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By Sloughter D.

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1. Show that the open cover of (0, 1) given in the previous example does not have a finite subcover. 3. We say a set K ⊂ R is compact if every open cover of K has a finite subcover. 2. As a consequence of the previous exercise, the open interval (0, 1) is not compact. 2. Show that every finite subset of R is compact. 3. Suppose n ∈ Z+ and K1 , K2 , . . , Kn are compact sets. Show n n that i=1 Ki and i=1 Ki are compact. 1. If I is a closed, bounded interval, then I is compact. Proof. Let a ≤ b be finite real numbers and I = [a, b].

Then the sequence {αxi }i∈I converges and i→∞ lim αxi = αL. 33) i→∞ Proof. If α = 0, then {αxi }i∈I clearly converges to 0. So assume α = 0. 34) |α| whenever i > N . Then for any i > N we have |αxi − αL| = |α||xi − L| < |α| Thus lim αxi = αL. i→∞ |α| = . D. 1. 6. Suppose {xi }i∈I and {yi }i∈I are convergent sequences in R with L = lim xi and M = lim yi . Then the sequence {xi + yi }i∈I converges i→∞ i→∞ and lim (xi + yi ) = L + M. 14. Prove the previous proposition. 7. Suppose {xi }i∈I and {yi }i∈I are convergent sequences in R with L = lim xi and M = lim yi .

5) n xi , then, by the previous exercise, Proof. If sn = i=0 sn = Hence ∞ 1 − xn+1 . 6) 1 − xn+1 1 = . D. 1 Comparison tests The following two propositions are together referred to as the comparison test. 34 CHAPTER 2. 3. Suppose ∞ ai and i=m bi are infinite series for which there i=k ∞ exists an integer N such that 0 ≤ ai ≤ bi whenever i ≥ N . If ∞ then bi converges, i=k ai converges. i=m ∞ Proof. 3 We need only show that ∞ the nth partial sum of ai converges. Let sn be i=N ∞ ai and let tn be the nth partial sum of i=N bi .

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A primer of real analysis by Sloughter D.

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