SYSTEMATIC MATHEMATICS

Real Analysis

This course contains 40 visible knowledge chapters in nine content-sized units of lengths 4, 6, 4, 4, 4, 4, 4, 6, and 4. Why do the rules of calculus actually work, and exactly when do they fail? Start with the real number line, upper bounds, nested intervals, and ordinary sequences. The course reads ε, δ, N, limsup, closure, and every new symbol aloud before using them, then turns “gets close,” “stays bounded,” “has no gaps,” and “one error bound works everywhere” into statements that can be proved. It now proves the complete sequence bridge explicitly: bounded data yields a convergent subsequence by Bolzano–Weierstrass, and Cauchy control plus that subsequential limit forces the full real sequence to converge. From those foundations it rebuilds series with tail certificates, compactness, continuity, derivatives including Darboux’s theorem, Riemann integration beyond continuous functions, uniform function series, completeness of continuous-function space, Arzelà–Ascoli subsequence extraction, and combined error budgets. Every theorem shows which assumption does the work, and the final study proves a root exists and is unique, bounds a bisection error, and justifies an exact integral one dependency at a time.

Before this course: Completed Calculus and Proof, Logic & Set Theory. The course assumes limits, derivatives, Riemann-sum intuition, elementary series, quantified proof, sets, functions, countability, induction, and contradiction, but no topology, measure theory, complex analysis, or abstract algebra.

COURSE FACTSLevel, chapters, units, prerequisite, and outcome
Chapter 1

Order and absolute value encode distance inequalities

Objective: Why does division preserve both inequality directions?

Absolute value is distance from zero, and the triangle and reverse-triangle inequalities turn geometric separation into algebraic error control.

Objects and quantifiers: For real x, |x|=x when x≥0 and |x|=−x when x<0; distance is d(x,y)=|x−y|. The result to establish is: For all real x,y, ||x|−|y||≤|x−y|≤|x|+|y|. State who chooses each tolerance, what may depend on it, and where one bound must work for every point.

The worked question is: Solve |2x−3|<5 and express the result as an interval. Begin with “Rewrite as −5<2x−3<5.” and finish with “Divide by positive 2 to obtain −1<x<4.” Then compare the quantified result with this boundary: Squaring an inequality without first controlling signs can reverse or destroy equivalence.

For all real x,y, ||x|−|y||≤|x−y|≤|x|+|y|.

From x=(x−y)+y, triangle inequality gives |x|≤|x−y|+|y|.

Thus |x|−|y|≤|x−y|; exchange x,y for the opposite difference.

Combining yields the reverse inequality, while ordinary triangle inequality gives the upper bound.

Solve |2x−3|<5 and express the result as an interval.

  1. Rewrite as −5<2x−3<5.
  2. Add 3 to obtain −2<2x<8.
  3. Divide by positive 2 to obtain −1<x<4.

Result: The solution set is (−1,4).