# So x s x n and x m but this contradicts the fact m

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So x S ( x n ) and x > M . But this contradicts the fact M = sup S ( x n ). This proves (i). Next suppose (ii) is false. Then there exists ε > 0 such that there are only finitely many n ’s such that x n > M - ε . Equivalently, this means that there exists K N such that x n M - ε for all n K . Then it follows readily that all subsequential limits are M - ε . So M - ε is an upper bound for S ( x n ). But this again contradicts the fact that M = sup S ( x n ). This proves (ii). . . . . 1234 n x n . . . . . . . . . . . . . . . . . . . . . . . . M M- ε M + ε . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .................................................................................................................... . . . .. . . . Exercise 3.5.1: Prove that the converse of Theorem 3.5.1 is also true, that is, if M is a real number satisfying conditions (i) and (ii), then M = lim sup x n . 86
Theorem 3.5.2. Let ( x n ) be a bounded sequence and let m = lim inf x n . (i) For each ε > 0, there are at most only finitely many n ’s such that x n m - ε . Equivalently, there exists K N such that x n > m - ε for all n K . (ii) For each ε > 0, there are infinitely many n ’s such that x n < m + ε . Proof. Similar to Theorem 3.5.1. (Exercise.) . . . . 1234 n x n . . . . . . . . . . . . . . . . . . . . . . . . M M- ε M + ε m m- ε m + ε . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................................................................................................................................. ................................................................................................................................................. . . . ... . . . . .. . . . . . . .. Remark. Let ( x n ) be a bounded sequence, and let M = lim sup x n , m = lim inf x n . Roughly speaking, Theorem 3.5.1 and Theorem 3.5.2 means that for all su ffi ciently large n , x n will oscillate within any interval slighly larger than [ m , M ], but it will not oscillate within any interval slightly smaller than [ m , M ]. 87
Theorem 3.5.3. Let ( x n ) be a bounded sequence. Then ( x n ) converges if and only if lim sup x n = lim inf x n . [In short, lim n →∞ x n = x (exists) ⇐⇒ lim sup x n = lim inf x n = x .] Proof. ( = ): If x n x , then by Theorem 3.4.1, every subsequence of ( x n ) also converges to x . Consequently S ( x n ) = { x } and lim sup x n = lim inf x n = x . ( = ): Let M = lim sup x n = lim inf x n and ε > 0. Then by Theorem 3.5.1 and Theorem 3.5.2, there exist K 1 , K 2 N such that x n < M + ε for all n K 1 , and (1) x n > M - ε for all n K 2 . (2) Let K = max( K 1 , K 2 ) N . Then it follows from (1) and (2) that for all n K , one has M - ε < x n < M + ε = ⇒ | x n - M | < ε. Hence we have lim n →∞ x n = M . Theorem 3.5.4. Let ( x n ) and ( y n ) be bounded sequences such that x n y n for every n N . Then lim sup x n lim sup y n and lim inf x n lim inf y n . Proof. Let x be a subsequential limit of ( x n ) and x n k x . Consider the cor- responding subsequence ( y n k ) of ( y n ). Since it is bounded, it has a convergent subsequence ( y n k ). Then since x n k y n k for all , x = lim k →∞ x n k = lim →∞ x n k lim →∞ y n k lim sup y n . This shows that lim sup y n is an upper bound for S ( x n ). It follows that lim sup x n lim sup y n . The proof of the second inequality is left as an exercise. 88
Alternative definition of lim sup and lim inf.