Infimum

Definition

Set $B\subset A$ where $(A,<)$ is an ordered set has an infimum $\alpha$ if

1. $B\ne \varnothing$
2. $(\forall a\in A,\alpha <a)(\exists b\in B,b<a)$

An ordered set $(A,<)$ is said to have the greatest lower bound property if $\forall B\subset A,B\ne \varnothing$ and $B$ is bounded below, we have

$$\inf B$$

which exists in $A$.

Example

Recall that we saw $(\mathbb{Q},<)$ does not have the least upper bound property. Indeed,

$$B=\{x\in \mathbb{Q}\mid x^2<2\}$$

has no Supremum in $\mathbb{Q}$.

Property (LUB = GLB)

Let $(A,<)$ be an ordered set. Then $A$ has the least upper bound property iff $A$ has the greatest lower bound property.

Proof:

$(⟹)$ Let $B\subset A,B\ne \varnothing$. $B$ must be bounded below because the properties exist if at least one subset exists that satisfies it. We want to show $\inf (B)$ exists.

           x      B ---->
<----------|------|---------> A

Let

$$C=\{a\in A\mid a\text{ is a lower bound of }B\}$$

• $C\ne \varnothing$ since $B$ is bounded below.
• $C$ is bounded above. Indeed, let $b\in B$ and $c\in C$. Then since $c\in C$, $c$ is a lower bound for $B$ and so $\forall x\in B,c\le x$. In particular, $\forall c\in C,c\le b$ and $C$ is bounded above.

Then $A$ has the least upper bound property, $C\ne \varnothing ,C$ bounded above. Therefore $\sup (C)$ exists. Let $d=\sup (C)$.

We claim that $d=\inf (B)$. To show this, we need to prove:

1. $d$ is a lower bound
2. $d$ is the greatest lower bound. I.e. $(\forall c\in A,d<c)(\exists b\in B,b<c)$.

Suppose $d$ is not a lower bound for $B$. That is, suppose $\exists b_0\in B,b_0<d$. Since $d=\sup (C)$ and $d>b_0$. This implies $\exists c_0\in C$ such that $d\ge c_0>b$.

However, $\forall c\in C,\forall b\in B$ we have $c\le b$ (by definition of the construction of $C$) which gives a contradiction

$$c=c_0,b=b_0$$

Now, for property $(2)$, suppose $d<c$. We want to show that $\exists b_0\in B$ such that $b_0<c$. Suppose by contradiction it is not. Then, $\forall b_0\in B,b_0\ge c$. Then $c$ is a lower bound for $B$. On the other hand, $d=\sup (C)$. So, $\forall c^{\prime }\in C,c^{\prime }\le d$ we can plug in $c^{\prime }=c$ to get a contradiction (conflicts with the supposition $d<c$).

$(⟸)$ TODO