proof of embedding theorem for ordered abelian groups of rank one
To prove the theorem, we shall use the characterization of ordered groups of rank one. Using this result, the theorem to be be proved can be reformulated as follows: (Note that we are now using additive notation for groups, so the formulation of the Archimedean property appears slightly differently than it did in the entry characterizing the ordered groups of rank one.)
Theorem. Let be an Abelian ordered group with order (http://planetmath.org/OrderGroup) at least 2. Then enjoys the Archimedean property
Proof One of the two implications is rather easy. Since the real numbers enjoy the Archimedean property and it is obvious that any subset of a set enjoying the Archimedean property also enjoys this property, it follows that every subgroup of the additive group of the real numbers enjoys the Archimedean property. It only remains to show that every group enjoying this property is isomorphic to a subgroup of additive group of the real numbers.
To accomplish this, we shall use the technique of Dedekind cuts to map our group into the real numbers. Since the order of our group is at least 2, there exists an element such that . By the basic theorem on ordered groups, it follows that either or and that, if then . Thus there exists an element such that — if , set , else set .
We shall now construct a map . Define the sets and as follows:
By Corollary 2 to the basic theorem on ordered groups, it follows that, for any strictly positive integer , we have if and only if . Likewise, if and only if . This means that and may be regarded as sets of rational numbers, and we can rewrite the defining equations as
We shall now show that is a Dedekind cut. This requires us to verify the three defining properties of a Dedekind cut. First, we need to check that neither nor are empty. We shall only present the proof for since the proof for can easily be obtained from it by reversing the inequality signs suitably. By the basic theorem, either or or . In the first case, . In the second case, . In the third case, the Archimedean property implies that there exists such that , so . Second, we need to check that every element of is less than every element of . Suppose that and , so that and . Then, by Corollary 2 of the basic theorem, and . By conclusion 2 of the basic theorem, this implies that . By Corollary 2, this means that . Dividing by , we conclude that . Third, we need to show that at most one rational number does not belong to either or to . Suppose that does not belong to either or . Then it cannot be the case that either
By the first conclusion of the basic theorem, it follows that . Now suppose that also does not belong to either or to . By the same line of reasoning, we must have . This would imply that . Similarly, . Combining these two facts using conclusion 2 of the basic theorem, we would have , which implies that .
Since is a Dedekind cut, it defines a real number. We shall define to be this number.
We will now show that is a homomrphism. To accomplish this, it suffices to show that for every . Suppose that and (as before, we assume . Then, by definition,
By Corollary 2 of the basic theorem, we obtain
By conclusion 4 of the basic theorem, we have
Therefore, . Thus, we have shown that, if and , then . Likewise, if and , then . By the definition of subtraction for Dedekind cuts, this implies that .
Next, we shall show that preserves the order. Since we already know that is a homomorphism, it suffices to show that when . By definition of as a Dedekind cut, this is equivalent to demanding that, whenever , then for all . This follows readily from the fact that, if and and , then , so and hence could not possibly belong to .
To the proof, we must show that is in fact an isomorphism. This may be accomplished by showing that the kernel of is trivial. Suppose that . Then, by the definition of Dedekind cut, consists of the negative rational numbers and consists of the positive rational numbers. By the definitions of and this, in turn, means that whenever and and whenever and . By the basic theorem, either or or . If then, by the Archimedean property, there must exist an such that . However, this would contradict the assertion that whenever and , so it is not possible to have . Likewise, if , then there must exist an such that , contradicting the assertion that whenever and . The only remaining possibility is to have . Therefore, the homomorphism is, in fact, an isomorphism, so is isomorphic to a subset of the additive group of the real numbers.
|Title||proof of embedding theorem for ordered abelian groups of rank one|
|Date of creation||2013-03-22 14:55:33|
|Last modified on||2013-03-22 14:55:33|
|Last modified by||rspuzio (6075)|