This note concerns the equivalence between the three properties usually taken as an axiom in synthetic constructions of the real numbers. We start with the least upper bound property, call **L**, which is usually appeared in construction of the real numbers.

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L, least upper bound axiom): If is a non-empty subset of , and if has an upper bound, then has a least upper bound , such that for every upper bound of , there holds .

The second property, call **C**, is the completeness of reals.

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C, completeness axiom): If and are non-empty subsets of with the property for any and , then there is some such that for any and .

The third, also last, property, call **AC**, is the set of two results: the Archimedean property and Cantor’s intersection theorem. These two results often appear as consequences of the construction of reals.

Archimedean property: For any real numbers and with , there exists some natural number such that .

Cantor’s intersection theorem: A decreasing nested sequence of non-empty, closed intervals in has a non-empty intersection.

Our aim is to prove that in fact the above three properties (L), (C), and (AC) are equivalent. Our strategy is to show the following direction:

(L) ⟶ (C) ⟶ (AC) ⟶ (L).

In the last part of this note, we present a simple application.

**PROOF OF (L) ⟶ (C)**

Let and be non-empty subsets of with the property for any and . Clearly, any element of is an upper bound for . This tells us that exists in . We denote

and prove that is the desired real number. The inequality for all is obvious by means of sup. The only possibility for which the inequality for all fails to hold is that is not a lower bound for , namely there is some such that . But this violates the definition of because there always exists some in such a way that .

**PROOF OF (C) ⟶ (AC)**

First, we prove Cantor’s Intersection Theorem. Let for be a decreasing nested sequence of non-empty, closed intervals in , namely and

for all . To make use of (C) we construct

Clearly, the above sets and fulfill the hypotheses of (C). Hence there is some such that

for all and . In particular, for all . Thus

This proves Cantor’s Intersection Theorem.

Second, we prove the Archimedean property. By way of contradiction, there exist and such that

for all . To make use of (C) we construct

Then there is some such that

for any and any . But then this implies that is also an upper bound for . Hence, belongs to . This extra information tells us that is a lower bound for , hence is the least upper bound for . As , there must exist some element in , say for some , in such a way that

which is equivalent to

Thus, we have just shown that

This is a contradiction.

**PROOF OF (AC) ⟶ (L)**

Given the (AC) property, we now prove the L property. Let be any non-empty, bounded from above subset of . We show that exists. (If is bounded from below, then exists, and the proof follows similarly.) Our strategy is to construct a sequence of nested closed intervals whose intersection is . Initially, we choose some and some with . Such points and exist because is non-empty and bounded from above. From the closed interval we construct via the rule

Obviously, and

Repeating the above process we can construct a sequence of nested closed intervals with with the following properties

and and is an upper bound of . By Cantor’s Intersection Theorem, there exists

We show . First, is an upper bound for . If this were not true, then we would have some such that . By the Archimedean property, there is some such that

As we deduce that which immediately implies that .

This contradict the choice of . Now we show that is the least upper bound for . Again if this were not true, then we would have some such that

Again by the Archimedean property, there is some such that

which implies , otherwise .

This contradict the choice of . Hence, we complete the proof.

**APPLICATION**

As application, we show that the equation has solution in , denoted by , given one of the above three equivalent properties.

**Assuming (L)**. We let

Clearly the set is non-empty because and is bounded from above by . Then the desired solution to is nothing but .

**Assuming (C).** We construct

Then the desired solution to is nothing but the real number , namely the one satisfying

for all and all .

**Assuming (AC).** We construct the sequence of nested closed intervals as in the above proof staring from . Then the desired solution is .

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