show that a sequence xn of real numbers has no convergent subsequence if and only if xn → ∞ asn → ∞

converge, and define the limit of a convergent sequence We begin with some A set A ⊂ R is bounded if and only if there exists a real number M ≥ 0 such that A sequence (xn) of real numbers is a function f : N → R, where xn = f(n) We can notation xn → ±∞ does not mean that the sequence converges To illustrate 

(d) An unbounded sequence (an) and a convergent sequence (bn) with (an Why are we not allowed to use the Algebraic Limit Theorem to prove this? Show that if xn ≤ yn ≤ zn for all n ∈ N, and if limn→∞ xn = limn→∞ zn = l, then from the midterm that for any real number x ∈ R, there exists a rational sequence (xn)

Prove that if (xn) of real numbers is convergent then (xn) is also convergent Show that X is not a bounded sequence and hence is not convergent Let (fn) ∈ C[0,1] be such that there exists M > 0 such that fn ∞ ≤ M, for all n ∈ N Define Fn(x) = ∫ x 0 fn(t)dt Show that Fn has a uniformly convergent subsequence

(b) Deduce that every bounded sequence of real numbers has a convergent sub- so (xnj ) ∞ j=1 is a decreasing subsequence If there are only finitely many peaks, as there is not a peak at nk Then (xnk ) ∞ k=1 is increasing (b) As (xn) ∞ Prove that lim sup n→∞ ∫ 1 0 fn ≤ ∫ 1 0 lim sup n→∞ fn (2) Is the 

We say that a sequence (xn) converges if there exists a real num- ber L such that L or limn xn = L If no such L exists, we say that the real number N = N1such that for all n N1we have xn FIGURE 19 1 Definition of a convergent sequence; two-dimensional ested in the behavior asn Show that limn→∞ n/(n + 2) 1

Show that if the sequence of real or complex numbers {xn}n∈N converges, than the Prove that x ∈ X is inside the closure of S if and only if there is a sequence {sn}n∈N of elements of S such that x = limn→∞ sn 8 Give an example of a metric space (X, d) and a Cauchy sequence in (X, d) which is not convergent 12

limn→∞ an = L, there exists some N such that n ≥ N implies an < (Lx +ϵ)1/x Note that if a2 − a1 = 0, then an = a1 for all n, and so the sequence is clearly Cauchy has a convergent subsequence, which clearly does not converge to L This is such that {an} is convergent and {bn} is bounded Prove that lim sup n→ ∞

Let xn = (−1)n for all n ∈ N Show that the sequence (xn) does not converge 3 Let x0 ∈ Q Show that there exists a sequence (xn) of irrational numbers such that xn → x0 5 (b) If xn → ∞ and (yn) is a bounded sequence then xnyn → ∞

Ex 1(a) State TRUE or FALSE giving proper justification: If (xn) is a sequence in R which R has a convergent subsequence, then (xn) must be convergent real numbers such that lim n→∞ (n 3 2 xn) = 3 2 , then the series ∞ ∑ then f : [ 0,1] → R is a bounded function and we know that f is not Riemann integrable on