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Skullpatrol
Skullpatrol
10:10 AM
@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?
 
anon
anon
not really
 
Skullpatrol
Skullpatrol
@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.
 
anon
anon
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.
 
Skullpatrol
Skullpatrol
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.
 
anon
anon
What are "the linear term" and "the midde"?
 
Skullpatrol
Skullpatrol
10:20 AM
ax^2 + bx + c, where bx is the linear term in the middle.
 
anon
anon
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.
 
Skullpatrol
Skullpatrol
c + bx + ax^2 is what I mean.
 
anon
anon
no duh that's the same, you can change the order in any (finite) sum.
reviewing suggested edits is such a thankless chore
 
Skullpatrol
Skullpatrol
@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.
 
anon
anon
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.
 
Skullpatrol
Skullpatrol
10:33 AM
@PaulSlevin Hi
 
Paul Slevin
Paul Slevin
Hello @Skullpatrol
 
Skullpatrol
Skullpatrol
@PaulSlevin How's it going?
 
Paul Slevin
Paul Slevin
@Skullpatrol I'm ok. Just learning about trees now. I feel like a biologist
how are you
 
Matt N.
Matt N.
@AsafKaragila What's the east coast of the solar system?
 
Asaf Karagila
Asaf Karagila
@MattN The entire solar system.
 
Skullpatrol
Skullpatrol
10:37 AM
@PaulSlevin Fine thanks.
 
Paul Slevin
Paul Slevin
does anyone here know about trees? I just have one super-mega-quick question
 
Asaf Karagila
Asaf Karagila
?
 
Paul Slevin
Paul Slevin
as in, set-theory trees
not the leafy ones
 
Asaf Karagila
Asaf Karagila
Shoot.
 
Skullpatrol
Skullpatrol
so much for biology ;-)
 
Paul Slevin
Paul Slevin
10:39 AM
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?
 
Asaf Karagila
Asaf Karagila
Yes.
 
Paul Slevin
Paul Slevin
haha
 
Asaf Karagila
Asaf Karagila
If $x$ is on level $\alpha$ then $\{y\in T\mid y<x\}$ has order type $\alpha$.
 
Paul Slevin
Paul Slevin
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
 
Asaf Karagila
Asaf Karagila
So if you had someone on level $\alpha+1$ you have someone on level $\alpha$.
 
Paul Slevin
Paul Slevin
10:43 AM
ok
Thanks
if you have a dude on a level$\alpha$ then surely there is a dude on every level $\beta < \alpha$?
 
Jonas Teuwen
Jonas Teuwen
The Dude.
 
Paul Slevin
Paul Slevin
That is my way of saying "someone" or "this guy" in maths
 
Skullpatrol
Skullpatrol
the null dude
occupies the empty level
 
Paul Slevin
Paul Slevin
haha
well he wouldnt be in it if it was empty
 
Skullpatrol
Skullpatrol
But he's the null-dude or he's nobody.
 
Paul Slevin
Paul Slevin
10:50 AM
brain melting
 
Skullpatrol
Skullpatrol
sorry
 
Ilya
Ilya
@Jonas: hi
 
Paul Slevin
Paul Slevin
its ok my brain melts 2-3 times a day
 
Jonas Teuwen
Jonas Teuwen
@Ilya Hi!
 
Skullpatrol
Skullpatrol
@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
 
Paul Slevin
Paul Slevin
10:59 AM
excellent
 
Skullpatrol
Skullpatrol
@WillHunting You change gravatars faster than a speeding bullet ;-)
 
user19161
@Skullpatrol How is the factorisation? Your school homework?
 
Skullpatrol
Skullpatrol
@WillHunting Thanks for the help.
 
Asaf Karagila
Asaf Karagila
@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.
 
Jonas Teuwen
Jonas Teuwen
@AsafKaragila No, it is the building next to mine.
We also have a phone.
 
Asaf Karagila
Asaf Karagila
11:06 AM
I thought that Delft TU was like one huge building and some parking space.
 
Jonas Teuwen
Jonas Teuwen
We don't need to use cups and a string anymore.
 
Asaf Karagila
Asaf Karagila
Are you saying that Ilya is the guard in the parking lot?
 
Jonas Teuwen
Jonas Teuwen
No, it is like 12,5% of the area of Delft.
 
Asaf Karagila
Asaf Karagila
Have you measured it inch by inch?
 
Jonas Teuwen
Jonas Teuwen
So about 3km^2.
No, they own about that much of Delft.
But there are ~15 buildings.
 
Skullpatrol
Skullpatrol
11:08 AM
@JonasTeuwen Do you have law and medicine faculties there?
 
Jonas Teuwen
Jonas Teuwen
No.
Only engineering and some small things.
 
Jonas Teuwen
Jonas Teuwen
11:48 AM
@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?
 
robjohn
robjohn
@JonasTeuwen What does $S_a$ look like?
 
Asaf Karagila
Asaf Karagila
@robjohn Tall, slender, blonde hair, blue eyes, a huge nose and crooked teeth.
 
robjohn
robjohn
@AsafKaragila typical
 
Asaf Karagila
Asaf Karagila
Yeah.
I think I can finally trust you, Robert. (at least on the main site)
@robjohn Now go here and prove yourself trustworthy!
 
Jonas Teuwen
Jonas Teuwen
@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.
 
Ilya
Ilya
11:58 AM
@Jonas: what are cups and strings?
 
Jonas Teuwen
Jonas Teuwen
@Ilya If you take two cups, and you put a tensioned string between them, you can talk like that to eachother :-).
 
Asaf Karagila
Asaf Karagila
@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.
 
Ilya
Ilya
@JonasTeuwen are, that stuff. We tried it once when I was 6 years old
 
robjohn
robjohn
@JonasTeuwen $L$ is a differential operator?
 
Ilya
Ilya
@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
 
robjohn
robjohn
12:03 PM
@JonasTeuwen coffee cans work better than cups
 
Ilya
Ilya
this is the main part of the campus, but still the Architecture building, Chemical Engineering, Biological and Aerospace ones are not there
 
Asaf Karagila
Asaf Karagila
@Ilya Use optic fibers made of doom.
I am going to stalk my advisor around his office.
 
Jonas Teuwen
Jonas Teuwen
@robjohn :D.
We have phones.
 
Paul Slevin
Paul Slevin
anyone willing to talk some more biology with me?
(i.e. about trees)
 
robjohn
robjohn
@JonasTeuwen do they work better than coffee cans?
 
Jonas Teuwen
Jonas Teuwen
12:05 PM
@robjohn I haven't tried coffee cans, but I think so.
 
robjohn
robjohn
@PaulSlevin trees... they make paper from those, right?
 
Jonas Teuwen
Jonas Teuwen
So do you know what they mean with a homogeneous estimate here? :-).
 
Paul Slevin
Paul Slevin
Set-theory trees
they waste paper cos you write so much trying to understand them
 
robjohn
robjohn
@JonasTeuwen do constant functions apply since they are not in $L^1$?
 
Jonas Teuwen
Jonas Teuwen
@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!
 
robjohn
robjohn
12:08 PM
@JonasTeuwen and then they are not killed by the $\nabla$
 
Jonas Teuwen
Jonas Teuwen
Yes.
 
t.b.
t.b.
Hi all
 
Rajesh D
Rajesh D
@rob : Is $\frac{1}{x^{0.3}}$ integrable...i mean integration from $0$ to $\infty$
Hi
 
Ilya
Ilya
@tb: hi - do you have now time to take a look at my question?
 
t.b.
t.b.
@raj no.
@Ilya hi
 
Jonas Teuwen
Jonas Teuwen
12:10 PM
Inverse powers of $x$ are never integrable from $0$ to $\infty$.
 
robjohn
robjohn
@RajeshD nope near $0$ yes, near $\infty$, no.
 
Jonas Teuwen
Jonas Teuwen
Either they are around infinity or around $0$. For $1/x$ never.
You could have checked that easily right? 8-).
 
t.b.
t.b.
@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 $(0,1]$.
 
robjohn
robjohn
@tb Hi there! good afternoon
 
t.b.
t.b.
yeah, and a brilliant and sunny one at that!
 
Ilya
Ilya
12:12 PM
@tb the same here :)
 
t.b.
t.b.
@Ilya wanna have a peek at the webcam? :)
can you give me the link to your question again?
 
Ilya
Ilya
@tb why not, give me the link please
@tb sure
 
t.b.
t.b.
@Ilya here
Oh, it's foggier on the other side of the hill :)
 
robjohn
robjohn
$\frac{1}{x^{1+\epsilon}+x^{1-\epsilon}}$ is integrable.
 
Kannappan Sampath
Kannappan Sampath
Hi @tb
 
t.b.
t.b.
12:14 PM
Hi Kannappan
 
Kannappan Sampath
Kannappan Sampath
I would like your help with some linear algebra. Remember, we proved that nilpotent operators are necessarily not injective? @tb
 
Rajesh D
Rajesh D
@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
 
robjohn
robjohn
@KannappanSampath since they send something to $0$ :-) (other than $0$)
 
t.b.
t.b.
@RajeshD that one is integrable on $(1,\infty)$ but not on $(0,1)$.
 
Kannappan Sampath
Kannappan Sampath
@robjohn Yes. right, but I could not prove it the other day. : (
 
robjohn
robjohn
12:16 PM
@RajeshD that is not integrable on that range.
 
Rajesh D
Rajesh D
but 1 over $x^2$ is right ?
 
t.b.
t.b.
where? Yes on $(1,\infty)$ but no on $(0,1)$.
 
Rajesh D
Rajesh D
(0,1)
 
robjohn
robjohn
@RajeshD on $(1,\infty)$
 
Kannappan Sampath
Kannappan Sampath
Hmph, he neglected it! : (
 
robjohn
robjohn
12:18 PM
@RajeshD not integrable there
@KannappanSampath neglected what?
 
t.b.
t.b.
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 $(1,\infty)$.
 
Rajesh D
Rajesh D
Okay $\log(x)$ on (0,1) ?
 
Kannappan Sampath
Kannappan Sampath
@robjohn my ping... : (
 
Ilya
Ilya
rendering spell
 
t.b.
t.b.
2
 
Ilya
Ilya
12:19 PM
@tb: you've missed \text{} or messed up dollar signs
 
t.b.
t.b.
@KannappanSampath I haven't neglected you... I do remember.
But I was talking to Rajesh and Ilya already.
 
Rajesh D
Rajesh D
@tb @rob Thanks very much for the help
 
Kannappan Sampath
Kannappan Sampath
@tb : ) Ok. The third argument went something like: only for finite dimensional spaces do we know injectivity $\implies$ bijectivity.
 
robjohn
robjohn
@RajeshD That is integrable.
 
t.b.
t.b.
@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.
 
robjohn
robjohn
12:23 PM
@RajeshD $\int_0^1\log(x)\mathrm{d}x=-1$
 
Ilya
Ilya
@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
 
Kannappan Sampath
Kannappan Sampath
@tb Thank you. : )
You read my mind: You knew my question ahead of time!
Hi @Ben
 
Benjamin Lim
Benjamin Lim
@KannappanSampath You see that every nilpotent operator can be put into upper triangular form
 
t.b.
t.b.
@KannappanSampath I had to, in order to prove to you that I hadn't neglected you. :)
 
Benjamin Lim
Benjamin Lim
@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.
 
Kannappan Sampath
Kannappan Sampath
12:26 PM
@BenjaminLim Yes. I am going to prove that soon. : )
 
t.b.
t.b.
@Ilya Are you familiar with the notions of $G_\delta$ and $F_\sigma$ sets?
 
Kannappan Sampath
Kannappan Sampath
@tb : )
 
Benjamin Lim
Benjamin Lim
@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.
 
t.b.
t.b.
@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.
 
Benjamin Lim
Benjamin Lim
@tb Ah true
You need to know that the dimension of the kernel is greater than zero :D
Whoops that gives out the proof
 
Kannappan Sampath
Kannappan Sampath
12:28 PM
@BenjaminLim Sure! : )
 
Benjamin Lim
Benjamin Lim
@tb But in a complex vector space you can forget about knowing the size of the kernel and then from there take determinants
 
robjohn
robjohn
There are kernels in my popcorn!
I hate that.
 
Benjamin Lim
Benjamin Lim
@KannappanSampath I suggest looking at Ascending Chain of nullspaces:

$ 0 \subset \ker T \subset \ker T^2 \subset ....$
 
t.b.
t.b.
@robjohn Do you have nilpotent popcorn? :)
 
Benjamin Lim
Benjamin Lim
Eventually the chain stabilises. I believe at least when $n = \dim V$ the chain stabilises.
 
robjohn
robjohn
12:29 PM
@tb evidently :-(
 
Jonas Teuwen
Jonas Teuwen
@tb As in: you eat it again a finite number of times and eventually it is gone?
 
Benjamin Lim
Benjamin Lim
@tb My supervisor said that if I do direct limits and stuff properly we can look at some homological algebra
 
Jonas Teuwen
Jonas Teuwen
I wouldn't want to re-eat my popcorn 8-).
 
Kannappan Sampath
Kannappan Sampath
I am gone. : ) Later, bye.
 
Benjamin Lim
Benjamin Lim
@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.
 
t.b.
t.b.
12:31 PM
Where's Ilya?
@Ilya: ping!
 
Benjamin Lim
Benjamin Lim
i'm off, bye guys
:D
 
t.b.
t.b.
@BenjaminLim sounds nice. Good night, Ben!
 
Jonas Teuwen
Jonas Teuwen
@tb I have called him. He is in a meeting.
8-).
 
Asaf Karagila
Asaf Karagila
@PaulSlevin Still need help?
 
Paul Slevin
Paul Slevin
Yeah
 
t.b.
t.b.
12:32 PM
@JonasTeuwen thanks.
 
Paul Slevin
Paul Slevin
Ok I have a Suslin tree $T$
 
t.b.
t.b.
Okay, off to work for me. I'll check back later.
 
Paul Slevin
Paul Slevin
height(T) = $\omega_1$, every branch is at most countable, every antichain in T is at most countable
 
Asaf Karagila
Asaf Karagila
Aronsjzan or Suslin?
 
Paul Slevin
Paul Slevin
(can anyone give me the link to make a mathjax bookmark)?
 
Asaf Karagila
Asaf Karagila
12:33 PM
There's one in the chat rules, first link in the starred stuff.
 
Ilya
Ilya
@tb nope, sorry - me supervisor has just dropped by my office
 
Jonas Teuwen
Jonas Teuwen
8-).
I hope the phone was not too loud.
 
Ilya
Ilya
@JonasTeuwen it was not, but I picked is up
 
Jonas Teuwen
Jonas Teuwen
No, I mean through the horn.
 
Ilya
Ilya
so apparently he has mentioned that I had a phone call :)
ah, no
 
Paul Slevin
Paul Slevin
12:35 PM
so given such a tree
jech says every level is countable (which he defines to mean of size $\aleph_0$.
 
Ilya
Ilya
@tb: I don't know what are these sets - would they help?
 
t.b.
t.b.
@Ilya that ping came just in time :)
 
Asaf Karagila
Asaf Karagila
@PaulSlevin Yes.
Every level in a tree is an antichain.
 
Paul Slevin
Paul Slevin
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
 
t.b.
t.b.
@Ilya they would. But try this link!
 
Asaf Karagila
Asaf Karagila
12:38 PM
@PaulSlevin Set theorists often dismiss the finite case.
 
Paul Slevin
Paul Slevin
right, but it is possible?
 
Asaf Karagila
Asaf Karagila
If there is a level with finitely many points you can always just add "fictional" points to make sure it's infinite.
 
Paul Slevin
Paul Slevin
Ok.
how do you add these fictional points ? Do you jst pull them out of thin air and adjust the order $< $ on $T$
 
Asaf Karagila
Asaf Karagila
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.
 
Paul Slevin
Paul Slevin
I had one more question
 
Ilya
Ilya
12:40 PM
@tb I'll try it, thanks
 
Paul Slevin
Paul Slevin
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)
 
t.b.
t.b.
@Ilya I think Asaf's answer there gives the bottom-up description you seem to be after.
 
Paul Slevin
Paul Slevin
and let $T_1 = \{ x \in T \mid T_x \text{ is uncountable } \}$.
 
Asaf Karagila
Asaf Karagila
@tb Mmmm... Borel algebra...
 
Paul Slevin
Paul Slevin
my question is can I be sure to find a set $T_x$ which is uncountable?
 
Asaf Karagila
Asaf Karagila
12:42 PM
You assumed that $T$ is uncountable, right?
 
Paul Slevin
Paul Slevin
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
 
Asaf Karagila
Asaf Karagila
@PaulSlevin So the root has this property.
 
Paul Slevin
Paul Slevin
I haven't assumed a root
 
t.b.
t.b.
Asaf, can't you recommend a book that is maybe a little less rough on the reader than Jech?
 
Asaf Karagila
Asaf Karagila
@tb He's in a reading group which reads Jech.
 
Paul Slevin
Paul Slevin
12:43 PM
@tb my exam will be on Jech :(
 
Asaf Karagila
Asaf Karagila
@PaulSlevin There's always a root to a tree. It's the only level which has finitely many elements.
 
Paul Slevin
Paul Slevin
oh it's possibel that the root is not unique .. that;s what I meant
 
Asaf Karagila
Asaf Karagila
@PaulSlevin Trees in set theory have a unique root.
 
Paul Slevin
Paul Slevin
but that is one of the hypotheses in the definition of a normal $\alpha-$tree
 
Asaf Karagila
Asaf Karagila
If you don't have a unique root, take all the elements of level $0$.
 
Paul Slevin
Paul Slevin
12:46 PM
why would he mention that
 
Asaf Karagila
Asaf Karagila
These are roots of proper trees.
There are only countably (or less) many of those.
 
Ilya
Ilya
@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
 
Asaf Karagila
Asaf Karagila
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.
 
Paul Slevin
Paul Slevin
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
 
Asaf Karagila
Asaf Karagila
Come again?
 
Paul Slevin
Paul Slevin
12:47 PM
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 !!
 
Asaf Karagila
Asaf Karagila
No problem.
@Ilya Transfinite induction?
 
t.b.
t.b.
@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.
 
Paul Slevin
Paul Slevin
@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 $x < y$?#
 
Asaf Karagila
Asaf Karagila
@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?
 
Paul Slevin
Paul Slevin
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
 
Asaf Karagila
Asaf Karagila
12:53 PM
$T$ is uncountable, but every antichain is countable.
 
Ilya
Ilya
@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?
 
Asaf Karagila
Asaf Karagila
Let $A=\{x\in T\mid x\text{is of level }0\}$, that is all the roots.
This is an antichain, therefore countable.
 
Paul Slevin
Paul Slevin
yes
 
Asaf Karagila
Asaf Karagila
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.
 
Paul Slevin
Paul Slevin
ok we are getting to the root of my difficulty now - why is $T_x \cap T_y = \emptyset$ ?
 
Asaf Karagila
Asaf Karagila
12:56 PM
@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.
 
Paul Slevin
Paul Slevin
unique least element.
ok thanks so much!
 
Asaf Karagila
Asaf Karagila
Sure.
 
Ilya
Ilya
@Jonas: can you call @tb to ping him? :)
 
Jonas Teuwen
Jonas Teuwen
@Ilya I don't have his phone number.
8-).
 
t.b.
t.b.
@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.
 
Ilya
Ilya
1:02 PM
render
 
t.b.
t.b.
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 $\mathscr{C}$.
 
Asaf Karagila
Asaf Karagila
Actually: $$\bigcup_{\alpha<\omega_1}\mathbf{\Sigma}^0_\alpha = \bigcup_{\alpha<\omega_1}\mathbf{\Pi}^0_\alpha$$
 
Ilya
Ilya
render
 
t.b.
t.b.
@AsafKaragila better :)
 
Asaf Karagila
Asaf Karagila
I'm gonna go now. Enjoy the rest of the day.
 
t.b.
t.b.
1:06 PM
The induction is really straightforward.
Bye Asaf
 
Ilya
Ilya
see you
 
Jonas Teuwen
Jonas Teuwen
Bye Asaf.
I need a pillow.
 
Ilya
Ilya
@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 $S$?
 
t.b.
t.b.
exactly
 
Ilya
Ilya
wait
 
t.b.
t.b.
1:08 PM
But actually you can do it much easier...
 
Ilya
Ilya
@tb how? (forget for a while about my previous comment)
 
t.b.
t.b.
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?
 
Ilya
Ilya
@tb what is $\mathscr F$?
 
t.b.
t.b.
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.
 
Ilya
Ilya
@tb in the case when $\mathscr F$ is given then $S$ doesn't have to be a subset of it, does it?
 
t.b.
t.b.
1:14 PM
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.)
 
Ilya
Ilya
render
 
t.b.
t.b.
you could also just click the bookmark again :)
Are you with me so far?
 
Ilya
Ilya
@tb then I have to have the bookmarks panel activated, but this panel annoys me
 
t.b.
t.b.
I see, you prefer to annoy me by casting rendering spells :)
 
Ilya
Ilya
@tb eemmm. Let's say I am with you
@tb you didn't tell that they annoy you
 
t.b.
t.b.
1:16 PM
@Ilya they don't really annoy me. forget about it.
 
Ilya
Ilya
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
 
t.b.
t.b.
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
 
Ilya
Ilya
@MattN hi
 
t.b.
t.b.
Hi Matt
 
Ilya
Ilya
@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
 
Matt N.
Matt N.
1:18 PM
@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?
Hello there. I'll be back later. I need to be more productive. (as opposed to sitting around in chat all day)
Can't even write I'm so tired *_*
 
t.b.
t.b.
@MattN No, that requires a little bit of thinking.
But it is not that hard.
Note that $\varphi \circ \exp : \mathbb{R} \to \mathbb{C}^\times$ is a homomorphism and you know what those look like.
 
Matt N.
Matt N.
: )
Ok. Thanks. I'll think about this.
 
Ilya
Ilya
@tb: Are you with me so far?
 
t.b.
t.b.
@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
 
Ilya
Ilya
render
 
t.b.
t.b.
1:23 PM
Now if all sets of $\mathscr{C}$ satisfy $S$ then you have $\sigma(\mathscr{C}) \subset \mathscr{S}$.
 
Ilya
Ilya
so if $\mathscr C\subset S$ then $\sigma(\mathscr C)\subset S$
 
t.b.
t.b.
@Ilya exactly
 
Ilya
Ilya
ok, it's clear and it doesn't need the transfinite induction
 
t.b.
t.b.
easier, isn't it?
Now this does not immediately apply to your concrete situation, but that can be fixed very easily.
 
Ilya
Ilya
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?
 
t.b.
t.b.
1:25 PM
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)
 
Ilya
Ilya
I am reading :) give me about 5 minutes, please
 
t.b.
t.b.
sure
 
Matt N.
Matt N.
I wonder if it's ok to write a non-self-contained answer advertising your own text book / notes.
 
t.b.
t.b.
I'm getting tired of Pete always posting links to his notes
 
Matt N.
Matt N.
: D
 
Kannappan Sampath
Kannappan Sampath
1:41 PM
Well, I am off!
 
Ilya
Ilya
@tb: yes, there $B$ instead of $S$?
 
t.b.
t.b.
I think so.
 
Matt N.
Matt N.
@TeddyBear How many characters are there?
 
t.b.
t.b.
I know of at least three different concepts having that name.
 
Ilya
Ilya
@tb ok. But you didn't finish your argument, did you? :)
 
t.b.
t.b.
1:45 PM
No, you told me to wait.
 
Ilya
Ilya
@tb I'm ready now
 
Matt N.
Matt N.
That was another comment of yours making me blush btw. I hate being caught red handed while being confused and writing nonsense.
 
t.b.
t.b.
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 $F \cap \Omega_2 \in \mathscr{F}_2$.
 
Matt N.
Matt N.
But in this case, reading it again, I still think I used the "right characters".
 
t.b.
t.b.
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.
 
Norbert
Norbert
1:52 PM
Hello guys!
 
t.b.
t.b.
@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
 
Norbert
Norbert
I was trying to read your talk but my browser doesn't parse TeX. How can I fix it?
 
Matt N.
Matt N.
@tb : )
@tb You are ... a lemming!? 0_o
 
t.b.
t.b.
@Norbert the ChatJaX applet is here
@MattN yep.
 
Matt N.
Matt N.
And I thought you were a stubborn donkey : )
 
t.b.
t.b.
1:54 PM
Okay, gotta go, the electrician is going to turn the power supply off.
 
Matt N.
Matt N.
Byee!
 
Ilya
Ilya
render
 
t.b.
t.b.
@MattN yes, an unholy cross-breeding of the two.
 
anon
anon
@t.b. How is @MattN's answer wrong? I don't think he was confusing anything.
 
Ilya
Ilya
@tb you mean, $\mathscr F$ is the smallest one here?
 
anon
anon
1:56 PM
Oh, nevermind, it's undeleted. Also I think joriki was confusing group characters with Dirichlet characters.
 
Norbert
Norbert
@tb Thanks
 
robjohn
robjohn
2:25 PM
@tb: we meet again. ;-)
 
gentmatt
gentmatt
2:50 PM
Hello!
Can you guys help somebody out with a question about Support Vector Machines?
 
Paul Slevin
Paul Slevin
does anyone here know much about grad school admissions?
 
t.b.
t.b.
@Ilya yes.
@robjohn indeed, we have a bad timing...
 
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