Supremum and Infimum

Q: Why does the set of rationals with x^2 < 2 have no supremum in Q?

The set S = {x in Q : x^2 < 2} is bounded above in Q (for example, by 2), but its least upper bound would be the square root of 2, which is irrational. Since the square root of 2 does not belong to Q, there is no rational number that serves as the least upper bound. This is the classic example showing that Q lacks the Completeness Property, and it is exactly why we need the real numbers.

Q: How do you prove that a number is the supremum of a set?

You need to show two things. First, show the number s is an upper bound of the set A, meaning x is less than or equal to s for every x in A. Second, show that s is the smallest such bound using the epsilon characterization: for every epsilon greater than 0, there exists an element x in A such that x is greater than s minus epsilon. This two-step approach works in virtually every proof involving the supremum.

Q: What happens if a set is not bounded above?

If a set is not bounded above, it has no supremum in the real numbers. We sometimes write sup(A) = positive infinity as a shorthand, but this means the supremum does not exist as a real number. The Completeness Axiom only guarantees the existence of a supremum for non-empty sets that are bounded above.

Q: Is the Completeness Axiom the same as the Archimedean Property?

No, they are different, though related. The Completeness Axiom states that every non-empty set of real numbers bounded above has a least upper bound. The Archimedean Property states that for any real number x, there exists a natural number n greater than x. The Archimedean Property is actually a consequence of the Completeness Axiom, not the other way around. The Completeness Axiom is strictly stronger.

Supremum and infimum are two of the most fundamental concepts you will encounter in real analysis. They generalize the familiar ideas of maximum and minimum to situations where a set may not actually contain its largest or smallest element. Without these concepts, you cannot state the Completeness Axiom, and without the Completeness Axiom, most of the theorems in analysis (convergence of sequences, existence of limits, the Intermediate Value Theorem) simply fall apart. If you want to understand why the real numbers behave differently from the rationals, this is where it starts.

Upper and Lower Bounds

Before we can define supremum and infimum, we need to understand what it means for a set to be bounded. The idea is straightforward: you are looking for numbers that “fence in” a set from above or below.

Upper Bound

A set $ A \subset \mathbb{R} $ of real numbers is bounded from above if there exists a real number $ a \in \mathbb{R} $, called an upper bound of $ A $, such that $ x \le a $ for every $ x \in A $.

Notice that an upper bound does not need to belong to the set $ A $ itself. It just needs to be greater than or equal to every element of $ A $. Also notice that upper bounds are never unique. If $ a $ is an upper bound of $ A $, then so is $ a + 1 $, $ a + 100 $, and every real number greater than $ a $.

Example: Consider the set $ A = \{1, 2, 3, 4, 5\} $. The number 5 is an upper bound, and so is 6, 10, or 1000. The number 4 is not an upper bound because $ 5 \in A $ and $ 5 > 4 $.

Lower Bound

Similarly, $ A $ is bounded from below if there exists $ b \in \mathbb{R} $, called a lower bound of $ A $, such that $ x \ge b $ for every $ x \in A $.

Again, the lower bound does not need to be in the set. And just like upper bounds, lower bounds are not unique. If $ b $ is a lower bound, then so is $ b – 1 $, $ b – 50 $, and every real number less than $ b $.

Example: For the open interval $ A = (0, 1) $, the number 0 is a lower bound even though $ 0 \notin A $. So is $ -5 $ or $ -100 $. Any non-positive real number works as a lower bound for this set.

Supremum (Least Upper Bound)

Among all the upper bounds of a set, there is a natural question: is there a smallest one? That smallest upper bound, if it exists, is called the supremum.

Formal Definition

Let $ A \subset \mathbb{R} $ be a non-empty set that is bounded above. A real number $ s $ is called the supremum (or least upper bound) of $ A $, written $ s = \sup(A) $, if:

$ s $ is an upper bound of $ A $: for every $ x \in A $, we have $ x \le s $.
$ s $ is the least such upper bound: if $ t $ is any upper bound of $ A $, then $ s \le t $.

An equivalent and extremely useful characterization of the supremum is this: $ s = \sup(A) $ if and only if $ s $ is an upper bound of $ A $ and for every $ \epsilon > 0 $, there exists some $ x \in A $ such that $ x > s – \epsilon $. In plain language, you can always find elements of $ A $ arbitrarily close to $ s $ from below.

This $ \epsilon $-characterization is the version you will use most often in proofs. It captures the idea that the supremum is “tight” against the set: you cannot push it down, not even by the tiniest amount, without some element of the set poking above.

Example: Consider $ A = (0, 1) $. Every number greater than or equal to 1 is an upper bound. The smallest such number is 1 itself. So $ \sup(A) = 1 $, even though $ 1 \notin A $. For any $ \epsilon > 0 $, the number $ 1 – \epsilon/2 $ belongs to $ A $ (assuming $ \epsilon < 2 $) and satisfies $ 1 - \epsilon/2 > 1 – \epsilon $.

Infimum (Greatest Lower Bound)

The infimum is the mirror image of the supremum. Among all the lower bounds of a set, we ask: is there a greatest one?

Formal Definition

Let $ A \subset \mathbb{R} $ be a non-empty set that is bounded below. A real number $ l $ is called the infimum (or greatest lower bound) of $ A $, written $ l = \inf(A) $, if:

$ l $ is a lower bound of $ A $: for every $ x \in A $, we have $ x \ge l $.
$ l $ is the greatest such lower bound: if $ t $ is any lower bound of $ A $, then $ l \ge t $.

The $ \epsilon $-characterization for the infimum is: $ l = \inf(A) $ if and only if $ l $ is a lower bound of $ A $ and for every $ \epsilon > 0 $, there exists some $ x \in A $ such that $ x < l + \epsilon $. You can always find elements of $ A $ arbitrarily close to $ l $ from above.

Example: For the set $ A = (0, 1) $, the infimum is 0. The number 0 is a lower bound (every element of $ A $ is positive), and no larger number can serve as a lower bound (for any $ \epsilon > 0 $, the element $ \epsilon/2 \in A $ satisfies $ \epsilon/2 < 0 + \epsilon $). Note that $ 0 \notin A $, so the set has no minimum, but it still has an infimum.

The supremum of a set is its least upper bound, and the infimum is its greatest lower bound. Both are unique when they exist, and neither needs to belong to the set.

Bounded Sets

A set $ A \subset \mathbb{R} $ is said to be bounded if it is both bounded above and bounded below. When a set is bounded, there exist real numbers $ m $ and $ M $ such that $ m \le x \le M $ for all $ x \in A $. In this case, both $ \sup(A) $ and $ \inf(A) $ exist, and we have:

$$ \inf(A) \le x \le \sup(A) \quad \text{for all } x \in A $$

A set that is bounded above but not below (like $ (-\infty, 5] $) has a supremum but no infimum in $ \mathbb{R} $. We sometimes write $ \inf(A) = -\infty $ in such cases, though this is a notational convention rather than claiming $ -\infty $ is a real number.

Similarly, a set bounded below but not above (like $ [3, \infty) $) has an infimum but no finite supremum. We write $ \sup(A) = +\infty $.

The empty set is a special case. By convention, $ \sup(\emptyset) = -\infty $ and $ \inf(\emptyset) = +\infty $, since every real number is vacuously an upper bound and a lower bound of the empty set.

Worked Examples

Let us work through several concrete examples. Finding the supremum and infimum of specific sets is one of the best ways to build intuition for these definitions.

Example 1: The Half-Open Interval (0, 1]

Let $ A = (0, 1] $.

Finding $ \sup(A) $: The number 1 belongs to $ A $, and every element of $ A $ satisfies $ x \le 1 $. So 1 is an upper bound. Is there a smaller upper bound? No, because $ 1 \in A $, and any number less than 1 would fail to be an upper bound. Therefore $ \sup(A) = 1 $. Since 1 actually belongs to $ A $, this set also has a maximum: $ \max(A) = 1 $.

Finding $ \inf(A) $: Every element of $ A $ satisfies $ x > 0 $, so 0 is a lower bound. Is there a larger lower bound? Suppose $ b > 0 $. Then $ \min(b/2, 1) \in A $ (since $ b/2 > 0 $) and $ b/2 < b $, so $ b $ is not a lower bound. Therefore $ \inf(A) = 0 $. Since $ 0 \notin A $, this set has no minimum.

Example 2: The Set {1/n : n in N}

Let $ A = \{1/n : n \in \mathbb{N}\} = \{1, 1/2, 1/3, 1/4, \ldots\} $.

Finding $ \sup(A) $: The largest element is $ 1/1 = 1 $, and every element satisfies $ 1/n \le 1 $. So 1 is an upper bound, and since $ 1 \in A $, no smaller upper bound exists. Therefore $ \sup(A) = 1 = \max(A) $.

Finding $ \inf(A) $: Every element is positive, so 0 is a lower bound. Is 0 the greatest lower bound? Yes: for any $ \epsilon > 0 $, by the Archimedean Property, there exists $ n \in \mathbb{N} $ with $ 1/n < \epsilon $, which means $ 1/n \in A $ and $ 1/n < 0 + \epsilon $. Therefore $ \inf(A) = 0 $. Since $ 0 \notin A $ (there is no natural number $ n $ with $ 1/n = 0 $), this set has no minimum.

Example 3: The Closed Interval [-3, 5]

Let $ A = [-3, 5] $.

This is the simplest case. Both endpoints belong to the set. Every $ x \in A $ satisfies $ -3 \le x \le 5 $. The number 5 is the least upper bound, and $ -3 $ is the greatest lower bound. So:

$$ \sup(A) = 5 = \max(A), \quad \inf(A) = -3 = \min(A) $$

When a set is a closed and bounded interval, the supremum equals the right endpoint and the infimum equals the left endpoint, and both are attained as maximum and minimum.

Example 4: The Set {x in R : x^2 < 2}

Let $ A = \{x \in \mathbb{R} : x^2 < 2\} = (-\sqrt{2}, \sqrt{2}) $.

Finding $ \sup(A) $: Since $ x^2 < 2 $ implies $ |x| < \sqrt{2} $, we have $ x < \sqrt{2} $ for all $ x \in A $. So $ \sqrt{2} $ is an upper bound. For any $ \epsilon > 0 $, the number $ \sqrt{2} – \epsilon/2 $ has square $ 2 – \epsilon\sqrt{2} + \epsilon^2/4 < 2 $ (for small enough $ \epsilon $), so it belongs to $ A $. Therefore $ \sup(A) = \sqrt{2} $.

Finding $ \inf(A) $: By symmetry, $ \inf(A) = -\sqrt{2} $.

This example is historically important. In the rational numbers $ \mathbb{Q} $, the set $ \{x \in \mathbb{Q} : x^2 < 2\} $ has no supremum within $ \mathbb{Q} $, because $ \sqrt{2} $ is irrational. This failure is precisely what the Completeness Axiom fixes in $ \mathbb{R} $.

Important Properties

Here are the key properties of supremum and infimum that you should know for any real analysis course.

Uniqueness

If the supremum of a set exists, it is unique. The proof is a one-liner: suppose $ s_1 $ and $ s_2 $ are both suprema of $ A $. Since $ s_1 $ is a least upper bound and $ s_2 $ is an upper bound, $ s_1 \le s_2 $. By the same reasoning with roles reversed, $ s_2 \le s_1 $. Therefore $ s_1 = s_2 $. The same argument applies to the infimum.

Relationship to Maximum and Minimum

If $ A $ has a maximum (a largest element that belongs to $ A $), then $ \sup(A) = \max(A) $. But the converse is false: the supremum can exist even when the maximum does not. The set $ (0, 1) $ has $ \sup = 1 $ but no maximum.

Similarly, if $ A $ has a minimum, then $ \inf(A) = \min(A) $, but the infimum can exist without a minimum. In short, every maximum is a supremum, but not every supremum is a maximum. This distinction is the entire reason the supremum concept exists.

The Completeness Axiom

The Completeness Axiom (also called the Least Upper Bound Property) states: every non-empty subset of $ \mathbb{R} $ that is bounded above has a supremum in $ \mathbb{R} $. This is not a theorem you prove; it is an axiom that defines the real numbers and distinguishes them from the rationals.

A consequence of the Completeness Axiom is that every non-empty subset of $ \mathbb{R} $ that is bounded below has an infimum in $ \mathbb{R} $. You can prove this by considering the set $ -A = \{-x : x \in A\} $ and applying the axiom to $ -A $.

Almost every major theorem in real analysis depends on this axiom, either directly or indirectly. The Monotone Convergence Theorem, the Bolzano-Weierstrass Theorem, the Nested Intervals Theorem, and the existence of limits for Cauchy sequences all trace back to the Completeness Axiom.

Supremum and Infimum of Subsets

If $ B \subset A $ and both sets are non-empty and bounded, then:

$$ \inf(A) \le \inf(B) \le \sup(B) \le \sup(A) $$

This makes intuitive sense: a smaller set is “squeezed” within the bounds of the larger set, so its supremum cannot exceed that of the larger set, and its infimum cannot be smaller.

Common Mistakes to Avoid

Students frequently make the following errors when working with supremum and infimum. Being aware of them will save you marks on exams and confusion in proofs.

Confusing Supremum with Maximum

The supremum does not need to belong to the set. If someone asks you for $ \sup(0, 1) $, the answer is 1, even though 1 is not in the open interval $ (0, 1) $. The maximum of a set, by definition, must belong to the set. The set $ (0, 1) $ has a supremum but no maximum.

Assuming “Bounded” Means the Set is Finite

The set $ \{1/n : n \in \mathbb{N}\} $ is infinite but bounded (it sits inside $ [0, 1] $). Boundedness is about the values of the elements, not the number of elements.

Forgetting the Epsilon Characterization

Many students can state the definition of supremum as “least upper bound” but struggle with proofs because they do not use the $ \epsilon $-characterization. In practice, showing $ s = \sup(A) $ almost always comes down to two steps: (1) show $ s $ is an upper bound, and (2) show that for every $ \epsilon > 0 $, there exists $ x \in A $ with $ x > s – \epsilon $. Master this technique early.

Applying Sup to Unbounded Sets

The supremum only exists (as a real number) for sets that are bounded above. The set $ \mathbb{N} = \{1, 2, 3, \ldots\} $ has no supremum in $ \mathbb{R} $. Writing $ \sup(\mathbb{N}) = \infty $ is a notational shorthand, not a claim that the supremum is a real number.

Mixing Up Inf and Sup in Proofs

The infimum is the greatest lower bound, and the supremum is the least upper bound. Students sometimes write the inequalities backwards. A helpful mnemonic: the supremum is “small for an upper bound” (least), and the infimum is “big for a lower bound” (greatest). They are extreme in opposite directions from what the words “upper” and “lower” might suggest.

FAQs

What is the difference between supremum and maximum?

The maximum of a set is the largest element that actually belongs to the set. The supremum is the least upper bound, which may or may not belong to the set. If a set has a maximum, then the supremum equals the maximum. But a set can have a supremum without having a maximum. For example, the open interval (0, 1) has sup = 1 but no maximum, because 1 is not in the set.

Can the supremum and infimum of a set be equal?

Yes, but only when the set contains exactly one element. If A = {c} for some real number c, then sup(A) = inf(A) = c. If a set has two or more distinct elements, the infimum is strictly less than the supremum.

Why does the set of rationals with x^2 < 2 have no supremum in Q?

The set S = {x in Q : x^2 < 2} is bounded above in Q (for example, by 2), but its least upper bound would be the square root of 2, which is irrational. Since the square root of 2 does not belong to Q, there is no rational number that serves as the least upper bound. This is the classic example showing that Q lacks the Completeness Property, and it is exactly why we need the real numbers.

How do you prove that a number is the supremum of a set?

You need to show two things. First, show the number s is an upper bound of the set A, meaning x is less than or equal to s for every x in A. Second, show that s is the smallest such bound using the epsilon characterization: for every epsilon greater than 0, there exists an element x in A such that x is greater than s minus epsilon. This two-step approach works in virtually every proof involving the supremum.

What happens if a set is not bounded above?

If a set is not bounded above, it has no supremum in the real numbers. We sometimes write sup(A) = positive infinity as a shorthand, but this means the supremum does not exist as a real number. The Completeness Axiom only guarantees the existence of a supremum for non-empty sets that are bounded above.

Is the Completeness Axiom the same as the Archimedean Property?

No, they are different, though related. The Completeness Axiom states that every non-empty set of real numbers bounded above has a least upper bound. The Archimedean Property states that for any real number x, there exists a natural number n greater than x. The Archimedean Property is actually a consequence of the Completeness Axiom, not the other way around. The Completeness Axiom is strictly stronger.