Nested Interval Theorem

Why Intervals Make Nesting Visible

The Nested Compact Sets Theorem says that a decreasing sequence of nonempty compact sets in \(\mathbb{R}\) has a common point. Closed intervals are a particularly useful case: instead of tracking arbitrary sets, we can track their left and right endpoints. The endpoints move in opposite directions, and their limiting positions describe how the intervals are narrowing.

This interval version is also a practical construction tool. Repeatedly choose one half of an interval, then the half of that half, and continue. The resulting nested intervals can encode a real number. We will use this idea to prove that every number in \([0,1]\) has a binary expansion.

If \(I_{n+1}\subseteq I_n\), the left endpoint of \(I_{n+1}\) cannot lie to the left of \(a_n\), and its right endpoint cannot lie to the right of \(b_n\). Thus nesting is equivalent to

for every \(n\). In particular, the left endpoints move weakly to the right, while the right endpoints move weakly to the left.

The Interval Form of Nested Compactness

Each closed bounded interval is compact by the Heine–Borel Theorem. Consequently, the Nested Compact Sets Theorem immediately gives the interval statement below. It guarantees a common point; it does not, by itself, say that the common point is unique.

The interval structure adds useful information that is not visible from the general compact-set statement alone. Because the endpoints are monotone and remain bounded, they each approach a limiting position. The following result makes that endpoint behavior precise without needing to identify the intersection.

Proof. The containment inequalities show that \((a_n)\) is increasing. Since \(a_n\leq b_n\leq b_1\), it is bounded above. Let \(\alpha=\sup\{a_n:n\geq1\}\). For any \(\varepsilon>0\), the definition of supremum gives an index \(N\) such that \(a_N>\alpha-\varepsilon\). For every \(n\geq N\), monotonicity gives \(a_N\leq a_n\leq\alpha\), so \(|a_n-\alpha|<\varepsilon\). Therefore \(a_n\to\alpha\).

The sequence \((b_n)\) is decreasing, and \(b_n\geq a_n\geq a_1\), so it is bounded below. Let \(\beta=\inf\{b_n:n\geq1\}\). Given \(\varepsilon>0\), the definition of infimum gives an index \(M\) with \(b_M<\beta+\varepsilon\). For \(n\geq M\), \(\beta\leq b_n\leq b_M\), and hence \(|b_n-\beta|<\varepsilon\). Thus \(b_n\to\beta\).

It remains to check the order of the limits. For any positive integers \(n\) and \(m\), we have \(a_n\leq b_m\). If \(n\geq m\), then \(a_n\leq b_n\leq b_m\); if \(n<m\), then \(a_n\leq a_m\leq b_m\). Thus every \(b_m\) is an upper bound for all the \(a_n\), so \(\alpha\leq b_m\) for every \(m\). Taking the infimum over \(m\) gives \(\alpha\leq\beta\). \(\square\)

The inequalities also imply that the interval lengths have a limit: \(b_n-a_n\to\beta-\alpha\), by the algebra of limits. This observation helps distinguish intervals that settle around a nontrivial interval from intervals that narrow to a single location. The Shrinking-Diameter Theorem from the previous tutorial gives the singleton conclusion when these lengths tend to zero.

Worked Examples: Following the Endpoints

Bisection and Binary Expansions

A binary digit is either \(0\) or \(1\). A binary expansion of \(x\in[0,1]\) is a representation of the form \(\sum_{j=1}^{\infty}\varepsilon_j/2^j\), where each \(\varepsilon_j\) is a binary digit. The nested interval idea constructs such digits one at a time: at each step, keep the half containing \(x\).

Proof. Start with \(I_0=[0,1]\). Suppose the interval at stage \(n-1\) is \(I_{n-1}=[k/2^{n-1},(k+1)/2^{n-1}]\) and contains \(x\). Bisect it at its midpoint. If \(x\) is at or to the left of the midpoint, retain the left half and set \(\varepsilon_n=0\). Otherwise retain the right half and set \(\varepsilon_n=1\). In either case, the retained interval \(I_n\) is closed, has length \(2^{-n}\), and contains \(x\).

Write \(I_n=[k_n/2^n,(k_n+1)/2^n]\). The choice of half gives \(k_n=2k_{n-1}+\varepsilon_n\), with \(k_0=0\). Dividing by \(2^n\) and applying this relation successively yields

Since \(x\in I_n\), its distance from the left endpoint is between zero and the interval length. Consequently,

The right side tends to zero, so the partial sums converge to \(x\). This proves the claimed binary expansion. \(\square\)

Why Binary Expansions Can Be Ambiguous

The bisection theorem guarantees at least one expansion, not exactly one. The familiar ambiguity occurs at endpoints of dyadic intervals. For example, the number \(3/4\) has both the terminating expansion \(0.11000\ldots\) and the expansion \(0.10111\ldots\), because

In general, a tail of all \(1\)s has sum \(\sum_{j=n+1}^{\infty}2^{-j}=2^{-n}\), exactly the size of one unit at the \(n\)th place. Thus a terminating expansion can be replaced by one that decreases an earlier digit by \(1\) and follows it with an infinite string of \(1\)s, whenever that earlier digit is \(1\). This is not a failure of the Nested Interval Theorem: the theorem concerns the point retained by the intervals, while different choices of endpoint representation can describe that same point.

The endpoint viewpoint and the bisection construction are useful for different purposes. Monotone endpoints reveal whether nested intervals retain a whole interval or narrow to one location. Bisection turns successive choices into a concrete numerical representation. In both settings, the closed, nested intervals keep the construction controlled: later stages refine earlier ones rather than abandoning them.