Pure Mathematics

## Introduction to Set Theory (International Series in Pure and by J. Donald Monk

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By J. Donald Monk

This e-book is inteded to be a self-contained advent to the entire set conception wanted by way of so much mathematicians. The method of set thought here's axiomatic. Logical symbolism is used, yet purely the place it's crucial, or the place it sort of feels to explain a state of affairs. Set thought can be in line with formal common sense, yet the following it's according to intuitive good judgment. We kingdom all of the set-theoretical axioms on the begining , in Sec.1. The axiomatic strategy used is that of Kelley and Morse, expounded within the appendix of Kelley 1955.

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Extra resources for Introduction to Set Theory (International Series in Pure and Applied Mathematics)

Example text

There is no 1000-digit number that is equal to the sum of the 1000th powers of its digits). 3, we gave a cunning geometrical √ construction that demonstrated the existence of the real number n for any positive integer n. However, proving the existence of a cube root and, more generally, an nth root of any positive real number x is much harder and requires a deeper analysis of the reals than we have undertaken thus far. We shall carry out such an analysis later, in Chapter 24. 2, and state it here.

Xn ∈ R, and suppose that k of these numbers are negative and the rest are positive. If k is even, then the product x1 x2 . . xn > 0. And if k is odd, x1 x2 . . xn < 0. PROOF Since the order of the xi s does not matter, we may as well assume that x1 , . . , xk are negative and xk+1 , . . , xn are positive. 1, −x1 , . . , −xk , xk+1 , . . , xn are all positive. By (4), the product of all of these is positive, so (−1)k x1 x2 , . . , xn > 0 . If k is even this says that x1 x2 , . . , xn > 0.

1 Let x be a real number. n) (i) If x = 1, then x + x2 + x3 + ∙ ∙ ∙ + xn = x(1−x 1−x . (ii) If −1 < x < 1, then the sum to infinity x + x2 + x3 + ∙ ∙ ∙ = x . 1−x 21 A CONCISE INTRODUCTION TO PURE MATHEMATICS 22 PROOF (i) Let sn = x + x2 + x3 + ∙ ∙ ∙ + xn . Then xsn = x2 + x3 + ∙ ∙ ∙ + xn + xn+1 . Subtracting, we get (1 − x)sn = x − xn+1 , which gives (i). (ii) Since −1 < x < 1, we can make xn as small as we like, provided we take n large enough. So we can make the sum in (i) as close as we like to x 1−x provided we sum enough terms.