Suppose that

(1.4)             x₁,x₂,...,x_n,...

is a sequence of numbers. Then the sequences x₁,x₃,x₅,... and x₂,x₅,x₈,x₁₁,... are examples of subsequences of (1.4). More generally, suppose that k₁,k₂,k₃,...,k_n,... is an increasing sequence of positive integers. Then we say that    

x_k1 ,x_k2 ,...,x_kn ,...

is a subsequence of (1.4). The choice k₁ = 1,k₂ = 3,k₃ = 5,k₄ = 7,... is an example of a subsequence of (1.4) above. To avoid double subscripts, which are cumbersome, we will frequently write y₁ = x_k1 ,y₂ =x_k2 ,...,y_n = x_kn ,..., in which case

y₁,y₂,...,y_n,...

is a subsequence of (1.4).

We easily prove by induction that if k₁,k₂,k₃,...,k_n,... is an increasing sequence of positive integers, then k_n ≥ n for all n .The sequence(1.5)

has the subsequences

which are obtained from (1.5) by taking the odd-numbered terms and the even-numbered terms, respectively. Each of these subsequences is convergent, but the original sequence (1.5) is not. The notion of a convergent subsequence of a given sequence occurs frequently in problems in analysis. The Bolzano–Weierstrass theorem is basic in that it establishes the existence of such convergent subsequences under very simple hypotheses on the given sequence.

Theorem 1.1 (Bolzano–Weierstrass Theorem)

Any bounded infinite sequence of real numbers contains a convergent subsequence.

Proof

We shall use the Nested intervals theorem (Theorem 3.1). Let {x_n} be a given bounded sequence. Then it is contained in some closed interval I ={x : a ≤ x ≤ b}. Divide I into two equal subintervals by the midpoint (a + b)/2. Then either the left subinterval contains an infinite number of the {x_n} or the right subinterval does (or both). Denote by I1 ={x : a₁ ≤ x ≤ b₁} the closed subinterval of I that contains infinitely many {x_n}. (If both subintervals do, choose either one.) Next, divide I1 into two equal parts by it smidpoint. Either the right subinterval or the left subinterval of I₁ contains infinitely many {x_n}. Denote by I₂ the closed subinterval that does. Continue this process, obtaining the sequence

with the property that each In contains xp for infinitely many values of p. Since b_n − a_n =(b − a)/2_n → 0 as n →∞, we may apply the Nested intervals theorem to obtain a unique number x0 contained in every I_n.

We now construct a subsequence of {x_p} converging to x₀. Choose x_k1 to be any member of {x_p} in I₁ and denote x_k1 by y₁. Next choose x_k2 to be any member of {x_p} such that x_k2 is in I₂ and such that k₂ >k₁.We can do this because I₂ has infinitely many of the {x_p}. Set x_k2 =y₂. Next, choose x_k3 as any member of {x_p} in I3 and such that k₃ >k₂. We can do this because I₃ also has infinitely may of the {x_p}. Set x_k3 = y₃.We continue, and by induction obtain the subsequence y₁,y₂,...,y_n,....By the method of selection we have

Since an → x0, b_n → x0,as n →∞, we can apply the Sandwiching theorem to conclude that y_n → x0 as n →∞.

Problems

In Problems1 through 7 decide whether or not the given sequence converges to a limit. If it does not, find, in each case, at least one convergent subsequence. We suppose n = 1, 2, 3,...

Basic Elements of Real Analysis, Murray H. Protter, Springer, 1998 .Page(53 -55)