# Infimum

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.

## 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]

# 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

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]