1.11. SOME TOPICS FROM ANALYSIS 31

From the definition of sup{ak : k ≥ N} , there exists n1 ≥ N such that

sup{ak : k ≥ N} ≤ an1 + ε/3.

Similarly, there exists n2 ≥ N such that

inf{ak : k ≥ N} ≥ an2 − ε/3.

It follows that sup{ak : k ≥ N} − inf{ak : k ≥ N} ≤ |an1 −an2 |+2ε

3 < ε. Since the se-quence, {sup{ak : k ≥ N}}∞

N=1 is decreasing and {inf{ak : k ≥ N}}∞

N=1 is increasing, itfollows that

0≤ limN→∞

sup{ak : k ≥ N}− limN→∞

inf{ak : k ≥ N} ≤ ε

Since ε is arbitrary, this shows

limN→∞

sup{ak : k ≥ N}= limN→∞

inf{ak : k ≥ N} (1.11)

Next suppose 1.11 and both equal a ∈ R. Then

limN→∞

(sup{ak : k ≥ N}− inf{ak : k ≥ N}) = 0

Since sup{ak : k ≥ N} ≥ inf{ak : k ≥ N} it follows that for every ε > 0, there exists Nsuch that sup{ak : k ≥ N}− inf{ak : k ≥ N} < ε, and for every N, inf{ak : k ≥ N} ≤ a ≤sup{ak : k ≥ N}. Thus if n≥ N, |a−an|< ε which implies that limn→∞ an = a. In case

a = ∞ = limN→∞

sup{ak : k ≥ N}= limN→∞

inf{ak : k ≥ N}

then if r ∈ R is given, there exists N such that inf{ak : k ≥ N} > r which is to say thatlimn→∞ an = ∞. The case where a =−∞ is similar except you use sup{ak : k ≥ N}. ■

The significance of limsup and liminf, in addition to what was just discussed, is con-tained in the following theorem which follows quickly from the definition.

Theorem 1.11.10 Suppose {an} is a sequence of points of [−∞,∞] . Define λ byλ = limsupn→∞ an. Then if b> λ , it follows there exists N such that whenever n≥N,an≤ b.If c < λ , then an > c for infinitely many values of n. Let γ = liminfn→∞ an. Then if d < γ,it follows there exists N such that whenever n ≥ N,an ≥ d. If e > γ, it follows an < e forinfinitely many values of n.

The proof of this theorem is left as an exercise for you. It follows directly from the defi-nition and it is the sort of thing you must do yourself. Here is one other simple proposition.

Proposition 1.11.11 Let limn→∞ an = a > 0. Then limsupn→∞ anbn = a limsupn→∞ bn.

Proof: This follows from the definition. Let λ n = sup{akbk : k ≥ n} . For all n largeenough, an > a− ε where ε is small enough that a− ε > 0. Therefore,

λ n ≥ sup{bk : k ≥ n}(a− ε)

for all n large enough. Then

lim supn→∞

anbn = limn→∞

λ n ≥ limn→∞

(sup{bk : k ≥ n}(a− ε)) = (a− ε) lim supn→∞

bn

Similar reasoning shows limsupn→∞ anbn ≤ (a+ ε) limsupn→∞ bn. Now since ε > 0 isarbitrary, the conclusion follows. ■

A fundamental existence theorem is the nested interval lemma.