Completeness Axiom

Let $A$ be a non empty subset of $\mathbb{R}$.

• $u$ is the upper bound of $A$ if: $\forall a\in A;a\le u$
• $A$ is bounded above if $A$ has an upper bound
• Maximum element of $A$: $\max{A} = u$ if $u\in A$ and $u$ is an upper bound of $A$
• Supremum of $A$ $\sup{A}$, is the smallest upper bound of $A$
• Maximum is a supremum. Supremum is not necessarily a maximum.
• $l$ is the lower bound of $A$ if: $\forall a\in A;a\ge l$
• $A$ is bounded below if $A$ has a lower bound
• Minimum element of $A$: $\min{A} = l$ if $l\in A$ and $l$ is a lower bound of $A$
• Infimum of $A$ $\inf{A}$, is the largest lower bound of $A$
• Minimum is a infimum. Infimum is not necessarily a minimum.

Theorems

Let $A$ be a non empty subset of $\mathbb{R}$.

• Say $u$ is an upper bound of $A$. Then $u= \sup A$ iff: $\forall \epsilon \gt 0\;\exists a \in A;\;a + \epsilon \gt u$
• Say $l$ is a lower bound of $A$. Then $l= \inf A$ iff: $\forall \epsilon \gt 0\;\exists a \in A;\;a - \epsilon \lt l$

Proof Hint

Prove the contrapositive. Use $\epsilon=\frac{1}{2}(L-sup(A))$ for supremum proof. $$

Required proofs

• $\sup(a,b)=b$
• $\inf(a,b)=a$

Completeness property

A set $A$ is said to have the completeness property iff every non-empty subset of $A$:

• Which is bounded below has a infimum in $A$
• Which is bounded above has a supremum in $A$

Both $\mathbb{R}, \mathbb{Z}$ have the completeness property. $\mathbb{Q}$ doesn’t.

In addition to that:

• Every non empty subset of $\mathbb{Z}$ which is bounded above has a maximum
• Every non empty subset of $\mathbb{Z}$ which is bounded below has a minimum