Conversation started Sep 25, 2016 at 7:48.
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
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
SoumyoB
so I just drag the image to the typing space?
Balarka Sen
Balarka Sen
@MAFIA36790 I have forgotten, what's the defn of order-completeness?
SoumyoB
SoumyoB
where do you upload the pic though
user116211
@SoumyoB See, there is an upload... button next to the send bar.
s.harp
s.harp
@MAFIA36790 consider $\mathbb R -\{0\}$, here the set $(0,1)$ has a lower bound but no infimum
SoumyoB
SoumyoB
Sep 25, 2016 07:53
@MAFIA36790 I don't see any upload button :/
s.harp
s.harp
user image
SoumyoB
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
Danu
Sep 25, 2016 08:13
@BalarkaSen hi
user116211
internet connection disrupted
user116211
@SoumyoB Never knew of such things.
Danu
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
Danu
Sep 25, 2016 08:17
15 hours ago, by Ted Shifrin
Obviously a homogeneous polynomial of degree $k$ restricts to such.
s.harp
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
Danu
Obviously :'(
user116211
@s.harp Well, in defining the axiom, the set concerned was a set of real numbers.
Danu
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
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
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
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
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
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
Danu
Sep 25, 2016 09:01
@BalarkaSen Yea!
And I'm getting real confused
Do you see what's going on?
Balarka Sen
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
Danu
Right, so section gives a hom poly
But why hom poly's give a section...
Balarka Sen
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
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....
 
Conversation ended Sep 25, 2016 at 9:16.