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Suprema, Epsilon Bounds & Compactness

Useful preparation: Direct, Contrapositive & Contradiction Proofs

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Epsilon proofs guarantee requested accuracy. Suprema describe sharp bounds, even outside a set. Compactness prevents continuous functions from losing extrema through missing boundaries or infinity.

Supremum and maximum
An upper bound dominates every element of a set. Its least possible value is the supremum. A maximum is an upper bound that also belongs to the set.
Sequence limit
A sequence an\displaystyle a_{n} converges to L when every positive tolerance ϵ\displaystyle \epsilon eventually bounds all errors ∣an−L∣\displaystyle \left|a_{n} - L\right|. The threshold N may depend on ϵ\displaystyle \epsilon, but not on the later index n.
Compact set
A set is compact when every cover by open sets has a finite subcover. In finite-dimensional Euclidean space, this is equivalent to being closed and bounded.

Conditions before calculation

  • Real completeness guarantees a real supremum for a nonempty subset of R bounded above; neither emptiness nor an unbounded set meets this statement.
  • The closed-and-bounded criterion applies in Rᵈ with its usual distance and finite d; it is not a general criterion in arbitrary metric spaces.
  • The extreme value theorem needs a continuous real-valued function on a nonempty compact domain. A function-limit statement uses a cluster point of its domain.

Why the method works

1. Read the quantifiers in order

The challenger chooses ϵ\displaystyle \epsilon first; then choose one integer N that controls every later term. A threshold chosen separately for each n would not establish convergence.

∀ε>0 ∃N∈N ∀n≥N: ∣an−L∣<ε\forall\varepsilon>0\ \exists N\in\mathbb N\ \forall n\ge N:\ |a_n-L|<\varepsilon

2. Prove sharpness, not just boundedness

To prove s is a supremum, show it bounds the set and that every smaller candidate fails. An element within ϵ\displaystyle \epsilon below s supplies the second part.

(∀x∈S, x≤s)and∀ε>0 ∃x∈S: s−ε<x(\forall x\in S,\ x\le s)\quad\text{and}\quad\forall\varepsilon>0\ \exists x\in S:\ s-\varepsilon<x

3. Control a function locally

For x2\displaystyle x^{2} near 2, factor the error. First impose ∣x−2∣<1\displaystyle \left|x - 2\right| < 1 so ∣x+2∣<5\displaystyle \left|x + 2\right| < 5; then a delta independent of the approaching x forces the requested error.

δ=min⁡{1,ε/5},0<∣x−2∣<δ⇒∣x2−4∣<ε\delta=\min\{1,\varepsilon/5\},\quad 0<|x-2|<\delta\Rightarrow |x^2-4|<\varepsilon

Worked examples

EXAMPLE 1

For positive integers n, find the limit and supremum of an=4−3n+2\displaystyle a_{n} = 4 - \frac{3}{n + 2}, and decide whether a maximum exists.

  1. Every term is below 4 because the subtracted fraction is positive.

  2. Given ϵ>0\displaystyle \epsilon > 0, choose N=⌈3ϵ\displaystyle \frac{3}{\epsilon}⌉. For n≥N\displaystyle n \ge N, the denominator n+2\displaystyle n + 2 exceeds 3ϵ\displaystyle \frac{3}{\epsilon}.

    ∣an−4∣=3n+2<ε|a_n-4|=\frac3{n+2}<\varepsilon

  3. Thus terms get above every proposed upper bound smaller than 4. No term equals 4, and increasing n strictly increases the term.

Conclusion: The limit and supremum are 4; there is no maximum.

EXAMPLE 2

Find the extrema of f (t)=t2−4 t\displaystyle f\, \left(t\right) = t^{2} - 4\, t on the interval [−1,3].

  1. The interval is nonempty, closed, and bounded; the polynomial is continuous, so extrema exist.

  2. Complete the square. Its smallest possible square is zero, reached at t=2\displaystyle t = 2.

    f(t)=(t−2)2−4f(t)=(t-2)^2-4

  3. On this interval, ∣t−2∣≤3\displaystyle \left|t - 2\right| \le 3, with equality at −1. Therefore the largest square is 9.

Conclusion: The minimum is −4 at t=2\displaystyle t = 2; the maximum is 5 at t=−1\displaystyle t = -1.

Pitfalls & counterexamples

Tempting claim: A bounded domain guarantees that a continuous function attains its bounds.

Correction: The compactness hypothesis cannot be replaced by boundedness alone.

Test the claim: The continuous function f (t)=t\displaystyle f\, \left(t\right) = t on (−1,3) has supremum 3 but no maximum.

Check your understanding

Does the set {1−1n:n≥1\displaystyle 1 - \frac{1}{n} : n \ge 1} contain its supremum?

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Compare your reasoning

No; its supremum is 1.

Every term is smaller than 1, while arbitrarily large n make the difference arbitrarily small.

Connect this idea

Quantifier control prepares uniform convergence. Compactness links local continuity to global bounds and supports rigorous existence arguments in optimization.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

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