Infimum
Contents
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]:
- [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
- [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
- ↑ Which may be written:
- ↑ 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:
- [ilmath]p[/ilmath] is majorised by [ilmath]q[/ilmath] or
- [ilmath]q[/ilmath] majorises [ilmath]p[/ilmath]
- [ilmath]p\preceq q[/ilmath] then we may say:
References
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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
- 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:
- [ilmath]\forall a\in A[\text{inf}(A)\le a][/ilmath] (that [ilmath]\text{inf}(A)[/ilmath] is a lower bound)
- [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])
Proof of claims
The message provided is:
See also
Notes
- ↑ This would require [ilmath]A\ne\emptyset[/ilmath]
- ↑ 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
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