By Jiri Lebl

A primary path in mathematical research. Covers the genuine quantity procedure, sequences and sequence, non-stop capabilities, the spinoff, the Riemann crucial, sequences of features, and metric areas. initially built to coach Math 444 at collage of Illinois at Urbana-Champaign and later improved for Math 521 at college of Wisconsin-Madison. See http://www.jirka.org/ra/

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Xy is also positive and hence xy = |xy|. If x and y are both negative then xy is still positive and xy = |xy|, and |x| |y| = (−x)(−y) = xy. Next assume that x > 0 and y < 0. Then |x| |y| = x(−y) = −(xy). Now xy is negative and hence |xy| = −(xy). Similarly if x < 0 and y > 0. (iv): Obvious if x ≥ 0. If x < 0, then |x|2 = (−x)2 = x2 . (v): Suppose that |x| ≤ y. If x ≥ 0, then x ≤ y. Obviously y ≥ 0 and hence −y ≤ 0 ≤ x so −y ≤ x ≤ y holds. If x < 0, then |x| ≤ y means −x ≤ y. Negating both sides we get x ≥ −y.

Set nk+1 := m. The subsequence {xnk } is defined. Next we need to prove that it converges and has the right limit. ) and that ank ≥ xnk . Therefore, for every k > 1 we have |ank − xnk | = ank − xnk ≤ a(nk−1 +1) − xnk 1 < . k Let us show that {xnk } converges x. Note that the subsequence need not be monotone. Let ε > 0 be given. As {an } converges to x, then the subsequence {ank } converges to x. Thus there exists an M1 ∈ N such that for all k ≥ M1 we have ε |ank − x| < . 2 Find an M2 ∈ N such that 1 ε ≤ .

4 only applies to the real line, but Bolzano-Weierstrass applies in more general contexts (that is, in Rn ) with pretty much the exact same statement. As the theorem is so important to analysis, we present an explicit proof. The following proof generalizes more easily to different contexts. Alternate proof of Bolzano-Weierstrass. As the sequence is bounded, then there exist two numbers a1 < b1 such that a1 ≤ xn ≤ b1 for all n ∈ N. We will define a subsequence {xni } and two sequences {ai } and {bi }, such that {ai } is monotone increasing, {bi } is monotone decreasing, ai ≤ xni ≤ bi and such that lim ai = lim bi .