Maxima and Minima

A real number $a_{0}$ is a maximum of the set $A$ if $a_{0}$ is an element of $A$ and $a_{0} \geq a$ for all $a \in A$ . Similarly, a number $a_{1}$ is a minimum of $A$ if $a_{1} \in A$ and $a_{1} \leq a$ for every $a \in A$ .

Example

Consider the intervals $(0, 2)$ and $[0, 2]$ . Both sets are bounded above and below and have the same least upper bound, namely 2. But for $2$ to be a maximum of the set, we require it being in the set, so then the former set doesn't have it's maximum, while the latter does.

This goes to show that the supremum need not be a maximum, but if a maximum exists, then it is the supremum.

Thus, the axiom of completeness doesn't get a proof, as it's the foundation of how $R$ exists. But let's look at why $Q$ is itself incomplete:

Example

Consider $S = {r \in Q : r^{2} < 2}$ . Clearly $S$ is bounded above, such as by $2$ , or $3 / 2$ or similar. But looking for the least upper bound, we can find rational approximations (ex: $b = 142 / 100$ ) but then we can always find a smaller upper bound (ex: $b = 1415 / 1000$ ). In $Q$ , we lack the axiom of completeness to get this least upper bound into the set.

But in $R$ there is the axiom! The axiom of completeness says that $α = sup (S)$ which is a real number. We could prove that $α^{2} = 2$ and thus $α = \sqrt{2}$ but we know that $α$ isn't rational from the irrationality of $\sqrt{2}$ , so then for the rationals we can't use this search to find it. \

References

[[Abbott Real Analysis.pdf#page=29]]