 By Jiri Lebl

A primary direction in mathematical research. Covers the true quantity method, sequences and sequence, non-stop services, the by-product, the Riemann imperative, sequences of services, and metric areas. initially constructed to coach Math 444 at collage of Illinois at Urbana-Champaign and later superior for Math 521 at collage of Wisconsin-Madison. See http://www.jirka.org/ra/

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Additional resources for Basic analysis: Introduction to real analysis

Example text

In fact, the cardinality of R is the same as the cardinality of P(N), although we will not prove this claim. 1 (Cantor). R is uncountable. 36 CHAPTER 1. REAL NUMBERS We give a modified version of Cantor’s original proof from 1874 as this proof requires the least setup. Normally this proof is stated as a contradiction proof, but a proof by contrapositive is always easier to understand. Proof. Let X ⊂ R be a countable subset such that for any two numbers a < b, there is an x ∈ X such that a < x < b.

That means that there exists a B such that xn ≤ B for all n, that is the set {xn | n ∈ N} is bounded. Let x := sup{xn | n ∈ N}. Let ε > 0 be arbitrary. As x is the supremum, then there must be at least one n0 ∈ N such that xn0 > x − ε (because x is the supremum). As {xn } is monotone increasing, then it is easy to see (by induction) that xn ≥ xn0 for all n ≥ n0 . Hence |xn − x| = x − xn ≤ x − xn0 < ε. 1. SEQUENCES AND LIMITS 43 Hence the sequence converges to x. We already know that a convergent sequence is bounded, which completes the other direction of the implication.

Let x := lim xn and y := lim yn . Let z := xy. Let ε > 0 be given. As {xn } is convergent, it is bounded. Therefore, find a B > 0 such that ε |xn | ≤ B for all n ∈ N. Find an M1 such that for all n ≥ M1 we have |xn − x| < 2|y| . Find an M2 such ε . Take M := max{M1 , M2 }. For all n ≥ M we have that for all n ≥ M2 we have |yn − y| < 2B |zn − z| = |(xn yn ) − (xy)| = |xn yn − (x + xn − xn )y| = |xn (yn − y) + (xn − x)y| ≤ |xn (yn − y)| + |(xn − x)y| = |xn | |yn − y| + |xn − x| |y| ≤ B |yn − y| + |xn − x| |y| ε ε |y| = ε.