The problems below are designed to reinforce your understanding of the nested intervals theorem and its applications. In particular, we will learn how to check when a sequence of intervals satisfies the hypotheses of the theorem, how to compute its intersection, and how to appreciate the role played by the completeness of the real numbers.
The problems are arranged in order of increasing difficulty. In the first few we apply the theorem directly; later on we examine counterexamples, more elaborate constructions, and some of its most important theoretical consequences.
We will use both the general form of the theorem — according to which the intersection of a sequence of closed, bounded, nonempty, nested intervals is itself nonempty — and the stronger form: if the lengths of the intervals tend to zero, then the intersection consists of a single point.
Problem 1 — level ★☆☆☆☆
Consider the sequence of intervals
\[ I_n=\left[0,\frac{1}{n}\right]. \]
Find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
We have
\[ \bigcap_{n=1}^{+\infty} I_n=\{0\}. \]
Solution
The intervals are closed and bounded. Moreover,
\[ I_{n+1} = \left[0,\frac{1}{n+1}\right] \subseteq \left[0,\frac{1}{n}\right] = I_n, \]
since
\[ \frac{1}{n+1}<\frac{1}{n}. \]
Hence the intervals form a nested sequence.
Furthermore, the length of \(I_n\) is
\[ \frac{1}{n}-0=\frac{1}{n}, \]
and
\[ \frac{1}{n}\longrightarrow 0. \]
By the stronger form of the nested intervals theorem, when the lengths tend to zero the intersection collapses to a single point.
Observe that \(0\) belongs to every interval \(I_n\).
On the other hand, if \(x>0\), then choosing \(n\) large enough we obtain
\[ \frac{1}{n}<x. \]
It follows that \(x\notin I_n\), and therefore \(x\) cannot belong to the intersection of all the intervals.
The only point common to every interval is thus \(0\).
Therefore
\[ \bigcap_{n=1}^{+\infty} \left[0,\frac{1}{n}\right] = \{0\}. \]
Problem 2 — level ★☆☆☆☆
Consider the sequence of intervals
\[ I_n= \left[ -\frac{1}{n}, \frac{1}{n} \right]. \]
Find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
We have
\[ \bigcap_{n=1}^{+\infty} I_n=\{0\}. \]
Solution
The intervals are closed, bounded, and nonempty.
Moreover,
\[ \left[ -\frac{1}{n+1}, \frac{1}{n+1} \right] \subseteq \left[ -\frac{1}{n}, \frac{1}{n} \right], \]
for every \(n\in\mathbb N\).
This is therefore a sequence of nested intervals.
The length of \(I_n\) equals
\[ \frac{1}{n} - \left(-\frac{1}{n}\right) = \frac{2}{n}, \]
and
\[ \frac{2}{n}\longrightarrow 0. \]
The nested intervals theorem then guarantees that the intersection contains exactly one point.
Since
\[ -\frac{1}{n} \leq 0 \leq \frac{1}{n} \qquad \forall n, \]
the number \(0\) belongs to every interval.
If, on the other hand, \(x\neq0\), then \(|x|>0\). Choosing \(n\) large enough we have
\[ \frac{1}{n}<|x|. \]
From this it follows that \(x\notin I_n\).
Therefore
\[ \bigcap_{n=1}^{+\infty} \left[ -\frac{1}{n}, \frac{1}{n} \right] = \{0\}. \]
Problem 3 — level ★☆☆☆☆
Consider the sequence of intervals
\[ I_n=\left[1,1+\frac{1}{n}\right]. \]
Find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
We have
\[ \bigcap_{n=1}^{+\infty} I_n=\{1\}. \]
Solution
The intervals \(I_n\) are closed, bounded, and nonempty. Moreover, as \(n\) grows the right endpoint \(1+\displaystyle \frac{1}{n}\) decreases, while the left endpoint stays equal to \(1\). Hence the intervals are nested.
Indeed, for every \(n\in\mathbb N\),
\[ I_{n+1}\subseteq I_n. \]
The length of \(I_n\) is
\[ \left(1+\frac{1}{n}\right)-1=\frac{1}{n}. \]
Since \(\frac{1}{n}\to0\), the nested intervals theorem guarantees that the intersection contains a single point.
The point \(1\) belongs to every interval, since it is always the left endpoint of \(I_n\). Thus
\[ 1\in\bigcap_{n=1}^{+\infty} I_n. \]
As the intersection contains a single point and that point is \(1\), we conclude that
\[ \bigcap_{n=1}^{+\infty}\left[1,1+\frac{1}{n}\right]=\{1\}. \]
Problem 4 — level ★☆☆☆☆
Consider the sequence of intervals
\[ I_n=\left[2-\frac{1}{n},2+\frac{1}{n}\right]. \]
Find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
We have
\[ \bigcap_{n=1}^{+\infty} I_n=\{2\}. \]
Solution
The intervals are closed, bounded, and nonempty. Moreover, as \(n\) grows the left endpoint
\[ 2-\frac{1}{n} \]
increases toward \(2\), while the right endpoint
\[ 2+\frac{1}{n} \]
decreases toward \(2\). Hence the intervals are nested.
The length of \(I_n\) is
\[ \left(2+\frac{1}{n}\right)-\left(2-\frac{1}{n}\right)=\frac{2}{n}, \]
and therefore
\[ \frac{2}{n}\longrightarrow0. \]
By the nested intervals theorem, the intersection contains exactly one point.
Since
\[ 2-\frac{1}{n}\leq 2\leq 2+\frac{1}{n} \]
for every \(n\), the point \(2\) belongs to every interval.
Therefore
\[ \bigcap_{n=1}^{+\infty} \left[2-\frac{1}{n},2+\frac{1}{n}\right] = \{2\}. \]
Problem 5 — level ★☆☆☆☆
Consider the sequence of intervals
\[ I_n=\left[0,2+\frac{1}{n}\right]. \]
Find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
We have
\[ \bigcap_{n=1}^{+\infty} I_n=[0,2]. \]
Solution
The intervals are closed, bounded, and nonempty. Moreover, since
\[ 2+\frac{1}{n+1}<2+\frac{1}{n}, \]
we have
\[ I_{n+1}\subseteq I_n. \]
This is therefore a sequence of nested intervals.
In this case, however, the lengths of the intervals do not tend to zero. Indeed,
\[ \left(2+\frac{1}{n}\right)-0=2+\frac{1}{n}, \]
and so
\[ 2+\frac{1}{n}\longrightarrow2. \]
Consequently, the intersection need not reduce to a single point.
Observe that every point \(x\in[0,2]\) belongs to all the intervals, because
\[ 0\leq x\leq2<2+\frac{1}{n} \]
for every \(n\in\mathbb N\).
Thus
\[ [0,2]\subseteq\bigcap_{n=1}^{+\infty}I_n. \]
If, on the other hand, \(x<0\), then \(x\notin I_n\) for every \(n\), since all the intervals have left endpoint \(0\).
Conversely, if \(x>2\), then \(x-2>0\). By the Archimedean property there exists \(n\) such that
\[ \frac{1}{n}<x-2. \]
Hence
\[ 2+\frac{1}{n}<x. \]
Therefore \(x\notin I_n\), and so \(x\) does not belong to the intersection of all the intervals.
We conclude that
\[ \bigcap_{n=1}^{+\infty} \left[0,2+\frac{1}{n}\right] = [0,2]. \]
Problem 6 — level ★★☆☆☆
Consider the sequence of intervals
\[ I_n=\left[1-\frac{1}{n},3+\frac{1}{n}\right]. \]
Find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
We have
\[ \bigcap_{n=1}^{+\infty} I_n=[1,3]. \]
Solution
The intervals are closed, bounded, and nonempty. As \(n\) grows, the left endpoint \(1-\displaystyle \frac{1}{n}\) increases toward \(1\), while the right endpoint \(3+\displaystyle \frac{1}{n}\) decreases toward \(3\). Hence the intervals are nested.
The length of \(I_n\) is
\[ \left(3+\frac{1}{n}\right)-\left(1-\frac{1}{n}\right)=2+\frac{2}{n}. \]
Since \(2+\displaystyle \frac{2}{n}\to2\), the length does not tend to zero. Hence the intersection does not reduce to a single point.
The left endpoints have supremum \(1\), while the right endpoints have infimum \(3\). By the nested intervals theorem we obtain
\[ \bigcap_{n=1}^{+\infty} I_n=[1,3]. \]
Problem 7 — level ★★☆☆☆
Consider the sequence of intervals
\[ I_n=\left[-\frac{1}{n},1\right]. \]
Find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
We have
\[ \bigcap_{n=1}^{+\infty} I_n=[0,1]. \]
Solution
The intervals are closed, bounded, and nonempty. Moreover, the left endpoint \(-\displaystyle \frac{1}{n}\) increases toward \(0\), while the right endpoint stays equal to \(1\). Hence the intervals are nested.
The length of \(I_n\) is
\[ 1-\left(-\frac{1}{n}\right)=1+\frac{1}{n}. \]
Since \(1+\displaystyle \frac{1}{n}\to1\), the length does not tend to zero.
The supremum of the left endpoints is \(0\), while the infimum of the right endpoints is \(1\). Therefore
\[ \bigcap_{n=1}^{+\infty}\left[-\frac{1}{n},1\right]=[0,1]. \]
Problem 8 — level ★★☆☆☆
Consider the sequence of intervals
\[ I_n=\left[2-\frac{1}{n},5\right]. \]
Find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
We have
\[ \bigcap_{n=1}^{+\infty} I_n=[2,5]. \]
Solution
The intervals are closed, bounded, and nonempty. The left endpoint \(2-\displaystyle \frac{1}{n}\) increases toward \(2\), while the right endpoint stays constant and equal to \(5\). Hence the intervals are nested.
The length of \(I_n\) is
\[ 5-\left(2-\frac{1}{n}\right)=3+\frac{1}{n}. \]
Since \(3+\displaystyle \frac{1}{n}\to3\), the length does not tend to zero.
The supremum of the left endpoints is \(2\), while the infimum of the right endpoints is \(5\). Hence
\[ \bigcap_{n=1}^{+\infty}\left[2-\frac{1}{n},5\right]=[2,5]. \]
Problem 9 — level ★★☆☆☆
Consider the sequence of intervals
\[ I_n=\left[-\frac{1}{n},1+\frac{1}{n}\right]. \]
Find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
We have
\[ \bigcap_{n=1}^{+\infty} I_n=[0,1]. \]
Solution
The intervals are closed, bounded, and nonempty. The left endpoint \(-\displaystyle \frac{1}{n}\) increases toward \(0\), while the right endpoint \(1+\displaystyle \frac{1}{n}\) decreases toward \(1\). Hence the intervals are nested.
The length of \(I_n\) is
\[ \left(1+\frac{1}{n}\right)-\left(-\frac{1}{n}\right)=1+\frac{2}{n}. \]
Since \(1+\displaystyle \frac{2}{n}\to1\), the length does not tend to zero.
The supremum of the left endpoints is \(0\), while the infimum of the right endpoints is \(1\). Therefore
\[ \bigcap_{n=1}^{+\infty}\left[-\frac{1}{n},1+\frac{1}{n}\right]=[0,1]. \]
Problem 10 — level ★★☆☆☆
Consider the sequence of intervals
\[ I_n=\left[\frac{n}{n+1},2-\frac{1}{n+1}\right]. \]
Find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
We have
\[ \bigcap_{n=1}^{+\infty} I_n=\left[1,\frac{3}{2}\right]. \]
Solution
Let us rewrite the endpoints of the interval:
\[ \frac{n}{n+1}=1-\frac{1}{n+1}, \qquad 2-\frac{1}{n+1}. \]
Thus
\[ I_n=\left[1-\frac{1}{n+1},2-\frac{1}{n+1}\right]. \]
The intervals are closed, bounded, and nonempty. They are not, however, decreasingly nested: as \(n\) grows, both endpoints shift to the right.
Indeed,
\[ I_1=\left[\frac{1}{2},\frac{3}{2}\right], \qquad I_2=\left[\frac{2}{3},\frac{5}{3}\right]. \]
The interval \(I_2\) is not contained in \(I_1\), since its right endpoint is greater than that of \(I_1\). Hence the nested intervals theorem does not apply directly.
We compute the intersection all the same. A number \(x\) belongs to every interval if and only if
\[ 1-\frac{1}{n+1}\leq x\leq 2-\frac{1}{n+1} \]
for every \(n\in\mathbb N\).
From the first inequality, requiring it to hold for every \(n\), we obtain
\[ x\geq1. \]
Indeed, the left endpoints \(1-\frac{1}{n+1}\) increase toward \(1\).
From the second inequality, on the other hand, the most restrictive constraint comes from \(n=1\), since the right endpoints \(2-\frac{1}{n+1}\) increase as \(n\) grows. Hence we must have
\[ x\leq 2-\frac{1}{2}=\frac{3}{2}. \]
Therefore every point in the intersection must satisfy
\[ 1\leq x\leq\frac{3}{2}. \]
Conversely, if \(1\leq x\leq\frac{3}{2}\), then for every \(n\in\mathbb N\) we have
\[ 1-\frac{1}{n+1}\leq1\leq x \]
and also
\[ x\leq\frac{3}{2}\leq2-\frac{1}{n+1}. \]
Hence \(x\in I_n\) for every \(n\in\mathbb N\).
We conclude that
\[ \bigcap_{n=1}^{+\infty}\left[\frac{n}{n+1},2-\frac{1}{n+1}\right]=\left[1,\frac{3}{2}\right]. \]
Problem 11 — level ★★★☆☆
Consider the sequence of intervals
\[ I_n=\left(0,\frac{1}{n}\right). \]
Decide whether the nested intervals theorem applies, and find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
The nested intervals theorem does not apply, because the intervals are not closed. Moreover,
\[ \bigcap_{n=1}^{+\infty}\left(0,\frac{1}{n}\right)=\varnothing. \]
Solution
The intervals \(I_n\) are open, bounded, nonempty, and nested. Indeed, as \(n\) grows the right endpoint \(\displaystyle \frac{1}{n}\) decreases.
However, the nested intervals theorem requires the intervals to be closed and bounded. Here the intervals are not closed, so the theorem cannot be applied.
Let us now find the intersection. If \(x\) belonged to every interval, then we would need
\[ 0\lt x\lt\frac{1}{n} \]
for every \(n\in\mathbb N\).
But if \(x\gt0\), the Archimedean property gives an \(n\in\mathbb N\) such that
\[ \frac{1}{n}\lt x. \]
For this \(n\), the number \(x\) does not belong to \(I_n\).
Hence no real number belongs to every interval. Therefore
\[ \bigcap_{n=1}^{+\infty}\left(0,\frac{1}{n}\right)=\varnothing. \]
Problem 12 — level ★★★☆☆
Consider the sequence of intervals
\[ I_n=[n,+\infty). \]
Decide whether the nested intervals theorem applies, and find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
The nested intervals theorem does not apply, because the intervals are not bounded. Moreover,
\[ \bigcap_{n=1}^{+\infty}[n,+\infty)=\varnothing. \]
Solution
The intervals \(I_n=[n,+\infty)\) are closed and nonempty. They are also nested, since
\[ [n+1,+\infty)\subseteq[n,+\infty). \]
They are not bounded, however. The nested intervals theorem requires the intervals to be closed and bounded, so it does not apply here.
Let us find the intersection. If \(x\) belonged to every interval, then we would need
\[ x\geq n \]
for every \(n\in\mathbb N\).
This is impossible, since no real number is greater than or equal to every natural number.
Hence
\[ \bigcap_{n=1}^{+\infty}[n,+\infty)=\varnothing. \]
Problem 13 — level ★★★☆☆
Consider the sequence of intervals
\[ I_n=\left(-\infty,\frac{1}{n}\right]. \]
Decide whether the nested intervals theorem applies, and find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
The nested intervals theorem does not apply, because the intervals are not bounded. Moreover,
\[ \bigcap_{n=1}^{+\infty}\left(-\infty,\frac{1}{n}\right]=(-\infty,0]. \]
Solution
The intervals are closed and nonempty, but they are not bounded below. Hence the nested intervals theorem cannot be applied directly.
The intervals are nonetheless nested, since the right endpoint \(\displaystyle\frac{1}{n}\) decreases toward \(0\).
If \(x\leq0\), then
\[ x\leq0\lt\frac{1}{n} \]
for every \(n\in\mathbb N\). Hence every \(x\leq0\) belongs to all the intervals.
If, on the other hand, \(x\gt0\), then by the Archimedean property there exists \(n\in\mathbb N\) such that
\[ \frac{1}{n}\lt x. \]
For this \(n\), we have \(x\notin I_n\).
Therefore the points common to all the intervals are precisely the real numbers less than or equal to \(0\):
\[ \bigcap_{n=1}^{+\infty}\left(-\infty,\frac{1}{n}\right]=(-\infty,0]. \]
Problem 14 — level ★★★☆☆
Consider the sequence of intervals
\[ I_n=\left[0,1+\frac{(-1)^n}{2}\right]. \]
Decide whether the intervals are nested and whether the nested intervals theorem applies.
Answer
The intervals are not nested. The nested intervals theorem does not apply.
Solution
Let us compute the first few intervals. For \(n=1\) we have
\[ I_1=\left[0,\frac{1}{2}\right], \]
while for \(n=2\) we have
\[ I_2=\left[0,\frac{3}{2}\right]. \]
Hence \(I_2\) is not contained in \(I_1\). Indeed,
\[ \frac{3}{2}\in I_2, \qquad \frac{3}{2}\notin I_1. \]
The sequence of intervals is therefore not nested.
Even though the intervals are closed, bounded, and nonempty, the nesting hypothesis fails. Hence the nested intervals theorem does not apply.
Problem 15 — level ★★★☆☆
Consider the sequence of intervals
\[ I_n=\left(0,1+\frac{1}{n}\right). \]
Decide whether the nested intervals theorem applies, and find
\[ \bigcap_{n=1}^{+\infty} I_n. \]
Answer
The nested intervals theorem does not apply, because the intervals are not closed. Moreover,
\[ \bigcap_{n=1}^{+\infty}\left(0,1+\frac{1}{n}\right)=(0,1]. \]
Solution
The intervals are open, bounded, nonempty, and nested, since the right endpoint \(1+\displaystyle\frac{1}{n}\) decreases toward \(1\).
However, the nested intervals theorem requires the intervals to be closed and bounded. Since the intervals \(I_n\) are not closed, the theorem does not apply.
Let us now find the intersection. If \(0\lt x\leq1\), then
\[ 0\lt x\lt1+\frac{1}{n} \]
for every \(n\in\mathbb N\), so \(x\in I_n\) for every \(n\).
Therefore
\[ (0,1]\subseteq\bigcap_{n=1}^{+\infty} I_n. \]
Conversely, if \(x\leq0\), then \(x\notin I_n\) for every \(n\). If instead \(x\gt1\), then \(x-1\gt0\), and by the Archimedean property there exists \(n\in\mathbb N\) such that
\[ \frac{1}{n}\lt x-1. \]
Hence
\[ 1+\frac{1}{n}\lt x. \]
For this \(n\), the number \(x\) does not belong to \(I_n\).
We conclude that
\[ \bigcap_{n=1}^{+\infty}\left(0,1+\frac{1}{n}\right)=(0,1]. \]
Problem 16 — level ★★★★☆
Construct a sequence of closed, bounded, nested intervals whose lengths tend to zero and whose intersection is \(\{\sqrt{2}\}\).
Answer
One possible example is
\[ I_n=\left[\sqrt{2}-\frac{1}{n},\sqrt{2}+\frac{1}{n}\right]. \]
In this case
\[ \bigcap_{n=1}^{+\infty} I_n=\{\sqrt{2}\}. \]
Solution
Consider
\[ I_n=\left[\sqrt{2}-\frac{1}{n},\sqrt{2}+\frac{1}{n}\right]. \]
Each \(I_n\) is a closed, bounded, nonempty interval.
As \(n\) grows, the left endpoint increases toward \(\sqrt{2}\), while the right endpoint decreases toward \(\sqrt{2}\). Hence the intervals are nested.
The length of \(I_n\) is
\[ \left(\sqrt{2}+\frac{1}{n}\right)-\left(\sqrt{2}-\frac{1}{n}\right)=\frac{2}{n}. \]
Since \( \displaystyle \frac{2}{n}\to0\), the nested intervals theorem guarantees that the intersection contains a single point.
The point \(\sqrt{2}\) belongs to every interval, since it always lies between the endpoints \(\sqrt{2}-\displaystyle \frac{1}{n}\) and \(\sqrt{2}+\displaystyle \frac{1}{n}\).
Hence the only point common to all the intervals is \(\sqrt{2}\), that is,
\[ \bigcap_{n=1}^{+\infty}\left[\sqrt{2}-\frac{1}{n},\sqrt{2}+\frac{1}{n}\right]=\{\sqrt{2}\}. \]
Problem 17 — level ★★★★☆
Let \(f(x)=x^2-2\). Take the initial interval \([1,2]\), where \(f(1)\lt0\) and \(f(2)\gt0\). Using bisection, construct a sequence of closed, nested intervals each of which contains a solution of the equation
\[ x^2=2. \]
Explain why the intersection contains exactly one point.
Answer
The bisection method produces a sequence of closed, bounded, nested intervals whose lengths tend to zero. The intersection contains a single point, namely \(\sqrt{2}\).
Solution
We start from the interval
\[ I_1=[1,2]. \]
Since \(f(1)=1^2-2=-1\) and \(f(2)=2^2-2=2\), the function changes sign between \(1\) and \(2\).
We split \(I_1\) into two equal halves and choose one of them on which the function again changes sign. Call this interval \(I_2\). Repeating the procedure, we obtain a sequence of intervals
\[ I_1\supseteq I_2\supseteq I_3\supseteq\cdots. \]
By construction, each \(I_n\) is closed, bounded, and nonempty. Moreover, the intervals are nested.
At each step the length is halved. Since the initial length is \(1\), the length of \(I_n\) is
\[ \frac{1}{2^{n-1}}. \]
As
\[ \frac{1}{2^{n-1}}\to0, \]
the nested intervals theorem guarantees that the intersection contains a single point.
Let \(x_0\) denote the unique point belonging to all the intervals \(I_n\). By construction, each interval \(I_n\) contains at least one solution of the equation \(x^2=2\).
On the other hand, the intersection of all the intervals consists of a single point. Since \(f\) is continuous and, by construction, changes sign on each interval \(I_n\), the common point must be a root of \(f\). As this point lies between \(1\) and \(2\), it coincides with the positive solution of the equation \(x^2=2\), namely \(\sqrt{2}\).
Hence
\[ \bigcap_{n=1}^{+\infty} I_n=\{\sqrt{2}\}. \]
Problem 18 — level ★★★★☆
Let \((a_n)\) and \((b_n)\) be two real sequences such that
\[ a_n\leq a_{n+1}\leq b_{n+1}\leq b_n \]
for every \(n\in\mathbb N\). Suppose, in addition, that
\[ b_n-a_n\to0. \]
Prove that the two sequences converge to the same limit.
Answer
The sequences \((a_n)\) and \((b_n)\) converge to the same limit.
Solution
Consider the intervals
\[ I_n=[a_n,b_n]. \]
The condition
\[ a_n\leq a_{n+1}\leq b_{n+1}\leq b_n \]
implies that
\[ I_{n+1}\subseteq I_n \]
for every \(n\). Hence \((I_n)\) is a sequence of nested intervals.
Moreover, the intervals are closed, bounded, and nonempty. Since \(b_n-a_n\to0\), the nested intervals theorem guarantees that there exists a unique point \(x_0\) such that
\[ \bigcap_{n=1}^{+\infty} I_n=\{x_0\}. \]
Since \(x_0\in I_n\) for every \(n\), we have
\[ a_n\leq x_0\leq b_n \]
for every \(n\).
From this double inequality it follows that
\[ 0\leq x_0-a_n\leq b_n-a_n \]
and also
\[ 0\leq b_n-x_0\leq b_n-a_n. \]
Since \(b_n-a_n\to0\), the squeeze theorem yields
\[ a_n\to x_0 \qquad\text{and}\qquad b_n\to x_0. \]
Therefore the two sequences converge to the same limit.
Problem 19 — level ★★★★★
Show that the nested intervals theorem can fail in \(\mathbb Q\).
Answer
In \(\mathbb Q\) there exist sequences of closed, bounded, nested rational intervals whose lengths tend to zero and whose intersection is empty.
Solution
We construct rational intervals that close in on the irrational number \(\sqrt{2}\).
Let \(a_n\) and \(b_n\) be rational numbers such that
\[ a_n\lt\sqrt{2}\lt b_n \]
and such that
\[ b_n-a_n\to0. \]
For instance, one may take \(a_n\) and \(b_n\) to be rational decimal approximations of \(\sqrt{2}\) from below and from above, respectively.
Let us also choose them so that the intervals
\[ [a_n,b_n] \]
are nested.
Now consider the sets
\[ I_n=[a_n,b_n]\cap\mathbb Q. \]
Within \(\mathbb Q\), the sets \(I_n\) are closed and bounded rational intervals, with respect to the usual order and the relative topology of \(\mathbb Q\). They are moreover nested, and their lengths tend to zero.
In \(\mathbb R\), the intersection of the intervals \([a_n,b_n]\) is the single point \(\sqrt{2}\):
\[ \bigcap_{n=1}^{+\infty}[a_n,b_n]=\{\sqrt{2}\}. \]
However,
\[ \sqrt{2}\notin\mathbb Q. \]
Hence, working inside \(\mathbb Q\), no rational number belongs to all the intervals \(I_n\).
Therefore
\[ \bigcap_{n=1}^{+\infty} I_n=\varnothing. \]
This shows that the nested intervals theorem hinges on the completeness of \(\mathbb R\) and may fail in \(\mathbb Q\).
Problem 20 — level ★★★★★
Prove the following form of the nested intervals theorem: if
\[ I_n=[a_n,b_n], \qquad I_{n+1}\subseteq I_n \]
for every \(n\in\mathbb N\), and if
\[ b_n-a_n\to0, \]
then there exists a unique point belonging to all the intervals \(I_n\).
Answer
There exists a unique \(x_0\in\mathbb R\) such that
\[ x_0\in I_n \]
for every \(n\in\mathbb N\).
Solution
Since the intervals are nested, we have
\[ a_n\leq a_{n+1}\leq b_{n+1}\leq b_n \]
for every \(n\).
Hence the sequence \((a_n)\) is increasing, while the sequence \((b_n)\) is decreasing.
We show that every \(b_n\) is an upper bound for the set \(\{a_k:k\in\mathbb N\}\). Indeed, if \(k\leq n\), then
\[ a_k\leq a_n\leq b_n. \]
If instead \(k>n\), then
\[ a_k\leq b_k\leq b_n. \]
In either case \(a_k\leq b_n\). So every \(b_n\) is an upper bound for \(\{a_k:k\in\mathbb N\}\).
By the completeness of \(\mathbb R\), there exists
\[ x_0=\sup\{a_n:n\in\mathbb N\}. \]
Consequently,
\[ x_0\leq b_n \]
for every \(n\). Moreover, by the definition of supremum, we have
\[ a_n\leq x_0 \]
for every \(n\).
Hence
\[ a_n\leq x_0\leq b_n \]
for every \(n\), and therefore \(x_0\in I_n\) for every \(n\).
This proves that the intersection is nonempty.
We now prove uniqueness. Suppose that \(x\) and \(y\) belong to all the intervals \(I_n\), with \(x\leq y\). Then, for every \(n\),
\[ a_n\leq x\leq y\leq b_n. \]
It follows that
\[ 0\leq y-x\leq b_n-a_n. \]
Since \(b_n-a_n\to0\), we get \(y-x=0\), that is, \(x=y\).
Hence the point common to all the intervals is unique.