@anon When you mentioned " only the order of the "constant terms" relative to the "quadratic terms" (as you call them) is important. " With reference to the parenthetical statement (as you call them) Is there another name for these terms?

not really

@anon It's just a matter of labels and it doesn't matter what actual term in the quadratic polynomial that goes there right? excluding the linear term of course.

I can't parse what you're saying.

When you're trying to find the form (Ax-B)(Cx-D) and you have candidate sets of factors {A,C} and candidate sets of factors {B,D}, you can fix the order of the two items in the first while letting the order vary in the second set, or vice-versa.

The linear term must stay in the middle and we can switch the order of the other two. As you say vice-versa with terms the other way round, right? For factoring purposes.

What are "the linear term" and "the midde"?

ax^2 + bx + c, where bx is the linear term in the middle.

you cannot rewrite ax^2+bx+c as cx^2+bx+a. If that's what you mean, you're taking my comments and placing them in a different context, making them invalid.

c + bx + ax^2 is what I mean.

no duh that's the same, you can change the order in any (finite) sum.

reviewing suggested edits is such a thankless chore

@anon So that is why you could, if you wanted, fix the order of -1, -5 and then go over the different orders of x,14x and 14x,x and 2x,7x and 7x,2x, when factoring 14x^2 -17x + 5.

No that is not why. It is because (Ax-B)(Cx-D) is the same as (Cx-D)(Ax-B) - they come out the same way regardless of which set of candidate factors order you fix and which you let vary. I'm going to go make myself some breakfast now.

@PaulSlevin Hi

Hello @Skullpatrol

@PaulSlevin How's it going?

@Skullpatrol I'm ok. Just learning about trees now. I feel like a biologist

how are you

@AsafKaragila What's the east coast of the solar system?

@MattN The entire solar system.

@PaulSlevin Fine thanks.

does anyone here know about trees? I just have one super-mega-quick question

?

as in, set-theory trees

not the leafy ones

Shoot.

so much for biology ;-)

i have defined the height of a tree to be the least ordinal $\alpha$ such that the $\alpha$th level of the tree is empty. If you have the $\alpha$th level of the tree being empty, is every higher level empty?

Yes.

haha

If $x$ is on level $\alpha$ then $\{y\in T\mid y<x\}$ has order type $\alpha$.

if i had height (T) = $\alpha$ then if there was some $\beta > \alpha$ where there was some element $x \in T$ with height $\beta$, then I guess I can shrink the set of predecessors to get an element of height$\alpha$, which is bad

ok thanks, i just wanted to be sure

So if you had someone on level $\alpha+1$ you have someone on level $\alpha$.

ok

Thanks

if you have a dude on a level$\alpha$ then surely there is a dude on every level $\beta < \alpha$?

The Dude.

That is my way of saying "someone" or "this guy" in maths

the null dude

occupies the empty level

haha

well he wouldnt be in it if it was empty

But he's the null-dude or he's nobody.

brain melting

sorry

@Jonas: hi

its ok my brain melts 2-3 times a day

@Ilya Hi!

@PaulSlevin Just trying to come up with a mnemonic for you.

I find emptiness hard to remember

like nothing = not a thing or no thing.

so nobody would be not any body = the null-dude

excellent

@WillHunting You change gravatars faster than a speeding bullet ;-)

@WillHunting Thanks for the help.

@Jonas, @Ilya: I don't get it. You guys are several floors from one another, just run a string with two cups and use that.

@AsafKaragila No, it is the building next to mine.

We also have a phone.

I thought that Delft TU was like one huge building and some parking space.

We don't need to use cups and a string anymore.

Are you saying that Ilya is the guard in the parking lot?

No, it is like 12,5% of the area of Delft.

Have you measured it inch by inch?

So about 3km^2.

No, they own about that much of Delft.

But there are ~15 buildings.

@JonasTeuwen Do you have law and medicine faculties there?

No.

Only engineering and some small things.

@robjohn Are you there? :-). I have been asked to control a maximal function by a square conical function in the $L^1$ norm. So we already have $\|S_a f\| \lesssim \|T_a f\|$ ($T$ is the maximal one, $S$ is the conical one) and they want $\|T_a f\| \lesssim \|S_a f\|$, so a homogeneous estimate in the other direction as well. But now I'm confused as $S$ maps all constants to $0$, how can this ever work?

@JonasTeuwen What does $S_a$ look like?

@robjohn Tall, slender, blonde hair, blue eyes, a huge nose and crooked teeth.

@AsafKaragila typical

Yeah.

I think I can finally trust you, Robert. (at least on the main site)

@robjohn Now go here and prove yourself trustworthy!

@robjohn $$S_a u(x) = \left ( \int_{\Gamma_x^a} |t \nabla e^{t^ 2 L} u(y)|^2 \, \mathrm{d}\gamma(y) \, \frac{\mathrm{d}t}{t} \right )^{\frac12}.$$

$e^{t^2 L}$ is a semigroup, and if you feed it a constant if spits back that constant.

And then the $\nabla$ kills it.

@Jonas: what are cups and strings?

@Ilya If you take two cups, and you put a tensioned string between them, you can talk like that to eachother :-).

@Ilya CUPS is a printing protocol and strings are these thing in the 23 extra-dimensions which we don't see because we're not tiny strings.

@JonasTeuwen are, that stuff. We tried it once when I was 6 years old

@JonasTeuwen $L$ is a differential operator?

@Asaf: here is the part of TU Delft's campus where me and Jonas are. He is where the blue oval, me is where the red one

in between of them the wind is very strong :D

@JonasTeuwen coffee cans work better than cups

this is the main part of the campus, but still the Architecture building, Chemical Engineering, Biological and Aerospace ones are not there

@Ilya Use optic fibers made of doom.

I am going to stalk my advisor around his office.

@robjohn :D.

We have phones.

anyone willing to talk some more biology with me?

(i.e. about trees)

@JonasTeuwen do they work better than coffee cans?

@robjohn I haven't tried coffee cans, but I think so.

@PaulSlevin trees... they make paper from those, right?

So do you know what they mean with a homogeneous estimate here? :-).

Set-theory trees

they waste paper cos you write so much trying to understand them

@JonasTeuwen do constant functions apply since they are not in $L^1$?

@robjohn We have a probability measure!

Hmm, but maybe they need to be compactly supported...

So, oops! 8-).

Oh, right then I am saved! Thank you!

@JonasTeuwen and then they are not killed by the $\nabla$

Yes.

Hi all

@rob : Is $\frac{1}{x^{0.3}}$ integrable...i mean integration from $0$ to $\infty$

Hi

@tb: hi - do you have now time to take a look at my question?

@raj no.

@Ilya hi

Inverse powers of $x$ are never integrable from $0$ to $\infty$.

@RajeshD nope near $0$ yes, near $\infty$, no.

Either they are around infinity or around $0$. For $1/x$ never.

You could have checked that easily right? 8-).

@RajeshD Among the $1/x^\alpha$ the critical case is $1/x$: you have to decay faster towards infinity (that is $x^\alpha$ with $\alpha \gt 1$) to be integrable on $[1,\infty)$ and explode slower towards zero that is $x^\alpha$ with $\alpha < 1$ to be integrable on $(0,1]$.

@tb Hi there! good afternoon

yeah, and a brilliant and sunny one at that!

@tb the same here :)

@Ilya wanna have a peek at the webcam? :)

can you give me the link to your question again?

@tb why not, give me the link please

@tb sure

@Ilya here

Oh, it's foggier on the other side of the hill :)

$\frac{1}{x^{1+\epsilon}+x^{1-\epsilon}}$ is integrable.

Hi @tb

Hi Kannappan

I would like your help with some linear algebra. Remember, we proved that nilpotent operators are necessarily not injective? @tb

@tb @rob : I intended $\frac{1}{x^{1.3}}$ in $(0,1)$...according to you it is...Am I right @tb ?

sorry for the double ping

@KannappanSampath since they send something to $0$ :-) (other than $0$)

@RajeshD that one is integrable on $(1,\infty)$ but not on $(0,1)$.

@robjohn Yes. right, but I could not prove it the other day. : (

@RajeshD that is not integrable on that range.

but 1 over $x^2$ is right ?

where? Yes on $(1,\infty)$ but no on $(0,1)$.

(0,1)

@RajeshD on $(1,\infty)$

Hmph, he neglected it! : (

@RajeshD not integrable there

@KannappanSampath neglected what?

Once again: if $\alpha \gt 1$ then $1/x^\alpha$ is integrable on $(1,\infty)$ but not on $(0,1)$ and if $\alpha \lt 1$ then $1/x^\alpha$ is integrable on $(0,1)$ but not on $(1,\infty)$.

Okay $\log(x)$ on (0,1) ?

@robjohn my ping... : (

rendering spell

2

@tb: you've missed \text{} or messed up dollar signs

@KannappanSampath I haven't neglected you... I do remember.

But I was talking to Rajesh and Ilya already.

@tb @rob Thanks very much for the help

@tb : ) Ok. The third argument went something like: only for finite dimensional spaces do we know injectivity $\implies$ bijectivity.

@RajeshD That is integrable.

@Ilya So, if I understand correctly the main issue is that given a collection $\mathscr{C}$ of sets you don't have a very explicit description of the $\sigma$-algebra generated by it, right?

@KannappanSampath Take a vector space with a countable basis $(e_n)_{n=1}^\infty$ and let $Te_n = e_{n+1}$ this is injective but not surjective.

@RajeshD $\int_0^1\log(x)\mathrm{d}x=-1$

@tb yes. So I don't have a nice representation of any set from it: like in topology when we generate it using the basis, we can defined open sets through basis sets like open balls. While for $\sigma$-algebra which relates to the information in probability for me it is important to understand the properties of elements of the generated sigma-algebra

@tb Thank you. : )

You read my mind: You knew my question ahead of time!

Hi @Ben

@KannappanSampath You see that every nilpotent operator can be put into upper triangular form

@KannappanSampath I had to, in order to prove to you that I hadn't neglected you. :)

@KannappanSampath Prove that an operator on a vector space over any field is nilpotent iff there is a basis for your space such that $T$ is upper triangular with all zeros on the diagonal.

@BenjaminLim Yes. I am going to prove that soon. : )

@Ilya Are you familiar with the notions of $G_\delta$ and $F_\sigma$ sets?

@tb : )

@KannappanSampath Proving that shows that the determinant of the matrix of a nilpotent operator is zero, so that you map is neither injective nor surjective.

@KannappanSampath You cannot use Jordan Canonical form for this one.

@BenjaminLim but this is a bit circular :) to prove this, you need to know that there is something non-zero in the kernel in order to do it by induction.

@tb Ah true

You need to know that the dimension of the kernel is greater than zero :D

Whoops that gives out the proof

@BenjaminLim Sure! : )

@tb But in a complex vector space you can forget about knowing the size of the kernel and then from there take determinants

There are kernels in my popcorn!

I hate that.

@robjohn Do you have nilpotent popcorn? :)

Eventually the chain stabilises. I believe at least when $n = \dim V$ the chain stabilises.

@tb evidently :-(

@tb As in: you eat it again a finite number of times and eventually it is gone?

@tb My supervisor said that if I do direct limits and stuff properly we can look at some homological algebra

I wouldn't want to re-eat my popcorn 8-).

I am gone. : ) Later, bye.

@KannappanSampath Nilpotent operators are extremely important. Primary decomposition tells you that you can always decompose your space into a direct sum of generalised eigenspaces, such that the operators $T - \lambda_i I$ are nilpotent.

Where's Ilya?

@Ilya: ping!

i'm off, bye guys

:D

@BenjaminLim sounds nice. Good night, Ben!

@tb I have called him. He is in a meeting.

8-).

@PaulSlevin Still need help?

Yeah

@JonasTeuwen thanks.

Ok I have a Suslin tree $T$

Okay, off to work for me. I'll check back later.

height(T) = $\omega_1$, every branch is at most countable, every antichain in T is at most countable

Aronsjzan or Suslin?

(can anyone give me the link to make a mathjax bookmark)?

There's one in the chat rules, first link in the starred stuff.

@tb nope, sorry - me supervisor has just dropped by my office

8-).

I hope the phone was not too loud.

@JonasTeuwen it was not, but I picked is up

No, I mean through the horn.

so apparently he has mentioned that I had a phone call :)

ah, no

so given such a tree

jech says every level is countable (which he defines to mean of size $\aleph_0$.

@tb: I don't know what are these sets - would they help?

@Ilya that ping came just in time :)

@PaulSlevin Yes.

Every level in a tree is an antichain.

does he mean that it should be "at most" countable? Because that's all that the hypotheses seem to imply. If we take any level of the tree, the elements form an antichain, so is at most countable

@Ilya they would. But try this link!

@PaulSlevin Set theorists often dismiss the finite case.

right, but it is possible?

If there is a level with finitely many points you can always just add "fictional" points to make sure it's infinite.

Ok.

how do you add these fictional points ? Do you jst pull them out of thin air and adjust the order $< $ on $T$

Yes.

You just take a point from a lower level and add $\omega$ successors in this level.

If the level is a limit ordinal, just add countably many non-splitting branches from the root.

I had one more question

@tb I'll try it, thanks

for $x \in T$, define $T_x = \{y \in T \mid y \ge x \}$

We dismiss all points where that set is at most countable (if such points exist)

@Ilya I think Asaf's answer there gives the bottom-up description you seem to be after.

and let $T_1 = \{ x \in T \mid T_x \text{ is uncountable } \}$.

@tb Mmmm... Borel algebra...

my question is can I be sure to find a set $T_x$ which is uncountable?

You assumed that $T$ is uncountable, right?

well it has height $\omega_1$ so i can easily get $\omega_1$ elements, by picking one from each lower level

than the $\omega_1$th level

@PaulSlevin So the root has this property.

I haven't assumed a root

Asaf, can't you recommend a book that is maybe a little less rough on the reader than Jech?

@tb He's in a reading group which reads Jech.

@tb my exam will be on Jech :(

@PaulSlevin There's always a root to a tree. It's the only level which has finitely many elements.

oh it's possibel that the root is not unique .. that;s what I meant

@PaulSlevin Trees in set theory have a unique root.

but that is one of the hypotheses in the definition of a normal $\alpha-$tree

If you don't have a unique root, take all the elements of level $0$.

why would he mention that

These are roots of proper trees.

There are only countably (or less) many of those.

@tb I read about this construction in wikipedia's article on Borel $\sigma$-algebra I believe. I'm ok of using such set-theoretical methods even though I don't feel comfortable with them. Nevertheless, I don't see a point how this construction will help to verify, say, the property that I've put in my question

As the entire tree is uncountable, one of them must have uncountable tree above it.

Now dismiss all the other points, and assume that your tree has a unique root.

ok. I just learned the definition of a tree. how can I be sure that given one of these roots, theres somthing int he level above it, and that there is a "connection"/ "edge" between that root and the element in the level above it

Come again?

sorry

IF i know the level above a given level is non empty, then by definition there must be an element in the lower level so there is an order relation between them

so I answered my own nonsensical question

ok. Thanks asaf you've saved the day !!

No problem.

@Ilya Transfinite induction?

@Ilya you could prove it by transfinite induction. It holds for the sets in the generating family, and the property is closed under countable unions and complementation.

@AsafKaragila oh wait, I was going backwards there. How do I know that if i have chosen a root $x$ in level 0, there is an element in level 1 such that $x < y$?#

@PaulSlevin You chose a root whose subtree is uncountable. If there is no one above it... what is the subtree? It's a singleton. Have you ever heard on an uncountable singleton?

I know that if I have $y$i can find such an $x$. But since I want an uncountable sequence, I have to start at the bottom

My question is I guess, how do I actually choose a root with an uncountable subtree though

$T$ is uncountable, but every antichain is countable.

@tb oh, ok. Can I ask you then for the reference with a brief insight in transfinite induction? Or do you mean that any property which holds for the generating family and is closed under countable unions and complementation holds for any set in the generated $\sigma$-algebra which can be proved by the transfinite induction?

Let $A=\{x\in T\mid x\text{is of level }0\}$, that is all the roots.

This is an antichain, therefore countable.

yes

For every $x,y\in A$ we have $T_x\cap T_y=\varnothing$ whenever $x\neq y$.

So $\{T_x\mid x\in A\}$ is a partition of $T$ into countably many parts.

If you take a partition of an uncountable set into countably many parts, one of these parts will be uncountable.

Therefore there is at least one $x\in A$ such that $T_x$ is uncountable.

Without loss of generality, now we assume that $T=T_x$, and we can continue as though the tree has a unique root.

ok we are getting to the root of my difficulty now - why is $T_x \cap T_y = \emptyset$ ?

@PaulSlevin Because the definition of a tree is a partially ordered set that if $x\in T$ then $\{y\in T\mid y<x\}$ is well ordered, and therefore linearly ordered.

So below each point there is at most one root.

unique least element.

ok thanks so much!

Sure.

@Jonas: can you call @tb to ping him? :)

@Ilya I don't have his phone number.

8-).

@Ilya Yes, that's what I'm saying. I can't think of a good reference right now. In the notation of Wikipedia $P(\alpha)$ is the statement that the sets in $\mathbf{\Sigma}_{\alpha}^{0}(\mathscr{C})$ and $\mathbf{\Pi}_{\alpha}^{0}(\mathscr{C})$ have your property. Now proceed to the next level by transfinite induction.

render

The point is that the union $\bigcup_{\alpha \lt \omega_1} \mathbf{\Sigma}_{\alpha}^{0} \cup \mathbf{\Pi}_{\alpha}^{0}$ is a $\sigma$-algebra, so you get this "hands-on" description of the $\sigma$-algebra generated by $\mathscr{C}$.

Actually: $$\bigcup_{\alpha<\omega_1}\mathbf{\Sigma}^0_\alpha = \bigcup_{\alpha<\omega_1}\mathbf{\Pi}^0_\alpha$$

render

@AsafKaragila better :)

I'm gonna go now. Enjoy the rest of the day.

The induction is really straightforward.

Bye Asaf

see you

Bye Asaf.

I need a pillow.

@tb so it means that if we have a property $S$ which is closed under countable unions and taking the complement, any $A\in \sigma(\mathscr C)$ satisfies $S$ if and only if any $B\in \mathscr C$ satisfies $S$?

exactly

wait

But actually you can do it much easier...

@tb how? (forget for a while about my previous comment)

If you have this property $S$ closed under countable unions and taking the complement then it defines a sub-$\sigma$-algebra of $\mathscr{F}$, right?

@tb what is $\mathscr F$?

well, you're in the setting of a measurable space $(\Omega, \mathscr{F})$. You can also take $\mathscr{F}$ to be the power set of $\Omega$, if you like.

@tb in the case when $\mathscr F$ is given then $S$ doesn't have to be a subset of it, does it?

Since $S$ is closed under countable unions and complementation (and the empty set fulfills it), the collection $\mathscr{S} = \{F \in \mathscr{F}\,:\,F \text{ satisfies }S\}$ is a $\sigma$-algebra (I'm not forgetting that your situation is slightly more complicated, but let's look at that instance first.)

render

you could also just click the bookmark again :)

Are you with me so far?

@tb then I have to have the bookmarks panel activated, but this panel annoys me

I see, you prefer to annoy me by casting rendering spells :)

@tb eemmm. Let's say I am with you

@tb you didn't tell that they annoy you

@Ilya they don't really annoy me. forget about it.

Yes, I agree - $S$ is $\sigma$-algebra, and so is $\mathscr F$ then $\mathscr S = S\cap \mathscr F$ is also

wait, I will fix it

I'm saying if $S$ is a property that is closed under countable unions and complementation and $\emptyset$ has $S$ then $\mathscr{S}$ is a $\sigma$ algebra

@MattN hi

Hi Matt

@tb here I agree. Also $S$ (identifying it with a satisfactory family) is a $\sigma$-algebra. So $\mathscr S = \mathscr F\cap S$ is a $\sigma$-algebra

@tb I think I'm starting to believe that. $\varphi \in \operatorname{Hom}{(S^1, S^1)}$ implies $\varphi (z) = z^n$ for some integer $n$ is not obvious. Is it?

@MattN No, that requires a little bit of thinking.

But it is not that hard.

: )

Ok. Thanks. I'll think about this.

@tb: Are you with me so far?

@Ilya The next step is to identify $$\sigma(\mathscr{C}) = \bigcap_{\substack{ \Sigma \supset \mathscr{C} \\ \Sigma \text{ is a } \sigma\text{-algebra}}} \Sigma$$

render

render

Now if all sets of $\mathscr{C}$ satisfy $S$ then you have $\sigma(\mathscr{C}) \subset \mathscr{S}$.

so if $\mathscr C\subset S$ then $\sigma(\mathscr C)\subset S$

@Ilya exactly

ok, it's clear and it doesn't need the transfinite induction

easier, isn't it?

Now this does not immediately apply to your concrete situation, but that can be fixed very easily.

ahhh, please don't write so fast :)

that's essentially a part of method that I've described in my question. The problem is: what if $S$ is not closed under countable unions and taking the complement, but $\sigma (\mathscr C)\cap S$ is?

I was just about to address this point

But let me look at the question again

In this particular instance you're interested in the measurable subsets of $S$ plus the null-sets. So you take the two subspaces $S$ and $\Omega \smallsetminus S$.

On the former you put the $\sigma$-algebra you called $\mathscr{F}_B$ together with the subsets of $B$ that are either null or co-null in $B$.

On the latter you put the $\sigma$-algebra of sets that are either null or co-null in $\Omega \smallsetminus S$.

(I'm slightly confused with the letters right now)

I am reading :) give me about 5 minutes, please

sure

I wonder if it's ok to write a non-self-contained answer advertising your own text book / notes.

I'm getting tired of Pete always posting links to his notes

: D

Well, I am off!

@tb: yes, there $B$ instead of $S$?

I think so.

@TeddyBear How many characters are there?

I know of at least three different concepts having that name.

@tb ok. But you didn't finish your argument, did you? :)

No, you told me to wait.

@tb I'm ready now

That was another comment of yours making me blush btw. I hate being caught red handed while being confused and writing nonsense.

To finish up, if $(\Omega_1, \mathscr{F}_1)$ and $(\Omega_2, \mathscr{F}_2)$ are two measurable spaces, then their union $\Omega_1 \cup \Omega_2$ has a $\sigma$-algebra $\mathscr{F}$ consisting of those sets $F$ such that $E \cap \Omega_1 \in \mathscr{F}_1$ and $F \cap \Omega_2 \in \mathscr{F}_2$.

But in this case, reading it again, I still think I used the "right characters".

Apply this to $\Omega_1 = B$ with the $\sigma$-algebra consisting $\mathscr{F}_B$ plus the null and conull sets and $\Omega_2 = \Omega \smallsetminus B$ with the null and co-null sets.

Hello guys!

@MattN actually you're right. I hadn't read the question properly (I was just following up on joriki's comment, which seems to have made the same goof as I). Orthogonality relations are for characters of group representations usually, if you restrict to abelian groups you get the same characters

I was trying to read your talk but my browser doesn't parse TeX. How can I fix it?

@tb : )

@tb You are ... a lemming!? 0_o

@Norbert the ChatJaX applet is here

@MattN yep.

And I thought you were a stubborn donkey : )

Okay, gotta go, the electrician is going to turn the power supply off.

Byee!

render

@MattN yes, an unholy cross-breeding of the two.

@t.b. How is @MattN's answer wrong? I don't think he was confusing anything.

@tb you mean, $\mathscr F$ is the smallest one here?

Oh, nevermind, it's undeleted. Also I think joriki was confusing group characters with Dirichlet characters.

@tb Thanks

@tb: we meet again. ;-)

Hello!

Can you guys help somebody out with a question about Support Vector Machines?

does anyone here know much about grad school admissions?

@Ilya yes.

@robjohn indeed, we have a bad timing...