NSA: proof of Sierpinski's theorem

This writeup uses the notation of Sierpinski's Theorem, and anyway is a proof of what's there, so why don't you read that first?

Suppose, to the contrary, that for some a₁,...,a_k,b there exist infinitely many solutions in natural numbers to the equation (*). Let F be the set of all solutions. By Robinson's overspill lemma, F^{^} (see NSA: some pseudo set theory for an explanation of "content") contains non-standard solutions of the equation. These are solutions where at least one x_j is non-standard: call N the set of natural numbers; the pseudo-elements N^{^}-N are the "non-standard natural numbers".

Let's examine one such non-standard solution. It claims that a₁/x₁+...+a_k/x_k=b, where some of the x's are non-standard natural numbers.

Claim 1:

If x is a non-standard natural number, then x>n for any (standard) natural number n.

Proof of claim 1:

There are only finitely many (standard) natural numbers {t: t<=n}; by Robinson's overspill lemma, there are no such non-standard ones. Since ">" (on the real numbers, say) is a total order, a property we can define in first order predicate calculus, "*->" (the transferred predicate on the pseudo-real numbers) is a total order too.

It directly follows that for any (positive) real number a, the *-number a/x is positive and smaller than any (standard) real number ε>0. Call such a number a positive infinitesimal.

Claim 2:

The sum of j (>0) positive infinitesimals is a positive infinitesimal.

Proof of claim 2:

In the standard world, it's true that for all x₁,...,x_j and ε/j, if 0<x₁,...,x_j<ε/j then 0<x₁+...+x_j<ε. Transferring to the non-standard world, taking ε to be standard real numbers, shows the claim.

Proof of the theorem:

We've seen that a non-standard solution contains some non-standard elements. Suppose without loss of generality that x₁,...,x_j are non-standard and x_j+1,...,x_k are standard (0<j<=k). Then

a₁/x₁+...+a_j/x_j = b - (a_j+1/x_j+1+...+a_k/x_k).

Now the LHS is a positive infinitesimal, but the RHS is some expression composed entirely of standard real numbers! Thus the RHS is some standard real number ε. It cannot equal a positive infinitesimal: it's positive, so it cannot be less than ε/2. But ε/2 is a standard real number too, thus larger than any positive infinitesimal.

This contradiction shows that the premise (having an infinite number of solutions) is absurd!