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 4.5.1Let I_{k} =

[ak,bk]

and supposethat for all k = 1,2,

⋅⋅⋅

,

Ik ⊇ Ik+1.

Then there exists a point, c ∈ ℝ which is an element of every I_{k}. If the diameters (length) ofthese intervals, denoted asdiam

(Ik)

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

. Thus c ∈ I_{k} for every k and this proves the lemma. The reason for the
last inequality in 4.7 is that from 4.6, b^{k} is an upper bound to

{ }
al : l = k,k + 1,⋅⋅⋅

.
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

|x − y|

= r > 0 but eventually the diameter of I_{k} is less than r. Thus it cannot contain both
x,y. ■

This is really quite a remarkable result and may not seem so obvious. Consider the
intervals I_{k}≡

(0,1∕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.

What if the lengths of the intervals converge to 0? In this case, there is a unique point in
the intersection of the closed intervals.