76 CHAPTER 4. FUNCTIONS AND SEQUENCES

because the distance between the partial sum up to n and x is always no more than 1/10n.In case x = 1, just let each an = 9 and observe that the sum of the geometric series equals1.

An amusing application of the above is in the following theorem. It gives an easy wayto verify that the unit interval is uncountable.

Theorem 4.10.6 The interval [0,1) is not countable.

Proof: Suppose it were. Then there would exist a list of all the numbers in this interval.Writing these as decimals,

x1 ≡ .a11a12a13a14a15 · · ·x2 ≡ .a21a22a23a14a25 · · ·x3 ≡ .a31a32a33a34a35 · · ·

...

Consider the diagonal decimal, .a11a22a33a44 · · · . Now define a decimal expansion for an-other number in [0,1) as follows. y ≡ .b1b2b3b4 · · · where |bk−akk| ≥ 4. Then |y− xk| ≥

410k .Thus y is not equal to any of the xk which is a contradiction since y ∈ [0,1).

4.10.2 lim sup and lim inf

Sometimes the limit of a sequence does not exist. For example, if an = (−1)n , thenlimn→∞ an does not exist. This is because the terms of the sequence are a distance of 1apart. Therefore there can’t exist a single number such that all the terms of the sequenceare ultimately within 1/4 of that number. The nice thing about limsup and liminf is thatthey always exist. First here is a simple lemma and definition. First review the definition ofinf and sup on Page 27 along with the simple properties of these things. Also limn→∞ an =∞

means that if l ∈ R is given, then for large enough n,an > l. A similar definition holds forlimn→∞ an =−∞.

Definition 4.10.7 Denote by [−∞,∞] the real line along with symbols ∞ and −∞.It is understood that ∞ is larger than every real number and −∞ is smaller than every realnumber. Then if {An} is an increasing sequence of points of [−∞,∞] , limn→∞ An equals ∞ ifthe only upper bound of the set {An} is ∞. If {An} is bounded above by a real number, thenlimn→∞ An is defined in the usual way and equals the least upper bound of {An}. If {An} isa decreasing sequence of points of [−∞,∞] , limn→∞ An equals −∞ if the only lower boundof the sequence {An} is −∞. If {An} is bounded below by a real number, then limn→∞ An isdefined in the usual way and equals the greatest lower bound of {An}. More simply, if {An}is increasing, limn→∞ An ≡ sup{An}and if {An} is decreasing then limn→∞ An ≡ inf{An} .

Lemma 4.10.8 Let {an} be a sequence of real numbers and let Un ≡ sup{ak : k ≥ n} .Then {Un} is a decreasing sequence. Also if Ln ≡ inf{ak : k ≥ n} , then {Ln} is an increas-ing sequence. Therefore, limn→∞ Ln and limn→∞ Un both exist.

Proof: From the definition, if m ≤ n, Lm ≡ inf{ak : k ≥ m} ≤ inf{ak : k ≥ n} ≡ Ln.Thus the Ln are increasing. If you take inf of a smaller set, it will be as large as inf of thelarger set. Similarly the Un are decreasing. Thus their limits exist as in the above definition.

From the lemma, the following definition makes sense.