The nested intervals theorem — also known as Cantor's theorem on nested intervals — is one of the cornerstone results of real analysis: it states that a sequence of closed, bounded intervals, each contained in the previous one, always has at least one point in common.
This result is a consequence of the completeness of the real numbers and is an essential tool for proving a number of fundamental theorems, the Bolzano–Weierstrass theorem among them.
In the sections that follow we state the theorem, give its proof, discuss its geometric meaning, explain why the hypotheses cannot be dropped, and examine its link with the completeness of \(\mathbb{R}\).
Contents
- The nested intervals theorem
- Geometric interpretation
- The hypotheses are necessary
- Examples and applications
- Connection with the completeness of \(\mathbb{R}\)
The nested intervals theorem
Consider a sequence of closed, bounded intervals
\[ I_n=[a_n,b_n],\qquad a_n\leq b_n,\qquad n\in\mathbb{N}, \]
in which each interval is contained in the previous one:
\[ I_1\supseteq I_2\supseteq I_3\supseteq\cdots. \]
Such a sequence is called a sequence of nested intervals. The inclusion condition \(I_{n+1}\subseteq I_n\) is equivalent to
\[ a_n\leq a_{n+1}\leq b_{n+1}\leq b_n \qquad \forall n\in\mathbb{N}, \]
that is: the left endpoints form an increasing sequence and the right endpoints a decreasing one.
Theorem (Cantor, nested intervals). Let \((I_n)\) be a sequence of nonempty, closed, bounded intervals such that
\[ I_1\supseteq I_2\supseteq I_3\supseteq\cdots. \]
Then the intersection is nonempty:
\[ \bigcap_{n=1}^{+\infty} I_n\neq\varnothing. \]
More precisely, setting
\[ x_0=\sup_{n}\,a_n \qquad\text{and}\qquad y_0=\inf_{n}\,b_n, \]
we have
\[ \bigcap_{n=1}^{+\infty} I_n=[x_0,y_0]. \]
If, in addition, the length of the intervals tends to zero,
\[ b_n-a_n\longrightarrow 0, \]
then the intersection reduces to a single point:
\[ \bigcap_{n=1}^{+\infty} I_n=\{x_0\}. \]
Proof. We argue in steps.
1. Comparing the endpoints. For every pair of indices \(m,n\in\mathbb{N}\) we have
\[ a_m\leq b_n. \]
Indeed, if \(m\leq n\) then \(a_m\leq a_n\leq b_n\); whereas if \(m\gt n\) then \(a_m\leq b_m\leq b_n\). In either case \(a_m\leq b_n\). It follows that every \(b_n\) is an upper bound of the set \(\{a_m:m\in\mathbb{N}\}\) and, by symmetry, that every \(a_m\) is a lower bound of the set \(\{b_n:n\in\mathbb{N}\}\).
2. Existence of \(x_0\) and \(y_0\). The set \(\{a_n\}\) is nonempty and bounded above (for instance by \(b_1\)). By the completeness of \(\mathbb{R}\), its supremum exists,
\[ x_0=\sup_{n}\,a_n. \]
Likewise, \(\{b_n\}\) is nonempty and bounded below, so its infimum exists,
\[ y_0=\inf_{n}\,b_n. \]
3. The inequality \(x_0\leq y_0\). Since every \(b_n\) is an upper bound of \(\{a_m\}\) and \(x_0\) is the least upper bound, we have \(x_0\leq b_n\) for every \(n\). Hence \(x_0\) is a lower bound of \(\{b_n\}\) and, since \(y_0\) is the greatest lower bound,
\[ x_0\leq y_0. \]
4. Identifying the intersection. We show that \(\displaystyle\bigcap_{n} I_n=[x_0,y_0]\) by proving the two inclusions.
If \(x\in\bigcap_{n} I_n\), then \(a_n\leq x\leq b_n\) for every \(n\); thus \(x\) is an upper bound of \(\{a_n\}\) and a lower bound of \(\{b_n\}\), whence \(x\geq x_0\) and \(x\leq y_0\), that is \(x\in[x_0,y_0]\).
Conversely, if \(x\in[x_0,y_0]\), then for every \(n\)
\[ a_n\leq x_0\leq x\leq y_0\leq b_n, \]
and therefore \(x\in I_n\) for every \(n\), that is \(x\in\bigcap_{n} I_n\). The two inclusions together prove the equality
\[ \bigcap_{n=1}^{+\infty} I_n=[x_0,y_0], \]
which is nonempty, since it contains \(x_0\).
5. The case of vanishing length. Suppose now that \(b_n-a_n\longrightarrow 0\). From the relations \(x_0\geq a_n\) and \(y_0\leq b_n\) it follows, for every \(n\), that
\[ 0\leq y_0-x_0\leq b_n-a_n. \]
Letting \(n\to+\infty\), the right-hand side tends to \(0\), so \(y_0-x_0=0\), that is \(x_0=y_0\). The interval \([x_0,y_0]\) then collapses to a single point:
\[ \bigcap_{n=1}^{+\infty} I_n=[x_0,y_0], \qquad \bigcap_{n=1}^{+\infty} I_n=\{x_0\} \ \text{if}\ b_n-a_n\to0. \]
This completes the proof.
Geometric interpretation
The theorem says that, as we steadily shrink a sequence of closed intervals nested one inside the other, we cannot lose all of the points: at least one always survives, common to every interval of the sequence.
Geometrically, we may picture a sequence of ever shorter segments, each lying inside the previous one. If the lengths \(b_n-a_n\) do not tend to zero, the intersection is still an interval \([x_0,y_0]\) of positive length; if instead the lengths become arbitrarily small, the segments close in on a single location \(x_0\) on the real line, and the intersection is that one point.
The hypotheses are necessary
The assumptions that the intervals be closed and bounded are not superfluous: if even one of them is dropped, the conclusion may fail.
Boundedness is essential. Consider the unbounded intervals
\[ I_n=[\,n,+\infty\,). \]
They are closed, nonempty and nested, yet their intersection is empty:
\[ \bigcap_{n=1}^{+\infty} [\,n,+\infty\,)=\varnothing, \]
since no real number is greater than or equal to every natural number \(n\).
Closedness is essential. Consider the open intervals
\[ I_n=\left(0,\displaystyle \frac1n\right). \]
They are bounded, nonempty and nested, but here too
\[ \bigcap_{n=1}^{+\infty}\left(0,\displaystyle \frac1n\right)=\varnothing, \]
for any common point \(x\) would have to satisfy \(0\lt x\lt \displaystyle \frac1n\) for every \(n\), which is impossible since \(\displaystyle \frac1n\to 0\).
Examples and Applications
Example 1. Consider the intervals
\[ I_n=\left[0,\frac1n\right]. \]
They are closed, bounded, nonempty and nested, with length \(\displaystyle \frac1n\to 0\). Hence
\[ \bigcap_{n=1}^{+\infty}\left[0,\frac1n\right]=\{0\}. \]
Example 2. Consider the intervals
\[ I_n=\left[-\frac1n,\frac1n\right]. \]
Here too the length \(\displaystyle \frac2n\to 0\), and the intersection is
\[ \bigcap_{n=1}^{+\infty}\left[-\frac1n,\frac1n\right]=\{0\}. \]
Example 3. If, on the other hand, the length does not tend to zero, the intersection is a non-degenerate interval. For instance, with
\[ I_n=\left[-\frac1n,\,1+\frac1n\right] \]
we have \(x_0=\sup_n\!\left(-\displaystyle \frac1n\right)=0\) and \(y_0=\inf_n\!\left(1+\displaystyle \displaystyle \frac1n\right)=1\), so that
\[ \bigcap_{n=1}^{+\infty}\left[-\frac1n,\,1+\frac1n\right]=[0,1]. \]
Connection with the completeness of \(\mathbb{R}\)
The nested intervals theorem is a direct consequence of the completeness of the real numbers: in the proof we made essential use of the existence of the supremum \(x_0=\sup_n a_n\) (and of the infimum \(y_0=\inf_n b_n\)).
This property does not hold in the field of rational numbers. As an illustration, let us build a sequence of nested rational intervals that closes in on the irrational number \(\sqrt{2}\): let \(a_n\) and \(b_n\) be the decimal approximations of \(\sqrt{2}\) from below and from above,
\[ a_1=1.4,\ a_2=1.41,\ a_3=1.414,\ \ldots \qquad b_n=a_n+10^{-n}. \]
Take the sets
\[ I_n=[a_n,b_n]\cap\mathbb Q. \]
They are nested and have length \(10^{-n}\to 0\). In \(\mathbb R\) the intersection of the intervals \([a_n,b_n]\) is \(\{\sqrt2\}\); but in \(\mathbb Q\), where \(\sqrt2\notin\mathbb Q\), we obtain
\[ \bigcap_{n=1}^{+\infty} I_n=\varnothing. \]
The theorem thus reflects the absence of «gaps» in the real line.
Remark (a characterization of completeness). It should be noted that the nested intervals property on its own is not equivalent to completeness: it becomes so only when paired with the Archimedean property of \(\mathbb{R}\). In other words, in an Archimedean ordered field the nested intervals property is equivalent to the least upper bound property. It is precisely the Archimedean property (that is, \(\displaystyle \frac1n\to 0\)) that ensures, in our examples, that the lengths of the intervals genuinely tend to zero.
For this reason the nested intervals theorem is a fundamental tool in mathematical analysis and features in the proofs of many classical results, among them the Bolzano–Weierstrass theorem and the Heine–Borel theorem.