Computing Suprema and Infima: Standard Techniques

Knowing that a supremum exists is one thing — computing it with a rigorous proof is another. Whether you are tackling a CSIR NET problem or writing a formal analysis argument, you need a reliable toolkit: the two-condition verification test, the $\varepsilon$-characterisation, and disciplined algebraic manipulation. This post builds exactly that toolkit, step by step.

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2Conditions to verify

εε-Characterisation

4Solved Examples

1872Dedekind's Completeness

CSIR/GATE/JAMExam Relevance

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Core Definitions and the Standard Techniques

Before stating the verification techniques, we recall the central definitions.

📐 Definition — Supremum (Least Upper Bound)

Let $S$ be a non-empty subset of $\mathbb{R}$ bounded above. A number $\alpha \in \mathbb{R}$ is the supremum of $S$, written $\alpha = \sup S$, if and only if:

(S1) $x \leq \alpha$ for all $x \in S$ (α is an upper bound)

(S2) If $\beta < \alpha$, then $\beta$ is not an upper bound: $\exists\, x \in S$ with $x > \beta$.

Equivalently, (S2) is the ε-characterisation:

$$\textbf{(S2')} \quad \forall\, \varepsilon > 0,\; \exists\, x \in S \;\text{ such that }\; x > \alpha - \varepsilon.$$

📐 Definition — Infimum (Greatest Lower Bound)

Let $S$ be a non-empty subset of $\mathbb{R}$ bounded below. A number $\gamma \in \mathbb{R}$ is the infimum of $S$, written $\gamma = \inf S$, if and only if:

(I1) $x \geq \gamma$ for all $x \in S$ (γ is a lower bound)

(I2) If $\delta > \gamma$, then $\exists\, x \in S$ with $x < \delta$.

$$\textbf{(I2')} \quad \forall\, \varepsilon > 0,\; \exists\, x \in S \;\text{ such that }\; x < \gamma + \varepsilon.$$

📖 Reference: Rudin, W. — Principles of Mathematical Analysis, 3rd Ed., Ch. 1, §1.7–§1.8. | Bartle & Sherbert — Introduction to Real Analysis, 4th Ed., Ch. 2, §2.3.

The two-condition method translates these definitions directly into a proof strategy:

🔧 Two-Condition Method for sup S = α

Claim a candidate α. Compute or guess the value you believe is $\sup S$.

Verify (S1): Show $x \leq \alpha$ for every $x \in S$. This proves $\alpha$ is an upper bound.

Verify (S2): For arbitrary $\varepsilon > 0$, explicitly construct $x \in S$ with $x > \alpha - \varepsilon$. This proves no smaller number can be an upper bound.

Conclude: By the completeness of $\mathbb{R}$, $\sup S = \alpha$. $\blacksquare$

🔧 Two-Condition Method for inf S = γ

Claim a candidate γ.

Verify (I1): Show $x \geq \gamma$ for every $x \in S$.

Verify (I2): For arbitrary $\varepsilon > 0$, exhibit $x \in S$ with $x < \gamma + \varepsilon$.

Conclude $\inf S = \gamma$. $\blacksquare$

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Intuition, History and Visual Understanding

✍️ Mathematician Quote

"The way I look at it, continuity of the real line is exactly the property that every bounded set of rationals has a limit — and that limit belongs to the line itself."

— Richard Dedekind, paraphrased from Stetigkeit und irrationale Zahlen (1872)

🏛️ Historical Note: Dedekind and the Completeness of ℝ

Richard Dedekind (1831–1916) formalised the notion of completeness — the guarantee that bounded sets have a supremum in $\mathbb{R}$ — through his famous Dedekind cuts in 1872. While preparing a calculus lecture in Zürich, Dedekind was dissatisfied that the rigour of limits depended on geometric intuition. His cuts gave a purely arithmetic construction of $\mathbb{R}$ from $\mathbb{Q}$, placing the existence of suprema on solid ground.

Without completeness, $\{x \in \mathbb{Q} : x^2 < 2\}$ is bounded above in $\mathbb{Q}$ but has no supremum there — $\sqrt{2} \notin \mathbb{Q}$. In $\mathbb{R}$, it does. Dedekind's cut is what creates it.

Intuition for the ε-condition: Think of $\sup S$ as a wall. Condition (S1) says $S$ cannot pierce the wall. Condition (S2) says there is no gap between the wall and $S$ — you cannot slide even a sliver of space $\varepsilon$ between the wall and the nearest point of $S$. The number $\alpha - \varepsilon$ is always within reach of some element of $S$.

Fig. 1 — A bounded set $S$. The infimum $\gamma$ (green) is the greatest lower bound; the supremum $\alpha$ (gold) is the least upper bound. Any $\alpha - \varepsilon$ (red) is exceeded by some element of $S$.

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Solved Examples

Basic | Two-Condition Method

Find $\sup S$ and $\inf S$ for $S = \left\{\dfrac{n}{n+1} : n \in \mathbb{N}\right\}$.

Candidate values. Terms $\tfrac{1}{2}, \tfrac{2}{3}, \tfrac{3}{4}, \ldots$ are strictly increasing toward $1$, with minimum $\tfrac{1}{2}$ at $n=1$. Conjecture: $\sup S = 1$, $\inf S = \tfrac{1}{2}$.

Verify $\sup S = 1$. (S1): $\dfrac{n}{n+1} < 1$ since $n < n+1$. ✓ (S2): Given $\varepsilon > 0$, choose $n > \frac{1}{\varepsilon}-1$ (Archimedean property). Then $\dfrac{n}{n+1} > 1 - \varepsilon$. ✓

Verify $\inf S = \tfrac{1}{2}$. (I1): $\dfrac{n}{n+1} \geq \dfrac{1}{2}$ iff $n \geq 1$. ✓ (I2): $\tfrac{1}{2} \in S$ ($n=1$), so for any $\varepsilon>0$, $\tfrac{1}{2} < \tfrac{1}{2}+\varepsilon$. ✓

$$\boxed{\sup S = 1 \notin S, \qquad \inf S = \tfrac{1}{2} \in S.}$$

Intermediate | Algebraic Manipulation

Find $\sup S$ and $\inf S$ for $S = \{x \in \mathbb{R} : x^2 < 3\}$.

Identify $S$. $x^2 < 3 \Leftrightarrow -\sqrt{3} < x < \sqrt{3}$, so $S = (-\sqrt{3},\,\sqrt{3})$. Conjecture: $\sup S = \sqrt{3}$, $\inf S = -\sqrt{3}$.

Verify $\sup S = \sqrt{3}$. (S1): $x < \sqrt{3}$ for all $x \in S$. ✓ (S2): Let $\varepsilon > 0$. Take $x_0 = \sqrt{3} - \varepsilon/2$. Then $x_0^2 = 3 - \sqrt{3}\,\varepsilon + \varepsilon^2/4 < 3$ for $\varepsilon < 4\sqrt{3}$, so $x_0 \in S$ and $x_0 > \sqrt{3} - \varepsilon$. ✓

Verify $\inf S = -\sqrt{3}$. By symmetry (replace $x$ by $-x$), the identical argument applies. ✓

$$\boxed{\sup S = \sqrt{3} \notin S, \qquad \inf S = -\sqrt{3} \notin S.}$$

Intermediate | Exponential Set

Find $\sup S$ for $S = \left\{1 - \dfrac{1}{2^n} : n \in \mathbb{N}\right\}$.

Candidate. Terms $\tfrac{1}{2}, \tfrac{3}{4}, \tfrac{7}{8}, \ldots$ increase strictly toward $1$ but never reach it. Conjecture: $\sup S = 1$.

Verify (S1). Since $\dfrac{1}{2^n} > 0$, we have $1 - \dfrac{1}{2^n} < 1$ for all $n$. ✓

Verify (S2). Given $\varepsilon > 0$, need $2^n > \dfrac{1}{\varepsilon}$. Choose $n > \log_2(1/\varepsilon)$ (Archimedean property). ✓

Inf. Minimum attained at $n=1$: $\inf S = \dfrac{1}{2} \in S$.

$$\boxed{\sup S = 1 \notin S, \qquad \inf S = \tfrac{1}{2} \in S.}$$

CSIR NET Level | Density Argument

Let $A = \left\{\dfrac{p}{q} : p, q \in \mathbb{N},\; p^2 < 2q^2\right\}$. Prove that $\sup A = \sqrt{2}$.

Rewrite $A$. $p^2 < 2q^2 \Leftrightarrow \dfrac{p}{q} < \sqrt{2}$ (since $p,q > 0$). So $A = \mathbb{Q} \cap (0, \sqrt{2})$.

Verify (S1). Every $\tfrac{p}{q} \in A$ satisfies $\tfrac{p}{q} < \sqrt{2}$ by definition. ✓

Verify (S2). Let $\varepsilon > 0$. By density of $\mathbb{Q}$ in $\mathbb{R}$ (Rudin, Theorem 1.20), $\exists\, r \in \mathbb{Q}$ with $\sqrt{2} - \varepsilon < r < \sqrt{2}$. Writing $r = p/q$ gives $r^2 < 2 \Rightarrow p^2 < 2q^2$, so $r \in A$ and $r > \sqrt{2} - \varepsilon$. ✓

Note. Since $\sqrt{2}$ is irrational, $\sqrt{2} \notin A$. This illustrates why $\mathbb{Q}$ is incomplete: $A \subset \mathbb{Q}$ yet $\sup A \notin \mathbb{Q}$.

$$\boxed{\sup A = \sqrt{2} \notin A, \quad \text{and } \sup A \notin \mathbb{Q}.}$$

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Quick Revision Cards

📊 A — Key Definitions

(S1) $x \leq \alpha$ for all $x \in S$
(S2) $\forall\,\varepsilon>0,\;\exists\,x\in S: x > \alpha-\varepsilon$
Mirror for $\inf$: (I1) $x \geq \gamma$; (I2) $\exists\,x: x < \gamma+\varepsilon$
$\sup S \in S \Rightarrow S$ has a maximum.

⚙️ B — Useful Tools for (S2)

Archimedean: $\forall\,\varepsilon>0,\;\exists\,n: \frac{1}{n}<\varepsilon$
Density of $\mathbb{Q}$: between any two reals lies a rational
$2^n > n$ for all $n \geq 1$
Bernoulli: $(1+h)^n \geq 1+nh$, $h>-1$

🎯 C — Exam Tips

🔵 CSIR NET: Always verify both (S1) and (S2) — partial marks only for one
🟢 GATE: Know the ε-form of (S2) cold
🟠 IIT JAM: sup ≠ max — always check membership
🔴 All: $\mathbb{Q}$ is incomplete — sup of a rational set may be irrational

⚠️

Common Mistakes to Avoid

❌ Errors to Avoid

Error 1: Confusing sup with max.

Wrong: "$\sup S = \max S$ always." | Correct: $\sup S = \max S$ only when $\sup S \in S$. For $S=(0,1)$, $\sup S = 1 \notin S$, so no maximum exists.

Error 2: Proving only one condition.

Wrong: Showing $\alpha$ is an upper bound (S1) alone. | Correct: You must also verify (S2); otherwise you have only shown $\sup S \leq \alpha$.

Error 3: Working in ℚ instead of ℝ.

Wrong: Assuming $\sup$ always exists in $\mathbb{Q}$. | Correct: The completeness axiom guarantees existence only in $\mathbb{R}$. E.g., $\{x\in\mathbb{Q}: x^2<2\}$ has no sup in $\mathbb{Q}$.

Error 4: Incorrect ε-witness.

Wrong: Constructing $x_\varepsilon$ without checking $x_\varepsilon \in S$. | Correct: The constructed $x_\varepsilon$ must satisfy both $x_\varepsilon \in S$ and $x_\varepsilon > \alpha - \varepsilon$. Verify both explicitly.

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Real-World Applications

📈

Optimisation Theory

Suprema of objective functions over feasible sets underlie linear programming and convex optimisation — the backbone of machine learning algorithms.

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Probability & Statistics

The essential supremum ($\text{ess}\,\sup$) generalises the ordinary supremum to $L^\infty$ spaces in measure theory and probability.

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Computer Science

Worst-case running time of an algorithm is the supremum of computation time over all inputs of a given length — the very definition of Big-O bounds.

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Control Theory

The $H^\infty$ norm of a transfer function is a supremum over frequencies, used in robust control design to guarantee stability under bounded disturbances.

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Summary Table & Key Result

Set $S$	$\sup S$	$\sup \in S$?	$\inf S$	$\inf \in S$?
$(0,1)$	$1$	No	$0$	No
$[0,1]$	$1$	Yes	$0$	Yes
$\left\{\frac{n}{n+1}\right\}$	$1$	No	$\tfrac{1}{2}$	Yes
$\left\{1-\frac{1}{2^n}\right\}$	$1$	No	$\tfrac{1}{2}$	Yes
$(-\sqrt{3},\sqrt{3})$	$\sqrt{3}$	No	$-\sqrt{3}$	No
$\mathbb{Q}\cap(0,\sqrt{2})$	$\sqrt{2}$	No ($\notin\mathbb{Q}$)	$0$	No

Core ε-Characterisation (Rudin, Ch. 1, §1.8):

$$\alpha = \sup S \;\iff\; \Bigl[(\forall\,x\in S:\,x\leq\alpha)\;\text{ and }\;(\forall\,\varepsilon>0,\;\exists\,x\in S:\,x>\alpha-\varepsilon)\Bigr]$$

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Cross-References & Related Posts

📚 Prerequisites & Next Steps

← Prerequisites: Bounds and Extrema of Sets — defines upper/lower bounds, sup, inf; this post applies those definitions to computation | Order Relations and Ordered Sets — ordered field structure underlies every inequality | Algebraic Structures and Field Axioms — field operations used in manipulation | Intervals and the Real Line — open/closed intervals are the most common sup/inf examples | Foundations of Real Numbers — the completeness axiom guarantees every supremum exists.

→ Next Topic: Sequences of Real Numbers — the sup and inf of a sequence's range are the limsup and liminf, appearing throughout convergence theory.

📖 Further Reading: Rudin, Ch. 1, §§1.7–1.11; Bartle & Sherbert, Ch. 2, §2.3; Apostol, Ch. 1, §1.8.

📖 Download Academic PDF Notes Computing Suprema and Infima: Standard Techniques — Fractal Frontier Maths

Computing Suprema and Infima: Standard Techniques

Computing Suprema and Infima: Standard Techniques

Core Definitions and the Standard Techniques

Intuition, History and Visual Understanding

Solved Examples

Quick Revision Cards

📊 A — Key Definitions

⚙️ B — Useful Tools for (S2)

🎯 C — Exam Tips

Common Mistakes to Avoid

❌ Errors to Avoid

Real-World Applications

Optimisation Theory

Probability & Statistics

Computer Science

Control Theory

Summary Table & Key Result

Cross-References & Related Posts

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