Proof of the Nested Interval Theorem

Completeness Locates a Common Point

The Nested Interval Theorem was stated in the previous tutorial. Here we prove it directly from the least-upper-bound property of the real numbers. The central idea is to collect all left endpoints into one set and take its supremum. Nesting ensures that this supremum cannot lie to the right of any interval’s right endpoint, so it belongs to every interval.

This proof shows how completeness enters: the supremum exists because the left endpoints form a nonempty set bounded above. It also reveals more than mere existence. Once the endpoint limits are taken into account, they determine the entire intersection.

Theorem (Nested Interval Theorem): Let $I_n=[a_n,b_n]$ be nonempty closed intervals such that $I_{n+1}\subseteq I_n$ for every positive integer $n$. Then $\bigcap_{n=1}^{\infty}I_n\neq\varnothing$.

Proof. Let $S=\{a_n:n\geq1\}$, the set of all left endpoints. This set is nonempty. Because the intervals are nested, $a_n\leq b_n\leq b_1$ for every $n$, so $b_1$ is an upper bound for $S$. By the least-upper-bound property, $S$ has a supremum; write $\alpha=\sup S$.

Fix any positive integer $n$. We first show that $b_n$ is an upper bound for the whole set $S$, not just for the left endpoint $a_n$. If $k\geq n$, nesting gives $a_k\leq b_k\leq b_n$. If $k<n$, then $a_k\leq a_n\leq b_n$. Thus $a_k\leq b_n$ for every $k$, and the definition of supremum gives $\alpha\leq b_n$. Also, $a_n\leq\alpha$, since a supremum is an upper bound for its set. Therefore

a_n\leq\alpha\leq b_n.

This holds for every $n$, so $\alpha\in I_n$ for every $n$. Consequently, $\alpha\in\bigcap_{n=1}^{\infty}I_n$, and the intersection is nonempty. $\square$

The closed-interval condition matters in the final step: the inequalities $a_n\leq\alpha\leq b_n$ guarantee membership in $[a_n,b_n]$. The argument also makes a useful distinction: the supremum need not itself be one of the left endpoints. It is enough that it lies between the endpoints of every interval.

The Endpoints Describe the Whole Intersection

The Convergence of Nested Interval Endpoints theorem from the previous tutorial gives limits $\alpha$ and $\beta$ for the left- and right-endpoint sequences, respectively, with $\alpha\leq\beta$. In the proof above, $\alpha$ was defined as the supremum of the left endpoints. The same endpoint theorem identifies $\beta$ as the infimum of the right endpoints. These values give an exact description of the common points.

Theorem (Endpoint Description of the Intersection): For a nested decreasing sequence $I_n=[a_n,b_n]$, let $\alpha=\sup\{a_n:n\geq1\}$ and $\beta=\inf\{b_n:n\geq1\}$. Then $$ \bigcap_{n=1}^{\infty} I_n=[\alpha,\beta]. $$

Proof. The endpoint convergence theorem gives $\alpha\leq\beta$. If $x\in\bigcap_{n=1}^{\infty}I_n$, then $a_n\leq x\leq b_n$ for every $n$. Since $x$ is an upper bound for all the $a_n$, the least-upper-bound property gives $\alpha\leq x$. Since $x$ is a lower bound for all the $b_n$, the greatest-lower-bound property gives $x\leq\beta$. Thus $x\in[\alpha,\beta]$.

Conversely, suppose $x\in[\alpha,\beta]$. Every left endpoint satisfies $a_n\leq\alpha$, because $\alpha$ is an upper bound for the left endpoints. Every right endpoint satisfies $\beta\leq b_n$, because $\beta$ is a lower bound for the right endpoints. Hence

a_n\leq\alpha\leq x\leq\beta\leq b_n

for every $n$. It follows that $x\in I_n$ for every $n$, so $x\in\bigcap_{n=1}^{\infty}I_n$. Both inclusions hold, proving the stated equality. $\square$

This description makes the existence theorem especially transparent: the endpoint convergence theorem ensures $\alpha\leq\beta$, so $[\alpha,\beta]$ is nonempty. The direct supremum proof above, in turn, constructs a common point without first needing to describe every point in the intersection.

Worked Examples: Using the Supremum Argument

Worked Example: A Supremum That Is Not a Left Endpoint

For $n\geq1$, let $I_n=[1-1/n,3]$. The left endpoints increase because $1/(n+1)<1/n$, which gives $1-1/n<1-1/(n+1)$. The right endpoint remains $3$, so the intervals are nested. The left endpoints have supremum $1$, although $1-1/n<1$ for every $n$; none of them equals their supremum.

The supremum proof shows that $1$ belongs to every interval, since $1-1/n\leq1\leq3$. More precisely, any $x\in[1,3]$ belongs to every $I_n$. If $x<1$, choose $n$ large enough that $1/n<1-x$. Then $1-1/n>x$, so $x\notin I_n$. If $x>3$, it is outside every $I_n$ because each right endpoint is $3$. Therefore

\bigcap_{n=1}^{\infty}I_n=[1,3].

Worked Example: Intervals Whose Intersection Is a Single Point

Let $J_n=[6-1/2^n,6+1/3^n]$ for $n\geq1$. The left endpoint increases with $n$, and the right endpoint decreases, so the intervals are nested. Each contains $6$, since $6-1/2^n\leq6\leq6+1/3^n$. Thus $6$ is a common point.

If $x$ belongs to every $J_n$ and $x<6$, then $6-x\leq1/2^n$ for every $n$. But the powers $1/2^n$ tend to zero, so for some $n$ we have $1/2^n<6-x$, a contradiction. If $x>6$, membership would require $x-6\leq1/3^n$ for every $n$, which is impossible because $1/3^n\to0$. Hence no point other than $6$ is common, and

\bigcap_{n=1}^{\infty}J_n=\{6\}.

In endpoint terms, both limits are $6$, so the interval $[\alpha,\beta]$ reduces to one point.

Worked Example: A Nontrivial Interval of Common Points

For $n\geq1$, define $K_n=[2-1/(n+1),8+1/(n+1)]$. As $n$ increases, the left endpoint increases and the right endpoint decreases. Each interval contains $[2,8]$: its left endpoint is below $2$, and its right endpoint is above $8$.

If $x<2$, choose $n$ so large that $1/(n+1)<2-x$. Then $2-1/(n+1)>x$, so $x\notin K_n$. If $x>8$, choose $n$ so large that $1/(n+1)<x-8$. Then $8+1/(n+1)<x$, so $x\notin K_n$. Thus

\bigcap_{n=1}^{\infty}K_n=[2,8].

Here the left endpoints have supremum $2$, while the right endpoints have infimum $8$. The common points form the whole interval between those two limiting positions.

Why Nesting and Closedness Matter

Two features of the proof do distinct jobs. Nesting is what lets $b_n$ bound every left endpoint, including those from later intervals. Without nesting, the supremum of the left endpoints might not lie to the left of every right endpoint. Closedness is what turns the endpoint inequalities into membership: if the common candidate equals an endpoint, it is still in the interval.

For example, the sequence alternating between $[0,1]$ at odd indices and $[2,3]$ at even indices consists of nonempty closed intervals, but it is not nested and has empty intersection. Closedness alone is therefore not enough. Conversely, replacing closed intervals with intervals that omit an endpoint can also defeat the conclusion. The direct proof relies on the actual membership rule for $[a_n,b_n]$, not merely on inequalities that approach its endpoints.

When applying the theorem, first check that every interval is nonempty and that each later interval is contained in the preceding one. Then collect the left endpoints and use their supremum. This method is useful well beyond interval problems: it illustrates how a least upper bound can turn a sequence of compatible constraints into a point satisfying them all.

Check Your Understanding

Use the supremum proof and endpoint description to answer the following questions.

Why is the set of left endpoints bounded above in a nested sequence of nonempty closed intervals?
For a fixed $n$, why does $b_n$ bound every left endpoint, including $a_k$ when $k<n$ and when $k\geq n$?
Why does the supremum of the left endpoints belong to every closed interval?
How do the supremum of the left endpoints and the infimum of the right endpoints determine the full intersection?
Which hypothesis fails for the alternating sequence $[0,1],[2,3],[0,1],[2,3],\ldots$, and why does that matter?

Proof of the Nested Interval Theorem

What You'll Learn