Real analysis – Basics III

— 3. Completeness Axiom —

As was said in the previous post the description of the set of real numbers made so far allows one to do almost everything but to formally differentiate the set { \mathbb{Q}} and the set { \mathbb{R}}. To do that we need one more axiom; but first let us introduce some auxiliary notions.

For { a,b \in \mathbb{R}} one usually writes { \lbrack a,b\rbrack}, { \lbrack a,b \lbrack}, { \rbrack a,b\rbrack} and { \rbrack a,b\lbrack} for the real numbers { x} for which the following conditions are verified:

  • { a \leq x \leq b}
  • { a \leq x < b}
  • { a < x \leq b}
  • { a < x < b}

{ \lbrack , \rbrack} is called a closed interval and { \rbrack , \lbrack} is called an open interval.

We can also have another kind of intervals. These intervals are said to be infinite and have the form { \lbrack a, \infty \lbrack}, { \rbrack a, \infty \lbrack}, { \rbrack -\infty,a\rbrack}, and { \rbrack -\infty, a\lbrack}. They represent respectively:

  • { x \geq a}
  • { x > a}
  • { x \leq a }
  • { x < a }
Definition 5 Lets { \mathrm{K}\subset \mathbb{R}} and { a,b \in \mathbb{R}}. We’ll say that the real { b} is an upper bound of { \mathrm{K}} if and only if (for now on written as iff) { \forall x \in \mathrm{K} \,x \leq b}.

We’ll also say that { a} is a lower bound of { \mathrm{K}} iff { \forall x \in \mathrm{K} \,a\leq x}.

We’ll say that { \mathrm{K}\subset \mathbb{R}} is bounded from above if it has at least one upper bound, bounded from below if it has at least one lower bound, and { \mathrm{K} \subset \mathbb{R}} is said to be bounded if it has an upper and a lower bound.

Finally a set is said to be unbounded if it isn’t bounded.

Definition 6 Let { \mathrm{K} \subset \mathbb{R}}. If { \exists x \in \mathrm{K} \land \forall y \in \mathrm{K} x\geq y}, then {x} is the maximal element of { \mathrm{K}} and we denote it by {\max \mathrm{K}}

If { {\mathrm K} \subset \mathbb{R}: \exists u \in {\mathrm K} \land \forall v \in {\mathrm K} \,u \leq v}, { u} is said to be the the minimal element of { {\mathrm K}} and is denoted by { \min {\mathrm K}}.

Definition 7 Let { {\mathrm K} \subset \mathbb{R}} and { {\mathrm V}} the set of all of its upper bounds ({ {\mathrm V}=\emptyset} iff { {\mathrm K}} isn’t bounded from above). The minimal element of { {\mathrm V}}, { s}, is said to be the supremum of { {\mathrm K}}. {\sup {\mathrm K}} Formally:

\displaystyle   \forall x \in {\mathrm K}; x \leq s \ \ \ \ \ (6)

\displaystyle   \forall v \in {\mathrm V}; s\leq v \Leftrightarrow \forall \epsilon > 0 \exists x \in {\mathrm K}: x > s-\epsilon \ \ \ \ \ (7)

The notion of the infimum of a set can be introduced in an analogous way.

Definition 8 Let { {\mathrm K} \subset \mathbb{R}} and { {\mathrm U}} the set of lower bounds of { {\mathrm K}}. The infimum of { {\mathrm K}} is the maximal element of { {\mathrm U}}. If such a a maximal element exists the infimum of { {\mathrm K}} is denoted by { \inf {\mathrm K}} and is the real number { r} such as:

\displaystyle   \forall x \in {\mathrm K}; x\geq r \ \ \ \ \ (8)

\displaystyle   \forall u \in {\mathrm U}; u\leq r \Leftrightarrow \forall \epsilon >0 \exists x \in {\mathrm K}: \quad x < r+\epsilon \ \ \ \ \ (9)

Let us see some examples of all of these previous notions so that we can have a feel of what’s going on.

  • {\sup \lbrack a,b\rbrack = \max \lbrack a,b\rbrack = b}
  • {\inf \lbrack a,b\rbrack = \min \lbrack a,b\rbrack = a}
  • {\sup \rbrack a,b\lbrack = b}
  • {\inf \rbrack a,b\lbrack = a}

Notice that the two last sets have no maximal nor minimal element.

As a work out the reader can try to find {\min}, {\max}, {\sup}, and {\inf} of the empty set.

And now we’ll state the Completeness Axiom and with it our basic study of the real numbers will be complete.

Axiom 8 (Completeness Axiom) Any non-empty subset of { \mathbb{R}} with an upper bound has a real supremum.
Theorem 9 Any non-empty subset of { \mathbb{R}} which is bounded from below has an infimum.

Proof: Omitted. \Box

Theorem 10 { \mathbb{N}} isn’t bounded above.

Proof: Omitted. \Box

Theorem 11 (Archimedean property) { a,b \in \mathbb{R} \land a > 0 \Rightarrow \exists n \in \mathbb{N}: na > b}

Proof: Omitted. \Box

Theorem 11 tell us that in { \mathbb{R}} infinitesimals don’t exist: if we add any given quantity ( it doesn’t matter how small it is ) a sufficient number of times the value of the sum will always be greater than any other given quantity.

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