Calculus of One and Several Variables

3.3 The Nested Interval Lemma

In Russia there is a kind of doll called a matrushka doll. You pick it up and notice it comes apart in the center. Separating the two halves you find an identical doll inside. Then you notice this inside doll also comes apart in the center. Separating the two halves, you find yet another identical doll inside. This goes on quite a while until the final doll is in one piece. The nested interval lemma is like a matrushka doll except the process never stops. It involves a sequence of intervals, the first containing the second, the second containing the third, the third containing the fourth and so on. The fundamental question is whether there exists a point in all the intervals. Sometimes there is such a point and this comes from completeness.

Lemma 3.3.1 Let I_k =

and suppose that for all k = 1,2,

Ik \supseteq Ik+1.

Then there exists a point, c ∈ ℝ which is an element of every I_k. If the diameters (length) of these intervals, denoted as diam

converges to 0, then there is a unique point in the intersection of all these intervals.

Proof:Since I_k ⊇ I_k+1, this implies

k k+1 k k+1 a \leq a ,b \geq b . (3.4)

(3.4)

Consequently, if k ≤ l,

l l k a \leq b \leq b . (3.5)

(3.5)

Now define

c \equiv sup{al : l = 1,2,\cdot\cdot\cdot}

By the first inequality in 3.4, and 3.5

{ } ak \leq c = sup al : l = k,k+ 1,\cdot\cdot\cdot \leq bk (3.6)

(3.6)

for each k = 1,2

. Thus c ∈ I_k for every k and this proves the lemma. The reason for the last inequality in 3.6 is that from 3.5, b^k is an upper bound to

. Therefore, it is at least as large as the least upper bound.

For the last claim, suppose there are two points x,y in the intersection. Then

= r > 0 but eventually the diameter of I_k is less than r. Thus it cannot contain both x,y. If so, assuming y > x, you would have a^k ≤ x < y ≤ b^k and so 0 < r = y − x < b^k − a^k < r. ■

This is really quite a remarkable result and may not seem so obvious. Consider the intervals I_k ≡

. Then there is no point which lies in all these intervals because no negative number can be in all the intervals and 1∕k is smaller than a given positive number whenever k is large enough. Thus the only candidate for being in all the intervals is 0 and 0 has been left out of them all. The problem here is that the endpoints of the intervals were not included, contrary to the hypotheses of the above lemma in which all the intervals included the endpoints.