78 CHAPTER 4. FUNCTIONS AND SEQUENCES

Proposition 4.10.13 Let limn→∞ an = a > 0 and suppose each bn > 0. Then

lim supn→∞

anbn = a lim supn→∞

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 limsupn→∞ anbn = limn→∞ λ n ≡

lim supn→∞

anbn ≥ 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 is ar-bitrary, the conclusion follows.

4.10.3 Shrinking DiametersIt is useful to consider another version of the nested interval lemma. This involves a se-quence of sets such that set (n+1) is contained in set n and such that their diametersconverge to 0. It turns out that if the sets are also closed, then often there exists a uniquepoint in all of them. This is just a more general version of the nested interval theorem whichholds in the context that the sets are not necessarily intervals.

Definition 4.10.14 Let S be a nonempty set. Then diam(S) is defined as

diam(S)≡ sup{|x− y| : x,y ∈ S} .

This is called the diameter of S.

Theorem 4.10.15 Let {Fn}∞

n=1 be a sequence of closed sets in Fp such that

limn→∞

diam(Fn) = 0

and Fn ⊇ Fn+1 for each n. Then there exists a unique p ∈ ∩∞k=1Fk.

Proof: Pick pk ∈ Fk. This is always possible because by assumption each set is non-empty. Then {pk}∞

k=m ⊆ Fm and since the diameters converge to 0 it follows {pk} is aCauchy sequence. Therefore, it converges to a point, p by completeness of Fp. Since eachFk is closed, it must be that p ∈ Fk for all k. Therefore, p ∈ ∩∞

k=1Fk. If q ∈ ∩∞k=1Fk, then

since both p,q ∈ Fk, |p−q| ≤ diam(Fk). It follows since these diameters converge to 0,|p−q| ≤ ε for every ε. Hence p = q.

A sequence of sets, {Gn}which satisfies Gn ⊇Gn+1 for all n is called a nested sequenceof sets.

The next theorem is a major result called Bair’s theorem. In fact, you just need thecontext of a complete metric space but we are emphasizing Fp here.

Definition 4.10.16 An open set U ⊆ Fp is dense if for every x ∈ Fp and r >0,B(x,r)∩U ̸= /0.