By Hrbacek K., Lessmann O., O'Donovan R.

ISBN-10: 149870266X

ISBN-13: 9781498702669

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Pk : If p1 , . . , pk are observable and there exists some object p for which the statement is true, then there exists some observable object p for which the statement is true. 6 Analysis with Ultrasmall Numbers Sets and Induction The last principle we need deals with the way one usually defines sets and functions. If P(x) describes some property of x and A is a given set, then there exists a unique set X such that, for all x, x ∈ X if and only if x ∈ A and P(x); we denote it {x ∈ A : P(x)}.

Proof: Assume S = {n ∈ N : n is observable relative to p} is a set. Then (a) 0 ∈ S, because 0 ∈ N and 0 is observable relative to p. (b) If n ∈ S, then n ∈ N and n is observable relative to p, so n + 1 ∈ N and n + 1 is observable relative to p (the latter follows from the Closure Principle); hence n + 1 ∈ S. By the Principle of Mathematical Induction applied to the statement (of traditional mathematics) “n ∈ S” we conclude that N = S, that is, all n ∈ N are observable relative to p. This is a contradiction with Theorem 2.

Proof. Only item (3) requires some argument. Since x is neither ultrasmall nor 0, there is an observable r0 > 0 such that |x| ≥ r0 . We know 1 that h1 is ultralarge, hence, for any observable r > 0, |h| > rr0 and |x| |h| > r0 · r r0 = r. 12 Analysis with Ultrasmall Numbers Rule 3. Let a, b be real numbers. If a then a = b. b and a and b are observable, Proof. The assumptions imply that a − b 0 (by Definition 2) and a − b is observable (by Closure). Therefore a − b is not ultrasmall (by Theorem 1); hence a − b = 0 and a = b.

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Analysis with ultrasmall numbers by Hrbacek K., Lessmann O., O'Donovan R.


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