) |u|^q dx

PVAL

@Vrouvrou A norm and inner product are not synonyms...

DeNiSkA

20:48

@BalarkaSen woah!! this made sense, thank you :)

Vrouvrou

yes i know but that is all what i have @PVAL

Akiva Weinberger

@Michael A clopen set is one that's open and closed

It depends on the topological space. In $\Bbb R$ with the usual topology, the only clopen sets are $\Bbb R$ and $\varnothing$.

In the space is $[0,1]\cup[2,3]$ with the subspace topology induced by $\Bbb R$ (do you know what the subspace topology is?), $[0,1]$ is clopen.

litmus

Hey! Can anyone please give a simplified explanation how Gödel's first Incompleteness theorem shows us that there are infinitely many true statements which are unprovable within a formal system? Is it a case of "the more we know, the more questions we create"?

Akiva Weinberger

A topological space is connected if the only clopen sets are the whole space and the empty set.

@litmus Well, the theorem says that the formal system has at least one unprovable statement, right?

Say that $A$ is the set of axioms in our formal system, and suppose that $x_0$ is a statement unprovable in it.

Now, apply the theorem to the formal system whose axioms are $A\cup\{x_0\}$…

we get another unprovable statement, which I'll call $x_1$.

litmus

@AkivaWeinberger I have this definition in my text:

There exists no formal proof system satisfying all of the following
conditions.
(i) There is an algorithm for telling whether or not a given sequence of
sentences constitutes a formal proof of a given sentence.
(ii) Every true sentence is formally provable.
(iii) Every formally provable sentence is true.

Akiva Weinberger

20:56

So you want to show that a system satisfying (i) and (iii) has infinitely many true, unprovable statements, right?

Balarka Sen

(iii) is said to be the consistency condition and (i) is known as the completeness condition. By a formal system Akiva really means a consistent and complete proof system which is enriched over the Peano axioms.

Akiva Weinberger

Apply the theorem to the system to get one unprovable statement, add it to your list of axioms, apply the theorem to that to get another unprovable statement, add that to your list of axioms, etc. No?

Balarka Sen

And Godel is equivalent to saying such a system doesn't satisfy (ii), as Akiva said.

litmus

@AkivaWeinberger Well, first I want to understand why there are infinitely many because then I have to understand how we could get a "perfect" formal language if there where only a finite number of true statements that are unprovable

Akiva Weinberger

If you had a formal language with only finitely many true-unprovable statements, then you could add those statements to your list of axioms, and you'd get a formal system that satisfies (i), (ii), and (iii). Contradiction. (If you try this where there's an infinite number of true-unprovable statements, you could break (i), I think, because there wouldn't be an algorithm telling you if something is on your list of axioms or not.)

@litmus

Semiclassical

21:02

similarly to what others have said: start with a system, use GIT to get an unprovably true result, then add it to your axioms and use GIT again to get a different unprobably true statement. What's the status of this second statement in your original system?

litmus

@AkivaWeinberger thanks, it is starting to make sense now. But how does having a finite number of true sentences which are unprovable solve conditions (ii) and (iii)?

@BalarkaSen hmm, but what do you mean by 'enriched' over PA?

Balarka Sen

That you can do arithmetic with your formal theory, in a manner consistent with PA.

Akiva Weinberger

If you add those unprovable statements to your axiom system — and those really are all of the true unprovable statements — then you get condition (ii)

As for (iii), if you start off with a consistent system, adding true statements can't make it inconsistent

GPhys

hey @AkivaWeinberger

Akiva Weinberger

Hey

GPhys

21:10

did you see the IST->AC

litmus

@Semiclassical By status, do you mean how does it affect PA (which is my system)?

GPhys

@AkivaWeinberger it's really the same idea I originally had, or the same idea for the total order proof

Semiclassical

i meant, is the second statement--which was true but unprovable in the new system--also true and unprovable in the original system?

litmus

@Semiclassical yes, the new system should include all the axioms of the old system

Akiva Weinberger

2 hours ago, by GPhys

@AkivaWeinberger Let $X'$ be the finite set that has for standard elements the standard elements of $X$. Because it's finite, there exists a $B'$ such that $\lvert B'\cap A_i\rvert =1$ for all standard $A_i$. Define $B={^\mathsf{s}\{}x\in\bigcup A_i :x\in B' \}$

Semiclassical

21:14

right. then you've just shown that there at least two true but unprovable statements in the initial system

Vrouvrou

@PVAL can we say that it is a reflexiv Banach space ?

Semiclassical

and you can keep iterating that logic to get as many as you want.

litmus

@AkivaWeinberger @BalarkaSen @Semiclassical also, thank you guys for your answers, i am blown away by your level of help!

PVAL

@vrouvrou Not without showing it.

Akiva Weinberger

@litmus Yay

GPhys

21:15

@AkivaWeinberger yeah, this

Vrouvrou

@PVAL what is the conditions to obtain a reflexiv Banach space please

Akiva Weinberger

So, suppose $A_i$ is standard. We know that $B'\cap A_i$ is a singleton (it has one element). Is the same true of $B\cap A_i$, then?

litmus

@Semiclassical yes, thanks

Akiva Weinberger

Hm. Suppose not; then, $(B\cap A_i)\setminus(B'\cap A_i)$ would be nonempty

litmus

@Semiclassical So basically to create a "perfect" formal system, IF there are only a finite number of unprovable true sentences, all I have to do is to modify the PA- sytem by adding axioms until i have proven all true sentences?

Akiva Weinberger

21:19

@GPhys What if $B'\cap A_i$ is nonstandard?

Semiclassical

sounds right?

Akiva Weinberger

Then $B$ doesn't need to contain the element of $B'\cap A_i$

And then we don't know anything about $B\cap A_i$

On the other hand, if all $A_i$ were finite, this proof would work. Because standard finite sets only have standard elements.

litmus

ok, thanks. now i'm off to create said system

PVAL

@Vrouvrou Try these notes ma.utexas.edu/users/arbogast/appMath08c.pdf for looking up definitions or Conways Functional Analysis book.

Akiva Weinberger

Yeah, @GPhys, I think this doesn't work if $B'\cap A_i$ is nonstandard for any of the $A_i$.

GPhys

21:22

good point @AkivaWeinberger, let me examine to see if I can fix it (really, I was being lazy to make it look like the others; I should know by now to not be lazy with IST....)

@AkivaWeinberger I think it is fine maybe still, because you can guarantee it is standard by relativizing the statement used to construct it

Akiva Weinberger

@GPhys Relativizing just lets you assume that $A_i$ is standard.

GPhys

@AkivaWeinberger let me elaborate what I mean, just a sec

Akiva Weinberger

But standard sets can have nonstandard elements.

And $B'\cap A_i$ is the intersection of a possibly-nonstandard set with a possibly-infinite set…

GPhys

writing it out will make clearer if I'm wrong in my thinking anyway

Akiva Weinberger

If anything, you've at least proven the weaker version of choice in which all of the $A_i$ are finite.

GPhys

21:27

I'm suggesting adding something when I say relativize: sec

Mike Miller

Thought: while bounded harmonic functions on $\Bbb R^2$ are constant, there is a massive number of bounded harmonic functions on the unit disc w/ hyperbolic metric (the Laplacian is a conformal invariant, and it's conformally equivalent to the Euclidean metric on the unit disc). Can you make some sort of growth condition that will let you get a finite-dimensional space of harmonic functions?

Note that you can't demand they vanish at infinity, because the maximum principle would imply they would all be zero. Maybe something silly like sub-linear growth.

2nd question: is there an appropriate such thing so that you can recover hodge theory on noncompact, complete manifolds?

GPhys

@AkivaWeinberger Yeah it can't quite be fixed

You are right though, it is fine if $A_i$ is finite

Although I don't know what kind of weaker AC this would be equivalent to

Akiva Weinberger

Weaker than BPI, I think

(en.m.wikipedia.org/wiki/Boolean_prime_ideal_theorem)

GPhys

I'm not familiar enough to comment, really

@AkivaWeinberger Although I guess I should re-examine if the total order proof has the same issue

@AkivaWeinberger I'd have to reread some stuff :P

Balarka Sen

21:49

If I look at the variety $XY = 0$, is the local ring $\mathcal{O}_{X, p}$ where $p$ is the origin, a familiar object?

Mary Star

Let $M$ be a field and $G$ the multiplicative group of matrices of the form $\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}$ with $x,y,z\in M$.

I have shown that all the elements of the center $Z(G)$ are the matrices of the form $\begin{pmatrix}
1 & 0 & \tilde{y} \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$.

How could I show that $G/Z(G)$ is abelian?

We have that $G/Z(G)=\{gZ(G)\mid g\in G\}$.

Do we have to take $A=g_1Z(G)$ and $B=g_2Z(G)$ and show that $AB=BA$ ?
Or do we have to take the same $g$ just an other element of the center?

Balarka Sen

Oh, of course it is.

litmus

@AkivaWeinberger could you pls help me understand something you said: "As for (iii), if you start off with a consistent system, adding true statements can't make it inconsistent." What does it mean when a system is consistent and inconsistent? Is it that it will always give the same truth-value to the same sentence?

i.e. something that is true continues to be true even if we add axioms (to include the finitely many true unproven sentences)?

Semiclassical

if the axiom you added would make the statement false, then said axiom wouldn't be consistent with the system, no?

litmus

@Semiclassical that is true

skill patrol

22:02

Truth is the litmus paper test :P

litmus

@skillpatrol haha

skill patrol

:D

litmus

@Semiclassical but how will i know that the newly added axioms will not ruin my "perfect" system by including false statements? or can i solve this by condition (i), i.e. by algorithmically checking ?

Akiva Weinberger

@litmus If, assuming $x_0$, you could prove a false statement, then that's the same as proving $x_0\implies\rm False$ in the original system. That's equivalent to $\lnot x_0$

litmus

@Ak

Akiva Weinberger

22:10

So then $x_0$ wouldn't be true and unprovable; it would be probably false.

litmus

@AkivaWeinberger sorry but i dont understand those logic symbols. We only use basics: x ' 0 S + · ( ) = ∼ [ ] ∀

what does ¬x0 mean?

@AkivaWeinberger nevermind, I understand what you said!, thank you

Akiva Weinberger

That means "not"

$\implies$ is "implies"

litmus

thanks!

GPhys

22:25

The putnam results should be out any day now

skill patrol

Are you expecting a good result? @GPhys

GPhys

@skillpatrol Decent, I guess

skill patrol

Good luck :-)

Save me a good seat in math 55 :P

litmus

@skillpatrol @GPhys are you really taking math55?

skill patrol

22:41

Does Oxbridge have a math 55 equivalent? @litmus

litmus

i would assume so

@skillpatrol

Julian Rachman

0

Q: Noetherian Lemma Contridiction

There is a noetherian version of Higman's Lemma which says If $X$ is a noetherian poset, then $X^*$ is noetherian. Now I was thinking, given $X=\{1\}$ we get $X^*=\{1,11,111,\cdots\}$ by the concatenation of $X$. In addition, say $X=\mathbb{N}$; however $\mathbb{N}$ is not noetherian with t...

Anyone? ^

skill patrol

Hi @JulianRachman

Julian Rachman

@skillpatrol hello! :)

skill patrol

Are you ready for the real hockey season?

aka playoffs

Julian Rachman

22:49

Of course!

No Canadian teams made the playoffs this year

I guess it is a US showdown

skill patrol

Yup, and they call it their game.

They game belongs to the highest salary cap.

Julian Rachman

Yep

Lol go kings!

Julian Rachman

23:01

Ok so no one has a clue for a solution to my question? (above ^)

DemCodeLines

How exactly would you find "find all elements in Z143 of order 13"?

bwDraco

23:18

welp. So I came home with a copy of The Number Devil.

skill patrol

Don't open it and let the "Devil" out!

bwDraco

$\frac{1}{1}$

$\frac{1}{1+1}$

$\frac{1}{1+1+1+1}$

Page 21:

$1 \times 1 = 1$

$11 \times 11 = 121$

Page 23:

$11111 \times 11111 = 123454321$

It all starts with $1$.

yesterday, by Eric Stucky

"Just ask; don't ask to ask"

We have this as a canned response from the chatbot in Super User's main room Root Access:

in Root Access, 9 secs ago, by ChatBot John Cavil

@bwDraco Please don't ask to ask; if you simply ask your actual question, we will help you if we know the answer or can help you find it. This is much faster and simpler than asking if it's OK to ask. As a rule, it is always OK to ask in this channel. Please go ahead.

DemCodeLines

Umm, ok...? I posted my question already...

bwDraco

lol

DemCodeLines

I have no clue what's going on...

Transcript for

$Mathematics$ Mathematics