# Basic analysis: Introduction to real analysis by Jiri Lebl By Jiri Lebl

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

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Basic analysis: Introduction to real analysis

A primary direction in mathematical research. Covers the true quantity approach, sequences and sequence, non-stop capabilities, the by-product, the Riemann indispensable, sequences of services, and metric areas. initially constructed to educate Math 444 at college of Illinois at Urbana-Champaign and later greater for Math 521 at college of Wisconsin-Madison.

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Example text

2xn 2xn 2xn If xn > 0, then xn2 + 2 > 0 and hence xn+1 > 0. Next let us show that the sequence is monotone decreasing. If we can show that xn2 − 2 ≥ 0 for all n, then xn+1 ≤ xn for all n. Obviously x12 − 2 = 2 > 0. For an arbitrary n we have that 2 −2 = xn+1 xn2 + 2 2xn 2 xn2 − 2 x4 + 4xn2 + 4 − 8xn2 xn4 − 4xn2 + 4 −2 = n = = 4xn2 4xn2 4xn2 2 2 −2 ≥ 0 for all n. Therefore, Since xn > 0 and any number squared is nonnegative, we have that xn+1 {xn } is monotone decreasing and bounded, and therefore the limit exists.

Let x := lim an = lim bn . Let ε > 0 be given. Find an M1 such that for all n ≥ M1 we have that |an − x| < ε/3, and an M2 such that for all n ≥ M2 we have |bn − x| < ε/3. Set M := max{M1 , M2 }. Suppose that n ≥ M. We compute |xn − an | = xn − an ≤ bn − an = |bn − x + x − an | ≤ |bn − x| + |x − an | ε ε 2ε < + = . 3 3 3 48 CHAPTER 2. SEQUENCES AND SERIES Armed with this information we estimate |xn − x| = |xn − x + an − an | ≤ |xn − an | + |an − x| 2ε ε < + = ε. 3 3 And we are done. 2: A very simple example of how to use the squeeze lemma is to compute certain sequences using already known ones.

That is, we want to prove that lim sup xnk ≤ lim sup xn . Define a j := sup{xk | k ≥ j} as usual. Also define c j := sup{xnk | k ≥ j}. It is not true that c j is necessarily a subsequence of a j . However, as nk ≥ k for all k, we have that {xnk | k ≥ j} ⊂ {xk | k ≥ j}. A supremum of a subset is less than or equal to the supremum of the set and therefore c j ≤ a j. 3 to conclude that lim c j ≤ lim a j , j→∞ which is the desired conclusion. 3. LIMIT SUPERIOR, LIMIT INFERIOR, AND BOLZANO-WEIERSTRASS 61 Limit superior and limit inferior are in fact the largest and smallest subsequential limits.