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Real Analysis: First Rigorous Steps / LEVEL 1 · DIFFICULTY 4/5

Real Analysis: First Rigorous Steps

Read the prerequisite, study a worked example and justify each step in these selected university topics.

2 stages · 16 practice problems · two 4-question assessment forms

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MINI QUEST URA 1.1Suprema, Epsilon Bounds & CompactnessRead the lesson
MINI QUEST URA 1.2Uniform Continuity & Function SequencesRead the lesson
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STAGE URA 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Suprema, Epsilon Bounds & Compactness

Useful preparation: Direct, Contrapositive & Contradiction Proofs

Open this topic in the university library →

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 $\displaystyle a_{n}$ converges to L when every positive tolerance $\displaystyle \epsilon$ eventually bounds all errors $\displaystyle \left|a_{n} - L\right|$ . The threshold N may depend on $\displaystyle \epsilon$ , 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 $\displaystyle \epsilon$ first; then choose one integer N that controls every later term. A threshold chosen separately for each n would not establish convergence.

\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 $\displaystyle \epsilon$ below s supplies the second part.

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

\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 $\displaystyle a_{n} = 4 - \frac{3}{n + 2}$ , and decide whether a maximum exists.

Every term is below 4 because the subtracted fraction is positive.
Given $\displaystyle \epsilon > 0$ , choose N=⌈ $\displaystyle \frac{3}{\epsilon}$ ⌉. For $\displaystyle n \ge N$ , the denominator $\displaystyle n + 2$ exceeds $\displaystyle \frac{3}{\epsilon}$ .
$|a_n-4|=\frac3{n+2}<\varepsilon$
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 $\displaystyle f\, \left(t\right) = t^{2} - 4\, t$ on the interval [−1,3].

The interval is nonempty, closed, and bounded; the polynomial is continuous, so extrema exist.
Complete the square. Its smallest possible square is zero, reached at $\displaystyle t = 2$ .
$f(t)=(t-2)^2-4$
On this interval, $\displaystyle \left|t - 2\right| \le 3$ , with equality at −1. Therefore the largest square is 9.

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

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 $\displaystyle f\, \left(t\right) = t$ on (−1,3) has supremum 3 but no maximum.

Check your understanding

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

Write a reason before opening the explanation. This reflection is not graded or saved.

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.

Practise this topic →Review concept flashcards

Reference reading

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

Preview the eight practice prompts

Problem 1
Find the supremum (0,3). What is its supremum?
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Problem 2
For $\displaystyle a_{n} = \frac{2}{n}$ , find the smallest positive integer N such that $\displaystyle a_{n} < \frac{1}{10}$ for all $\displaystyle n \ge N$ .
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Problem 3
Consider the interval [−2,4] in R. Is it compact? Enter 1 for yes, 0 for no.
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Problem 4
The sequence $\displaystyle a_{n} = 3 - \frac{1}{n}$ for positive integers n approaches what limit?
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Problem 5New context
A feasible open interval is (0,4). What is its supremum?
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Problem 6New context
An error sequence is $\displaystyle a_{n} = \frac{3}{n}$ , find the smallest positive integer N such that $\displaystyle a_{n} < \frac{1}{10}$ for all $\displaystyle n \ge N$ .
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Problem 7New context
A domain is the interval [−3,5] in R. Is it compact? Enter 1 for yes, 0 for no.
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Problem 8New context
A sequence for a completeness example $\displaystyle a_{n} = 4 - \frac{1}{n}$ for positive integers n approaches what limit?
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Open stage URA 1.1 in the student workspace →

STAGE URA 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Uniform Continuity & Function Sequences

Useful preparation: Suprema, Epsilon Bounds & Compactness

Open this topic in the university library →

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

Meaning & notation

Uniform convergence sets one accuracy deadline for the whole domain. Uniform continuity similarly chooses one input tolerance that protects output accuracy everywhere.

Uniform continuity: For every $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta > 0$ such that all domain points x,y with $\displaystyle \left|x - y\right| < \delta$ satisfy $\displaystyle \left|f\, \left(x\right) - f\, \left(y\right)\right| < \epsilon$ . The choice of $\displaystyle \delta$ cannot depend on x or y.
Pointwise versus uniform convergence: Pointwise convergence chooses N after both x and $\displaystyle \epsilon$ . Uniform convergence chooses N after $\displaystyle \epsilon$ alone, and that N works for every x and every $\displaystyle n \ge N$ .
Uniform error: The uniform error $\displaystyle E_{n}$ is the extended supremum of $\displaystyle \left|f_{n}\, \left(x\right) - f\, \left(x\right)\right|$ over the entire domain; it may be $\displaystyle +\infty$ when those errors are unbounded. Uniform convergence is equivalent to $\displaystyle E_{n}$ tending to zero.

Conditions before calculation

Continuity on a compact domain implies uniform continuity. Continuity alone on an arbitrary domain does not supply this global guarantee.
A uniform limit of continuous functions on a fixed domain is continuous. On a finite closed interval, uniform convergence of Riemann-integrable functions also permits passage of the limit through the integral.
Uniform convergence alone does not justify differentiating limits. A sufficient condition on a finite closed interval is continuous differentiability, uniform convergence of the derivatives, and convergence of function values at one fixed point.

Why the method works

1. Put the deadline before the location

Uniformity is precisely this order of choices. In the pointwise version, the universal choice of x occurs before the existence of N.

\forall\varepsilon>0\ \exists N\ \forall x\in D\ \forall n\ge N:\ |f_n(x)-f(x)|<\varepsilon

2. Seek a bound without x

An estimate depending only on n gives a single deadline. An estimate still containing x may establish only pointwise convergence unless that dependence is bounded uniformly.

E_n=\sup_{x\in D}|f_n(x)-f(x)|

3. Understand the continuity transfer

Approximate f at two nearby points by one fixed continuous $\displaystyle f_{n}$ . Two approximation errors and one continuity error each below $\displaystyle \frac{\epsilon}{3}$ make the total below $\displaystyle \epsilon$ .

Worked examples

EXAMPLE 1

Show $\displaystyle f\, \left(x\right) = x^{2}$ is uniformly continuous on [−2,2], and explain why the same claim fails on R.

Factor the difference. Within the bounded interval, $\displaystyle \left|x + y\right| \le 4$ .
$|x^2-y^2|\le4|x-y|$
Choose $\displaystyle \delta = \frac{\epsilon}{4}$ . It works for every pair in the interval, independently of their locations.
On R choose $\displaystyle x_{n} = n$ and $\displaystyle y_{n} = n + \frac{1}{n}$ . Their distance tends to zero, but their squared outputs stay more than 2 apart.
$y_n^2-x_n^2=2+1/n^2$

Conclusion: The compact restriction is uniformly continuous; the function on all of R is not.

EXAMPLE 2

For $\displaystyle f_{n}\, \left(x\right) = 2\, x + \frac{3}{n + 1}$ on R, prove uniform convergence and find the first N making every later error below $\displaystyle \frac{1}{10}$ .

Subtract the candidate limit $\displaystyle f\, \left(x\right) = 2\, x$ . The error is independent of x.
$E_n=\frac3{n+1}$
The strict inequality $\displaystyle \frac{3}{n + 1} < \frac{1}{10}$ is equivalent to $\displaystyle n > 29$ , so choose $\displaystyle N = 30$ .
For $\displaystyle n \ge 30$ the denominator only increases. For any smaller tolerance, choosing N sufficiently large repeats the same argument.

Conclusion: Convergence to 2x is uniform on R, and the requested smallest threshold is 30.

Pitfalls & counterexamples

Tempting claim: Continuous functions converging at every point must converge uniformly.

Correction: The deadline may still depend on the point.

Test the claim: On [0,1], $\displaystyle x^{n}$ tends to zero below 1 and to 1 at 1. The limit is discontinuous, so the convergence cannot be uniform.

Check your understanding

Do $\displaystyle f_{n}\, \left(x\right) = \frac{x}{n}$ converge uniformly to zero on [−3,3]? What about on R?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

Yes on [−3,3]; no on R.

The first supremum error is $\displaystyle \frac{3}{n}$ . On R the error is unbounded for each fixed n.

Connect this idea

Uniform bounds explain when approximations preserve continuity or permit integration. Differentiation needs stronger control.

Practise this topic →Review concept flashcards

Reference reading

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

Preview the eight practice prompts

Problem 1
For $\displaystyle f\, \left(x\right) = 2\, x$ , choosing $\displaystyle \delta = \frac{\epsilon}{a}$ guarantees $\displaystyle \left|f\, \left(x\right) - f\, \left(y\right)\right| < \epsilon$ when $\displaystyle \left|x - y\right| < \delta$ . Find the smallest positive a that works for every positive tolerance and all real x and y.
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Problem 2
On $\displaystyle \left[0, \frac{1}{2}\right]$ uses $\displaystyle f_{n}\, \left(x\right) = x^{n}$ . At $\displaystyle n = 2$ , what is sup $\displaystyle \left|f_{n}\, \left(x\right)\right|$ ?
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Problem 3
For the sequence $\displaystyle f_{n}\, \left(x\right) = x^{n}$ on [0,1], what is the pointwise limit at $\displaystyle x = 1$ ? The displayed exponent $\displaystyle n = 4$ is just one term.
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Problem 4
For $\displaystyle f_{n}\, \left(x\right) = x + \frac{2}{n}$ on all real x, find the smallest positive integer N giving sup_x| $\displaystyle f_{n}\, \left(x\right) - x$ |<1 for all $\displaystyle n \ge N$ .
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Problem 5New context
A Lipschitz example uses $\displaystyle f\, \left(x\right) = 3\, x$ , choosing $\displaystyle \delta = \frac{\epsilon}{a}$ guarantees $\displaystyle \left|f\, \left(x\right) - f\, \left(y\right)\right| < \epsilon$ when $\displaystyle \left|x - y\right| < \delta$ . Find the smallest positive a that works for every positive tolerance and all real x and y.
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Problem 6New context
A function-sequence bound on $\displaystyle \left[0, \frac{1}{2}\right]$ uses $\displaystyle f_{n}\, \left(x\right) = x^{n}$ . At $\displaystyle n = 3$ , what is sup $\displaystyle \left|f_{n}\, \left(x\right)\right|$ ?
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Problem 7New context
A pointwise-convergence example is $\displaystyle f_{n}\, \left(x\right) = x^{n}$ on [0,1], what is the pointwise limit at $\displaystyle x = 1$ ? The displayed exponent $\displaystyle n = 5$ is just one term.
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Problem 8New context
A uniform approximation uses $\displaystyle f_{n}\, \left(x\right) = x + \frac{3}{n}$ on all real x, find the smallest positive integer N giving sup_x| $\displaystyle f_{n}\, \left(x\right) - x$ |<1 for all $\displaystyle n \ge N$ .
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Open stage URA 1.2 in the student workspace →

Browse another level →

Real Analysis: First Rigorous Steps

Suprema, Epsilon Bounds & Compactness

Meaning & notation

Conditions before calculation

Why the method works

1. Read the quantifiers in order

2. Prove sharpness, not just boundedness

3. Control a function locally

Worked examples

For positive integers n, find the limit and supremum of an=4−3n+2\displaystyle a_{n} = 4 - \frac{3}{n + 2}an​=4−n+23​, and decide whether a maximum exists.

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

Pitfalls & counterexamples

Check your understanding

Connect this idea

Reference reading

Uniform Continuity & Function Sequences

Meaning & notation

Conditions before calculation

Why the method works

1. Put the deadline before the location

2. Seek a bound without x

3. Understand the continuity transfer

Worked examples

Show f (x)=x2\displaystyle f\, \left(x\right) = x^{2}f(x)=x2 is uniformly continuous on [−2,2], and explain why the same claim fails on R.

For fn (x)=2 x+3n+1\displaystyle f_{n}\, \left(x\right) = 2\, x + \frac{3}{n + 1}fn​(x)=2x+n+13​ on R, prove uniform convergence and find the first N making every later error below 110\displaystyle \frac{1}{10}101​.

Pitfalls & counterexamples

Check your understanding

Connect this idea

Reference reading

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

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

Show $\displaystyle f\, \left(x\right) = x^{2}$ is uniformly continuous on [−2,2], and explain why the same claim fails on R.

For $\displaystyle f_{n}\, \left(x\right) = 2\, x + \frac{3}{n + 1}$ on R, prove uniform convergence and find the first N making every later error below $\displaystyle \frac{1}{10}$ .