Why the Interval Matters
The Extreme Value Theorem guarantees that a continuous function on a closed, bounded interval attains both an absolute maximum and an absolute minimum. Before using that result, you need to identify the interval correctly: whether its endpoints are included and whether its values have finite bounds.
This tutorial checks only the interval hypothesis of the theorem. A closed, bounded interval is necessary, but the function must also be continuous on that interval. Continuity is a separate condition.
What “Closed” and “Bounded” Mean
A set of real numbers is bounded if all its elements lie between some finite lower and upper bounds. In symbols, \(S\subseteq\mathbb R\) is bounded if there is a real number \(M>0\) such that \(|x|\le M\) for every \(x\in S\). An interval extending indefinitely to the left or right is unbounded.
A set is closed if it contains its limit points. For intervals on the real line, a useful practical test is to check whether finite endpoints that the interval approaches are included. Square brackets include an endpoint; parentheses exclude it. Infinity is never an endpoint that belongs to the interval.
| Interval | Closed? | Bounded? |
|---|---|---|
| \([a,b]\), with finite \(a\le b\) | Yes | Yes |
| \((a,b)\), \([a,b)\), or \((a,b]\), with finite \(a<b\) | No | Yes |
| \([a,\infty)\) or \((a,\infty)\) | Depends on whether the finite endpoint is included | No |
| \(\mathbb R\) | Yes | No |
The table separates two different questions. Boundedness concerns whether the interval extends arbitrarily far. Closedness concerns whether it includes its finite boundary points. An interval can satisfy one condition without satisfying the other.
A Practical Endpoint Test
For an interval with finite endpoints, inspect both brackets and parentheses. For boundedness, check whether both ends are finite. These checks are independent: including an endpoint does not make an interval bounded, and having finite endpoints does not make an interval closed.
Write down its finite endpoints and note whether it extends to \(+\infty\) or \(-\infty\).
The interval is bounded only if its values remain between finite numbers.
For a finite endpoint approached by the interval, verify that the endpoint is included.
The set must be a closed, bounded interval, typically \([a,b]\) with finite \(a<b\).
Worked Examples
Worked Example: Classifying a Closed Finite Interval
Classify \(I=[-3,4]\), then decide whether it meets the Extreme Value Theorem’s interval hypothesis.
The interval has finite endpoints \(-3\) and \(4\). Both endpoints are included because the notation uses square brackets. Every \(x\in I\) satisfies
Thus, the interval is bounded: for example, \(|x|\le 4\) for every \(x\in[-3,4]\). Since both finite endpoints are included, it is closed as well.
Therefore, \([-3,4]\) meets the interval hypothesis. The Extreme Value Theorem can be applied to a function continuous on this interval; this classification alone does not establish continuity.
Check: The interval is exactly of the form \([a,b]\), with \(a=-3\), \(b=4\), and \(-3<4\).
Worked Example: A Bounded Interval That Is Not Closed
Classify \(J=(-2,5]\) and determine whether it meets the interval hypothesis.
The interval has finite endpoints, so its values stay between \(-2\) and \(5\). More precisely,
It is bounded. But the left endpoint \(-2\) is excluded, as shown by the parenthesis. The interval approaches \(-2\) without containing it, so it is not closed.
Therefore, \(J\) is bounded but not closed, and it does not meet the Extreme Value Theorem’s interval hypothesis. Having two finite endpoints is not sufficient; both must be included for a finite interval to be closed.
Check: The right endpoint \(5\) is included, but the missing left endpoint still means the whole interval is not closed.
Worked Example: A Closed but Unbounded Ray
Classify \(K=[1,\infty)\) and determine whether it meets the interval hypothesis.
The finite endpoint \(1\) is included. The ray contains every real number \(x\) satisfying
It has no finite upper bound: given any real number \(M\), there is an \(x\in K\) greater than \(M\), such as \(x=\max(1,M)+1\). Hence \(K\) is unbounded.
The included finite endpoint and the absence of any missing finite endpoint mean that the ray is closed. It is not bounded, however, so it does not meet the Extreme Value Theorem’s interval hypothesis.
Check: A square bracket at \(1\) does not make the ray bounded; the notation \(\infty\) signals that the interval continues without an upper limit.
Worked Example: Classifying a Function’s Domain
Find the real domain of \(f(x)=\sqrt{9-x^2}\), classify it, and decide whether it meets the interval hypothesis.
For the square root to be real, its radicand must be nonnegative. Thus,
The real domain is therefore \([-3,3]\). Both endpoints are included because the radicand equals \(0\) at \(x=-3\) and \(x=3\), and \(\sqrt{0}\) is defined. The domain is bounded because every element lies between \(-3\) and \(3\); it is closed because it includes both endpoints.
The domain meets the interval hypothesis. For example, the function is continuous on this domain, so the Extreme Value Theorem applies to it there. The domain calculation and the endpoint check are both essential.
Check: Substituting either endpoint gives \(9-(-3)^2=0\) and \(9-3^2=0\), so neither endpoint needs to be removed.
Why the Endpoint Test Works
For a nonempty interval, the finite-endpoint test follows from the definitions of an interval, closedness, and boundedness. Here is the key characterization: a nonempty interval of real numbers is closed and bounded exactly when it has the form \([a,b]\) for finite real numbers \(a\le b\).
First, if \(I=[a,b]\) with finite endpoints, then every \(x\in I\) satisfies \(a\le x\le b\), so \(I\) is bounded. It contains its endpoints and all its limit points, so it is closed.
Conversely, suppose \(I\) is a nonempty, closed, bounded interval. Because it is bounded, it has a finite infimum \(a=\inf I\) and a finite supremum \(b=\sup I\). Every element of \(I\) lies between these values. If \(a<b\), then for any \(x\) with \(a<x<b\), the definitions of infimum and supremum give elements \(u,v\in I\) with \(u<x<v\). Since \(I\) is an interval, it contains every number between \(u\) and \(v\), so it contains \(x\).
The interval also contains \(a\) and \(b\). For example, the definition of infimum gives elements of \(I\) arbitrarily close to \(a\) from above, so \(a\) is a limit point of \(I\). Since \(I\) is closed, it contains \(a\). The same argument shows that \(b\in I\). Thus \(I=[a,b]\). If \(a=b\), the nonempty set \(I\) lies between \(a\) and \(b\), so \(I=\{a\}=[a,a]\). This proves the characterization.
Common Mistake: Confusing Bounded with Closed
A frequent error is to see finite endpoints and conclude that an interval is closed. For instance, the claim “\((1,4)\) is closed because it lies between \(1\) and \(4\)” confuses boundedness with closedness. The inequality \(1<x<4\) shows that the interval is bounded, but it excludes both endpoints. It is therefore not closed and does not meet the interval hypothesis.
The correction is to ask two separate questions: Are there finite bounds? Are the approached finite endpoints included? For \((1,4)\), the first answer is yes and the second is no. Consequently, it is bounded but not closed.
Domains May Need More Than an Endpoint Check
A domain is the set of all real inputs for which a function is defined. It may be an interval, but it may also have gaps. For example, \(g(x)=1/(x-2)\) has domain \(\mathbb R\setminus\{2\}\): every real input is allowed except \(2\). This domain is not an interval, since it contains numbers on both sides of \(2\) but does not contain every number between them. It is also unbounded and not closed, so it does not meet the interval hypothesis.
When classifying a domain, first find the full set of permitted inputs. Do not assume it is one interval just because a formula contains familiar functions. Then test the actual set for closedness and boundedness. Even a closed, bounded domain would only satisfy the interval condition if it is an interval of the required form.
Why This Check Matters
The Extreme Value Theorem has specific hypotheses. If a continuous function is considered on an open interval, a half-open interval, or an unbounded interval, the theorem does not supply a guarantee that both absolute extrema are attained. That does not prove that extrema cannot occur; it means only that this theorem’s guarantee is unavailable. The next step in a complete application is to check continuity on the closed interval.
Check Your Understanding
For each set, decide whether it is closed, bounded, and suitable as the interval in the Extreme Value Theorem. Give a brief reason.
- Classify \([-1,6)\).
- Classify \((-\infty,3]\).
- Find and classify the real domain of \(h(x)=\sqrt{x+4}\).
- Is \(\{2\}=[2,2]\) closed and bounded? How does it relate to the usual nondegenerate interval hypothesis?
- Explain why knowing that an interval is bounded does not establish that it is closed.
