Difference between revisions of "Infimum"

From Maths
Jump to: navigation, search
m
m
 
Line 1: Line 1:
 +
{{Stub page|msg=Fleshing out, make sure the caveat is known, proof of claim|grade=A}}
 +
__TOC__
 +
==Definition==
 +
Let {{M|(X,\preceq)}} be a [[poset]] and let {{M|A\subseteq X}} be any [[subset of]] {{M|X}}<ref group="Note">Which may be written:
 +
* {{M|A\in\mathcal{P}(X)}} where {{M|\mathcal{P}(S)}} denotes the [[power set]] of a [[set]] {{M|S}}</ref>. The ''infimum'' ({{AKA}}: ''greatest lower bound'', ''g.l.b'') of {{M|A}} is an element of {{M|X}}, written {{M|\text{Inf}(A)}} that satisfies the following two conditions{{rLTFGG}}:
 +
# {{M|1=\forall a\in A[\text{Inf}(A)\preceq a]}} - which states that {{M|\text{Inf}(A)}} is a [[lower bound]] of {{M|A}} - and
 +
# {{M|1=\forall b\in\underbrace{\left\{x\in X\ \vert\ (\forall a\in A[x\preceq a])\right\} }_{\text{the set of all lower bounds of }A }\Big[b\preceq\text{Inf}(A)\Big]}} - which states that for all lower bounds of {{M|A}}, that lower bound "is ''[[majorised by]]''"<ref group="Note">Recall that if for a [[poset]] {{M|(P,\preceq)}} and for {{M|p,q\in P}} if we have:
 +
* {{M|p\preceq q}} then we may say:
 +
*# {{M|p}} is ''majorised by'' {{M|q}} or
 +
*# {{M|q}} ''majorises'' {{M|p}}</ref><!-- <--this line is the end
 +
 +
END OF SECOND # FOR DEFINITION - there's a long note here which has a *, a *# and another *#, which makes it hard to tell what is where
 +
 +
--> {{M|\text{Inf}(A)}}
 +
#* '''Claim 1: ''' we have ''part 2'' of the definition {{iff}} {{M|1=\forall x\in X\Big[\underbrace{\left(\forall a\in A[x\preceq a]\right)}_{x\text{ is a lower bound of }A}\implies x\preceq\text{Inf}(A)\Big]}}
 +
#* '''Claim 2: ''' we ''claim 1'' {{iff}} {{M|1=\left(A=\emptyset\vee\Big(\forall x\in X\exists a\in A[x\succ\text{Inf}(A)\implies a\prec x]\Big)\right)}}
 +
Notice the {{M|1=A=\emptyset}} condition here, as in the case {{M|A}} is empty, {{M|\exists a\in A}} is ''always'' false. This is a very big caveat.
 +
==See also==
 +
* [[Passing to the infimum]]
 +
* [[Supremum]]
 +
** [[Passing to the supremum]]
 +
==Notes==
 +
<references group="Note"/>
 +
==References==
 +
<references/>
 +
{{Order theory navbox|plain}}
 +
{{Definition|Order Theory|Real Analysis|Set Theory}}
 +
=OLD PAGE=
 +
: {{Caution|Rather than trying to fix the old page (which was written with an erroneous claim) I shall instead re-write it and make the caveat known}}
 +
I got this slightly wrong initially, I was taught that an infimum is the greatest lower bound, that would mean that {{M|\text{Inf}(A)}} was a lower bound such that any value greater than {{M|\text{Inf}(A)}} would fail to be a lower bound (thus {{M|\text{Inf}(A)}} is the greatest one, as any bigger fail to be). This leads to the formulation of {{M|\text{Inf}(A)}} as:
 +
* {{M|1=\forall x\in X\exists a\in A[x>\text{inf}(A)\implies a<x]}} (If you pick a value greater than the inf, there exists an element in {{M|A}} less than what you picked) and
 +
* {{M|1=\forall a\in A[\text{inf}(A)\le a]}} (the inf is actually a lower bound)
 +
However there is a problem, the book I was reading speaks about {{M|\text{Inf}(\emptyset)}}, if {{M|1=A:=\emptyset}} then the expression:
 +
* {{M|\exists a\in A}}
 +
cannot be true (there does not exist anything in {{M|A}} at all! Let alone something that satisfies the rest of the statement!).
 +
 +
I must make this caveat very clear in the new version
 +
=OLD PAGE START=
 
{{Stub page|Needs fleshing out, INCOMPLETE PAGE|grade=A}}
 
{{Stub page|Needs fleshing out, INCOMPLETE PAGE|grade=A}}
 
: A closely related concept is the [[supremum]], which is the smallest upper bound rather than the greatest lower bound.
 
: A closely related concept is the [[supremum]], which is the smallest upper bound rather than the greatest lower bound.

Latest revision as of 08:55, 29 July 2016

Stub grade: A
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Fleshing out, make sure the caveat is known, proof of claim

Definition

Let [ilmath](X,\preceq)[/ilmath] be a poset and let [ilmath]A\subseteq X[/ilmath] be any subset of [ilmath]X[/ilmath][Note 1]. The infimum (AKA: greatest lower bound, g.l.b) of [ilmath]A[/ilmath] is an element of [ilmath]X[/ilmath], written [ilmath]\text{Inf}(A)[/ilmath] that satisfies the following two conditions[1]:

  1. [ilmath]\forall a\in A[\text{Inf}(A)\preceq a][/ilmath] - which states that [ilmath]\text{Inf}(A)[/ilmath] is a lower bound of [ilmath]A[/ilmath] - and
  2. [ilmath]\forall b\in\underbrace{\left\{x\in X\ \vert\ (\forall a\in A[x\preceq a])\right\} }_{\text{the set of all lower bounds of }A }\Big[b\preceq\text{Inf}(A)\Big][/ilmath] - which states that for all lower bounds of [ilmath]A[/ilmath], that lower bound "is majorised by"[Note 2] [ilmath]\text{Inf}(A)[/ilmath]
    • Claim 1: we have part 2 of the definition if and only if [ilmath]\forall x\in X\Big[\underbrace{\left(\forall a\in A[x\preceq a]\right)}_{x\text{ is a lower bound of }A}\implies x\preceq\text{Inf}(A)\Big][/ilmath]
    • Claim 2: we claim 1 if and only if [ilmath]\left(A=\emptyset\vee\Big(\forall x\in X\exists a\in A[x\succ\text{Inf}(A)\implies a\prec x]\Big)\right)[/ilmath]

Notice the [ilmath]A=\emptyset[/ilmath] condition here, as in the case [ilmath]A[/ilmath] is empty, [ilmath]\exists a\in A[/ilmath] is always false. This is a very big caveat.

See also

Notes

  1. Which may be written:
    • [ilmath]A\in\mathcal{P}(X)[/ilmath] where [ilmath]\mathcal{P}(S)[/ilmath] denotes the power set of a set [ilmath]S[/ilmath]
  2. Recall that if for a poset [ilmath](P,\preceq)[/ilmath] and for [ilmath]p,q\in P[/ilmath] if we have:
    • [ilmath]p\preceq q[/ilmath] then we may say:
      1. [ilmath]p[/ilmath] is majorised by [ilmath]q[/ilmath] or
      2. [ilmath]q[/ilmath] majorises [ilmath]p[/ilmath]

References

  1. Lattice Theory: Foundation - George Grätzer

OLD PAGE

Caution:Rather than trying to fix the old page (which was written with an erroneous claim) I shall instead re-write it and make the caveat known

I got this slightly wrong initially, I was taught that an infimum is the greatest lower bound, that would mean that [ilmath]\text{Inf}(A)[/ilmath] was a lower bound such that any value greater than [ilmath]\text{Inf}(A)[/ilmath] would fail to be a lower bound (thus [ilmath]\text{Inf}(A)[/ilmath] is the greatest one, as any bigger fail to be). This leads to the formulation of [ilmath]\text{Inf}(A)[/ilmath] as:

  • [ilmath]\forall x\in X\exists a\in A[x>\text{inf}(A)\implies a<x][/ilmath] (If you pick a value greater than the inf, there exists an element in [ilmath]A[/ilmath] less than what you picked) and
  • [ilmath]\forall a\in A[\text{inf}(A)\le a][/ilmath] (the inf is actually a lower bound)

However there is a problem, the book I was reading speaks about [ilmath]\text{Inf}(\emptyset)[/ilmath], if [ilmath]A:=\emptyset[/ilmath] then the expression:

  • [ilmath]\exists a\in A[/ilmath]

cannot be true (there does not exist anything in [ilmath]A[/ilmath] at all! Let alone something that satisfies the rest of the statement!).

I must make this caveat very clear in the new version

OLD PAGE START

Stub grade: A
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Needs fleshing out, INCOMPLETE PAGE
A closely related concept is the supremum, which is the smallest upper bound rather than the greatest lower bound.

Definition

An infimum or greatest lower bound (AKA: g.l.b) of a subset [ilmath]A\subseteq X[/ilmath] of a poset [ilmath](X,\preceq)[/ilmath][1]:

  • [ilmath]\text{inf}(A)[/ilmath]

such that:

  1. [ilmath]\forall a\in A[\text{inf}(A)\le a][/ilmath] (that [ilmath]\text{inf}(A)[/ilmath] is a lower bound)
  2. [ilmath]\forall x\in\underbrace{\{y\in X\ \vert\ \forall a\in A[y\le a]\} }_{\text{The set of all lower bounds} }\ \ [\text{inf}(A)\ge x][/ilmath] (that [ilmath]\text{inf}(A)[/ilmath] is an upper bound of all lower bounds of [ilmath]A[/ilmath])
    • Claim 1: , this is the same as [ilmath]\forall x\in X\exists a\in A[x>\text{inf}(A)\implies a<x][/ilmath][Note 1][Note 2]

Proof of claims

(Unknown grade)
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
Make a subpage and put the proof here

See also

Notes

  1. This would require [ilmath]A\ne\emptyset[/ilmath]
  2. Let some [ilmath]x\in X[/ilmath] be given, if [ilmath]x\le\text{inf}(A)[/ilmath] we can choose any [ilmath]a\in A[/ilmath] as for implies if the LHS of the [ilmath]\implies[/ilmath] isn't true, it matters not if we have the RHS or not.

References

  1. Lattice Theory: Foundation - George Grätzer