📂Analysis

Three Axioms of Analysis: 3 The Completeness Axiom

Axiom¹

If a set $E \subset \mathbb{R}$ is nonempty and $E$ is bounded above, then the supremum $\sup(E) < \infty$ exists.

Explanation

The field axioms and the order axioms merely rewrote, in a difficult way, what we already knew, but at first glance the completeness axiom is not like that. First of all, the words appearing here seem to require definitions.

Definition

If $a \le M$ holds for every element $a$ of $E$, then $E$ is said to be bounded above. Every $M$ satisfying this condition is called an upper bound of $E$. $\sup(E)$ denotes the smallest upper bound of $E$, that is, the number satisfying $\sup (E) \le M$ for every upper bound $M$ of $E$. This is called the supremum of $E$.

For the opposite inequality, it goes as follows.

If $a \ge m$ holds for every element $a$ of $E$, then $E$ is said to be bounded below. Every $m$ satisfying this condition is called a lower bound of $E$. $\inf(E)$ denotes the largest lower bound of $E$, that is, the number satisfying $\inf (E) \ge m$ for every lower bound $m$ of $E$. This is called the infimum of $E$.

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It may be confusing to have several definitions pour out all at once, but essentially they do not shake our concepts. They merely define when a set is said to have a limit, and what that limit is called in such cases.

Returning to the completeness axiom, the completeness axiom feels like a repetition of the definitions introduced above. The difference is simple. The definitions only tell us what to call something when it exists; they never say whether it really exists. It is the completeness axiom that speaks about that very ’existence.'

However, one may well wonder whether this really has to be an axiom. Is it such a basic fact that it must be made an axiom? Can it not be proved? Reading the definitions, it seems that such a supremum obviously exists by definition and could be proved, but that is not the case.

Refutation

Since $E$ is said to be bounded above, there will be upper bounds $M$ satisfying the condition, and among the upper bounds $M$ there will exist a smallest one, so that the supremum $\sup(E)$ will exist. But thinking of it the other way around, $\sup(E)$ is the largest value among the $-M$, that is, a supremum. The very claim that a smallest value exists is itself grounded on the existence of a supremum. Thus we inevitably fall into circular reasoning.

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Whether upper or lower, larger or smaller, only the direction is reversed and the argument keeps going around in circles. In the end, we simply cannot establish the existence of such a supremum or infimum. Therefore, there was no choice but to create a new axiom called the completeness axiom.

Theorem

If a subset $E$ of the set of integers $\mathbb{Z}$ has a supremum, then $\sup(E) \in E$

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Without the completeness axiom, even such an obvious fact cannot be believed, because its assumption remains questionable.

Complete?

The Korean word 완비 (完備) is a refined rendering of ‘complete’; when generalized to metric spaces beyond the real space $\mathbb{R}$, a space that contains the convergence points of Cauchy sequences is defined as a complete space. However, in everyday life the English word ‘complete’ is not used as an expression meaning to be fully (完) equipped (備); rather, it is often used together with the end of something that continues, as in ‘completion’ or ‘conclusion.’ As mentioned, in that it guarantees that the end of a sequence, namely its convergence point, exists (within the space), we can see that the expression ‘complete’ is appropriate.

Of course, taking a Cauchy sequence and taking $E \subset \mathbb{R}$ are different things, but explanations such as $\mathbb{R}$ having separability are still far too early. It is fine to move on, thinking that such things will be learned later on.