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By Vinogradov S.S., et al.

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**Extra info for Canonical problems in scattering and potential theory, part 2**

**Example text**

To show that Q is a field. It is plain from the definitions that NeZ c Q c lR. All of these subsets inherit the order structure of lR. 20 A Concise Approach to Mathematical Analysis Completeness axiom It looks as if we have found in Q a "satisfactory" system of numbers which suits all the required properties of a field and respects the order structure. We say that Q is an ordered field. However, we will see that Q is not perfect in a certain sense. First we will prove that one cannot find any rational number a such that a 2 = 2.

This shows that the range of any sequence in [0, 1] cannot cover the whole of the interval [0, 1] and proves that [0, 1] is uncountable. 5 Show that every infinite set contains an infinite countable subset. Solution It suffices to show that an infinite set A contains (the range of) an infinite sequence. Since A is not empty, there exists an element al E A. Since A is infinite the set Al = A", {xd is nonempty. nite A2 = A", {Xl, X2} is nonempty. There exists a3 E A 2 . Proceeding in this way, we obtain the infinite subset {Xl, X2, X3, ••• } of A, which by construction is the range of the sequence (xn).

Thus M could not be the supremum of A. This contradiction proves our claim. D Thus if M = supA exists as a rational number, then we must have M2 = 2. 38, such a number could not exist. The set A has upper bounds but no supremum in Q. If we only consider the system of rational numbers, then we come to the conclusion that there is a gap between the set A and its upper bounds B. We say that the ordered field Q is not cOInplete. This observation leads to a fundamental property, known as the cOInpleteness axiom, which distinguishes IR from Q.