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Although it is tempting, and commonly done, to define real numbers as infinite sequences of digits there are insurmountable logical difficulties with that construction. The fact that some real numbers have two representations, such as unity being 1.000... and 0.999..., poses no real problem. The insurmountable problem is that the sum of two infinite sequences of digits cannot always be defined. Likewise the product and difference of two infinite sequences is not always defined. Furthermore the inverse (reciprocal) of a nonzero inifinite sequence may not be defined.
The above problems may be eliminated by taking a real number to be the limit of an infinite sequence. Thus the square root of 2 should not be thought of as a particular infinite sequence of digits, but instead as the limit of a sequence; e.g.,
But this sequence is not the only one that has √2 as its limit. There are an infinitude of convergent sequences that correspond to √2. The important point is that if √2 and √3 are thought of as particular infinite sequences of digits there is no way to define (√2 + √3). On the other hand, it is elementary to construct a convergent sequence for (√2 + √3) from any of the convergent sequences for √2 and √3.
The conceptual difficulty in defining real numbers as convergent sequences is that there is no entities to be taken as the limits of convergent sequences. There is nothing there to be the limit of the sequence. As Gertrude Stein said, "There is no there, there." Real numbers have to be equivalence classes of convergent sequences.
Let {a_{n}: n=0, 1, 2, ...} be a sequence of rational numbers. A sequence converges if and only if for any rational ε>0 there is an integer N such that
It is essential to define equivalence classes of convergent sequences, but for now this task is postponed until some other constructions are introduced.
Proof: Suppose that for an unbounded sequence {a_{n}} and an ε>0 there existed an N such that for all m,n≥N a_{m}−a_{n}<ε. Then let n=N. By the unboundedness of {a_{m}} for any A>0 there exists an M a_{M}>A. Then choose A>a_{N}+ε and m a value M such that the bound is exceeded. This produces a contradiction. Thus an unbounded sequence cannot be convergent and a convergent sequence cannot be unbounded.
Let {a_{n}} and {b_{n}} be two converent sequences. Define {c_{n}} as {a_{n}+b_{n}}. It is necessary to show that {c_{n}} is convergent. Let ε be any positive rational number.
Consider for any n and m
Since {a_{n}} and {b_{n}} are both convergent there exists two integers N_{a} and N_{b}} such that
By choosing N=max(N_{a},N_{b}) the conditions for convergence of {c_{n}}={a_{n}+b_{n}} are assured. Thus {a_{n}+b_{n}} is a convergent sequence.
If {b_{n}} is convergent then so is {−b_{n}} and so the difference of two convergent sequences is a convergent sequence; i.e., {a_{n}−b_{n}}= {a_{n}+(−b_{n}}},
Again let {a_{n}} and {b_{n}} be two convergent sequences. Consider {c_{n}}={a_{n}*b_{n}}.
Let A and B be bounds on {a_{n}} and {b_{n}}, respectively. Then
By choosing N_{b} such that b_{m}−b_{n}<(ε/2)/A for all m,n>N_{b} and N_{a} such that a_{m}−a_{n}<(ε/2)/B for all m,n>N_{a}, and then N=max(N_{a}, N_{b}) it is assured that c_{m}c_{n}<ε. Thus the product of two convergent sequences is convergent.
Before extending the analysis to reciprocals (multiplicative inverses) it is necessary to define what it means for a sequence to be convergent to zero. Then it must be established that a convergent sequence not convergent to zero is bounded away from zero.
It is an essential part of the analysis that it is not necessary to specify what a convergent sequence converges to. If the sequence converges to a rational number there is no problem about dealing with limits. Zero is a special and important case.
A sequence {a_{n}} converges to zero if for any ε>0 there exists an N such that for all n≥N a_{n}<ε
Proof:
Let {a_{n}} be a convergent sequence that is not convergent to zero. There therefore exists a lower bound on the magnitude of a_{n}. Let A be such a lower bound.
For {a_{n}} define a reciprocal sequence {b_{n}} as follows:
where M is some arbitrary large number. Its choice does not affect the analysis because beyond some point there are no cases of a_{n}=0.
Consider
Therefore is N is chosen so that a_{n}−a_{m}<εA² then b_{m}−b_{n} will be less than ε. Thus {b_{n}} is a convergent sequence and hence any convergent sequence not convergent to zero will have an inverse. This means that the quotient of any two sequences is defined if the divisor sequence is not converent to zero.
Any sequence convergent to zero is in the equivalence class of zero, denoted as [0]. Two convergent sequences are equivalent; i.e., belong to the same equivalence class, is their difference is in the equivalence class of zero.
After defining equivalence classes it is necessary to define the sum, difference, product and reciprocal of equivalence classes. This is done using representives from the equivalence classes, but it also must be shown that it does not matter which element of the class is chosen as representative. This shows the sum, difference, product and reciprocal are well defined.
(To be continued.)
This approach to the real numbers uses eleven axioms to define a complete ordered field, the real numbers. Let (S, +, *, <) be a set and two function, + and *, called addition and multiplication, respectively, and an order relation <, which is a boolean function. The functions are usually called operations and represented in infix notation, but they are nothing more than special binary functions. The first eight axioms define a field:
The first seven conditions can be shown to be satisfied for the equivalence classes of convergent sequences as a result of the rational numbers satisfying those conditions. Order relations require more analysis.
Order for two convergent sequences of rational numbers {a_{n}} and {b_{n}} must be defined without any reference to the limits of the sequences. This is not too hard to do.
A convergent sequence {a_{n}} is greater than a convergent sequence {b_{n}} if there exists an interger N such that for all i>N
It is first necessary to demonstrate that convergent sequences obey the trichotomy law; i.e.; for any two convergent sequences {a_{n}} and {b_{n}} either {a_{n}} > {b_{n}}, {b_{n}} > {a_{n}} or {a_{n}} = {b_{n}}. Equality in this context means that {a_{n}} is not greater than {b_{n}} and {b_{n}} is not greater than {a_{n}}. Stated in this form trichotomy is a logical requirement.
Transitivity is very simple to prove. If {a_{n}} > {a_{n}} then there exists an N_{1} such that for all n≥N_{1} a_{n} > b_{n}. Likewise there is an N_{2} such that for all n ≥ N_{2} b_{n} > c_{n}. Thus for all n ≥ N=max(N_{1}, N_{2}) a_{n} > c_{n}.
Isotony is similarly simple to prove. If {a_{n}} > {a_{n}} then there exists an N such that for all n≥N a_{n} > b_{n}. The inequality holds true as well for
The existence of the necessary N is guaranteed and hence
For a convergent sequence {c_{n}} > {0} it is only necessary that for some M all c_{n} > 0 for all n≥M. If for {a_{n}} > {b_{n}} N_{1} is the number such that for all n≥N_{1} a_{n}>b_{n}, then by choosing N=max(N_{1}, M) it is guaranteed that for all n≥N
(To be continued.)
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