By G. R. Jensen

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**Sample text**

A subset A of R is closed if and only if whenever an ∈ A for all n ∈ N, and an → c, we also have c ∈ A. 2. Compact sets 41 Proof. ⇒ Say an ∈ A, n ∈ N, and an → c. If c ∈ / A, then, since R \ A is open, R \ A is a neighbourhood of c, so an ∈ R \ A for all but finitely many n, which is absurd. ⇐ We prove the contrapositive. Suppose A is not closed, that is, R \ A is not open. This means that there is c ∈ R \ A such that R \ A is not a neighbourhood of c. Thus, for each n ∈ N, R \ A does not contain the 1 n -neighbourhood of c, so there is an ∈ A in this neighbourhood, that is, |an − c| < n1 .

B − 1} with ∞ a ∞ c � � n n = . Prove that, after possibly interchanging (an ) and (cn ), n n b b n=1 n=1 there is m ∈ N such that an = cn for n < m, am = cm + 1, and an = 0 and cn = b − 1 for n > m. 34. (a) Let (xn ) be a bounded sequence. For each k ∈ N, let yk = sup xn = sup{xk , xk+1 , xk+2 , . . }. n≥k Show that the sequence (yk ) is decreasing and bounded below. Conclude that (yk ) converges. The limit of (yk ) is called the limit superior of the original sequence (xn ). In other words, lim sup xn = lim sup xn .

Theorem (squeeze theorem). If an → s, cn → s, and an ≤ bn ≤ cn for all but finitely many n ∈ N, then bn → s. Proof. Let � > 0 and I = (s − �, s + �). By assumption, for n suﬃciently large, an ∈ I. Also, for n suﬃciently large, cn ∈ I. Hence, since I is an interval and bn lies between an and cn for n suﬃciently large, we have bn ∈ I for n suﬃciently large. This shows that bn → s. 11. Theorem (algebraic limit theorem). If an → a and bn → b, then: (1) can → ca for all c ∈ R. (2) an + bn → a + b. (3) an bn → ab.