Latest revision as of 08:55, 29 July 2016

Definition

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

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

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

See also

• Passing to the infimum
• Supremum
  • Passing to the supremum

Notes

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

References

1. Jump up ↑ 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 $\text{Inf}(A)$ was a lower bound such that any value greater than $\text{Inf}(A)$ would fail to be a lower bound (thus $\text{Inf}(A)$ is the greatest one, as any bigger fail to be). This leads to the formulation of $\text{Inf}(A)$ as:

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

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

• $\exists a\in A$

cannot be true (there does not exist anything in $A$ 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

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 $A\subseteq X$ of a poset $(X,\preceq)$^[1]:

• $\text{inf}(A)$

such that:

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

Proof of claims

See also

• Passing to the infimum
• Supremum
  • Passing to the supremum

Notes

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

References

1. Jump up ↑ Lattice Theory: Foundation - George Grätzer