order-complete

user116211

Sep 25, 2016 07:48

@balarka, have you heard this theorem on order-complete:

user116211

> A set $X$ is order-complete relative to an ordering if and only if each non-void subset which has a lower bound has an infimum.

user116211

?

user116211

@SoumyoB You could have uploaded that.

user116211

Anyways, what I'm not getting is if a set has lower bound, doesn't the Completeness Axiom guarantee the existence of an infimum?

SoumyoB

@MAFIA36790 you mean upload to the chat on here?

wow you learn something new everyday

user116211

Sep 25, 2016 07:50

@SoumyoB That's your call, but I would have uploaded a pic instead of giving the link.

SoumyoB

so I just drag the image to the typing space?

Balarka Sen

@MAFIA36790 I have forgotten, what's the defn of order-completeness?

SoumyoB

where do you upload the pic though

user116211

@SoumyoB See, there is an upload... button next to the send bar.

s.harp

@MAFIA36790 consider $\mathbb R -\{0\}$, here the set $(0,1)$ has a lower bound but no infimum

SoumyoB

Sep 25, 2016 07:53

@MAFIA36790 I don't see any upload button :/

s.harp

SoumyoB

mine doesn't have an upload button there

perhaps it's an option unlocked when more reputation points are earned

I only have 46

Danu

Sep 25, 2016 08:13

@BalarkaSen hi

user116211

internet connection disrupted

user116211

@SoumyoB Never knew of such things.

Danu

I almost understood what Ted told me yesterday

I just don't see how homogeneous polynomials restrict to k-linear maps on a ray

user116211

@s.harp But what about the completeness axiom?

user116211

> A set $X$ is order-complete (relative to the ordering $\lt$) iff each non-void subset of $X$ which has an upper bound has a supremum.

Danu

Sep 25, 2016 08:17

15 hours ago, by Ted Shifrin

Obviously a homogeneous polynomial of degree $k$ restricts to such.

s.harp

@MAFIA36790 are you referring to the real numbers? this is not the only set on which one has a total (or partial) order, and you may want to speak about other sets that have an order that is order complete

Danu

Obviously :'(

user116211

@s.harp Well, in defining the axiom, the set concerned was a set of real numbers.

Danu

I'm thinking of this example, for instance: $f:\Bbb C^2\to\Bbb C$, $f(z_0,z_1)=z_0^3$. How does this restrict to a $3$-linear map on e.g. $\ell=\lambda(1,0)\subset\Bbb C^2$?

s.harp

@MAFIA36790 I don't quite understand your question

user116211

Sep 25, 2016 08:26

@s.harp Well, my query is that while reading about order-completeness, I noticed ... if a set which has an upper bound has a supremum; I thought by the completeness axiom, it is guaranteed that if the set has upper bound, it must have a supremum. But I think I may be wrong since the set cannot be always a set of real numbers.

s.harp

@MAFIA36790 so it would not be wrong to reformulate the question as: "Why is it not redundant to define "order-completeness", since it is automatically true for every totally ordered set?"

user116211

@s.harp yup!

s.harp

In this case you must note that it is not actually true for every totally ordered set, the completeness axiom is a statement about the reals. But it is easy to construct also subsets of the reals that are not order complete, for example $\mathbb R-\{0\}$ is one such set that is not order complete

as the set $(0,1)$ is bounded in $\mathbb R-\{0\}$ but it does not have a infimum in $\mathbb R-\{0\}$

user116211

@s.harp Yes, I was pondering over your example... but what is then a set of real numbers? A set whose elements belong to $\mathbb R- \{0\}$ must also be real numbers, right?

user116211

@s.harp So, does that mean the axiom only applies when you are working in $\mathbb R\,?$

Balarka Sen

Sep 25, 2016 08:35

@Danu By 3-linear he means homogeneous of degree 3, yes?

In which case it's obvious.

Oh, sorry, I see. You're identifying $O(3)$ with $O(1) \otimes O(1) \otimes O(1)$ and asking why a section aka homogeneous polynomial restrict on a fiber to a $3$-linear map.

s.harp

Every set of real numbers (ie every subset of $\mathbb R$) that is bounded from below (ie there exists an $a\in\mathbb R$ s.t. $a≤x$ for all $x$ in the set we are considering) has an infimum in $\mathbb R$ (ie there exists a $b\in\mathbb R$ so that $b≤x$ for all $x$ in the set and if $a≤x$ for all $x$ in the set then also $a≤b$).

This refers always to a space we are considering: the situation is that a set is a subset of the space and the condition is that the set is bounded in the space and then the consequence is that the infimum exists in the space

so while $(0,1)$ has lower bound in $\mathbb R$ and a lower bound in $\mathbb R-\{0\}$, it only has an infimum in one of them

also note that $(0,1)$ is not bounded from below if you consider it as a subset of $\mathbb R_{>0}$

so these notions are always relative to some space, the axiom of completeness says that $\mathbb R$ is order complete, if you consider a modification of $\mathbb R$ by removing some points or adding some points then the statement does not apply

Danu

Sep 25, 2016 09:01

@BalarkaSen Yea!

And I'm getting real confused

Do you see what's going on?

Balarka Sen

Haven't thought about it much. I am not really familiar with this language, I think topologically. One direction is clear: if you have a linear functional $f$ on $O(1) \otimes O(1) \otimes O(1)$, on a fiber $f(\lambda x, \lambda y, \lambda z) = \lambda ^3 f(x, y, z)$ by linearity on each component, so that's homogeneous.

But that homogeneous ones are always multilinear...

Danu

Right, so section gives a hom poly

But why hom poly's give a section...

Balarka Sen

Not sure

@Danu OK, no, I see what's up. A homogeneous polynomial is a multilinear map $\Bbb C^{n+1} \otimes \cdots \otimes \Bbb C^{n+1} \to \Bbb C$, is the point.

The number of copies in that tensor product is the degree of the homogeneous polynomial.

Danu

So it takes k vectors as input??

And how is it linear in each argument (in my example, for instance)?

user116211

Sep 25, 2016 09:16

@s.harp Okay, somewhat got the point; re-reading your arguments again....

Balarka Sen

A homogeneous polynomial is of the form $\sum z_0^{i_0} \cdots z_n^{i_n}$ where $i_0 + \cdots + i_n = k$. Think of $z_p$'s as elements of $(\Bbb C^{n+1})^*$ (namely, the coordinate function).

@Danu Your example is not relevant because you seemed to have misunderstood the domain of the k-linear map. $z_0 \cdot z_0 \cdot z_0$ is a fine element of $(\Bbb C^2)^* \otimes (\Bbb C^2)^* \otimes (\Bbb C^2)^*$.

Danu

@BalarkaSen Hmmmm that seems sensible-ish

Balarka Sen

Think about it algebraically.

Danu

@BalarkaSen Yeah, I sorta see what you're saying

Thanks :)

I feel so bad about asking everybody for help all the time :(

Balarka Sen

It's fine. I can rarely help you anyway, so I'd probably do the same thing.

Had I been studying this stuff, I meant. Which I am not, so annoying troll grin towards you.

Danu

Sep 25, 2016 09:26

^^

Balarka Sen

(not really; I am kind of envious)

VermillionAzure

Hey yo yo yo

Is anyone here?

Danu

Just study with me

VermillionAzure

OMG PEOPLE AR EHERE!

Danu

I wrote a summary of what I did so far (about 1/3 the length)

VermillionAzure

Sep 25, 2016 09:27

Can I ask a combinatorics questions here? I don't think it merits asking on the SE

Balarka Sen

I'll read it if you'd send it.

Danu

Yeah, sure

VermillionAzure

@Danu Was that to me?

Danu

@VermillionAzure No. But don't ask to ask

VermillionAzure

@Danu Well okay

I have negative probability in my equation to solve a certain problem but that doesn't seem valid

I'm in an introductory level probability class and we're in distributions but I can't understand what distribution fits the problem

> Five distinct numbers are randomly distributed
to players numbered 1 through 5. Whenever two
players compare their numbers, the one with the
higher one is declared the winner. Initially, players
1 and 2 compare their numbers; the winner then
compares her number with that of player 3, and so
on. Let X denote the number of times player 1 is a
winner. Find P{X = i}, i = 0, 1, 2, 3, 4.

But I came up with a formula already

But the formula involves negative probabilities but it's canceled out by zeroes after being intersected with the probabilities of impossible/null conditional trials

Is that... okay?

...

yeah maybe I should just ask this

Danu

Sep 25, 2016 09:34

We can't help you if you don't give the formula

s.harp

@MAFIA36790 let me know if there is something dodgy or I'm sort of missing the point with my explanation, I'm not the best at constructive pedagogy

$Mathematics$ Mathematics

Participants