Real Analysis

Studying Real Analysis

The first analysis course asks you to justify operations that calculus made familiar: taking limits, passing a limit through a function, and integrating a sequence. Lebl organizes an entry through sequences and continuity before uniform convergence. Trench revisits calculus with detailed theorem statements and worked examples, placing sequences later. Choose the order that exposes your actual difficulty rather than repeating every computational topic.

What to work on

• Write epsilon proofs for limits of sequences.
• Use completeness to prove convergence of monotone bounded or Cauchy sequences.
• Distinguish continuity from uniform continuity by examples.
• Test uniform convergence before interchanging limits, integrals or derivatives.

A starting text

Basic Analysis I: Introduction to Real Analysis, Volume I — Jiří Lebl. Read sections 2.1-2.4 on sequences and completeness, 3.2-3.4 on continuity, and 6.1-6.2 on uniform convergence and interchange of limits. Review 1.2 on the real numbers if the least-upper-bound property is unfamiliar.

Practice

Sequence, continuity and convergence exercises. Exercises embedded in sections 2.1-2.4, 3.4 and 6.1-6.2; include proof and counterexample questions.. Exercises provided; public solutions not verified.

Try these before moving on

Try a proof or calculation before opening the hint. These solutions address the checks below, not the exercises in the linked textbooks.

1. Give an epsilon-N proof that (3n + 1)/(n + 2) converges to 3, with an explicit choice of N.

   A hint

   Subtract the proposed limit before trying to choose $N$. You only need one sufficient bound, not the smallest possible $N$.

   Work through the solution

   For $n\ge1$,

   $$\left|\frac{3n+1}{n+2}-3\right|=\frac5{n+2}.$$

   Given $\varepsilon>0$, choose $N=\lceil5/\varepsilon\rceil$. If $n\ge N$, then $n+2>5/\varepsilon$, so $5/(n+2)<\varepsilon$. This is exactly the required estimate. The choice of $N$ depends only on $\varepsilon$, and once chosen it works for every $n\ge N$.

2. Determine the pointwise limit of x to the nth power on [0,1] and prove convergence is not uniform. Explain why each function is continuous but the limit is not.

   A hint

   First fix $x$ and let $n$ vary. Then reverse your viewpoint: for a fixed $n$, how close to $1$ can you choose $x<1$? A supremum need not be attained.

   Work through the solution

   For each $0\le x<1$, $x^n\to0$, while $1^n=1$. The pointwise limit $f$ is therefore zero on $[0,1)$ and one at $1$.

   For every fixed $n$,

   $$\sup_{x\in[0,1]}|x^n-f(x)|=\sup_{0\le x<1}x^n=1.$$

   The supremum is $1$ because $x^n$ approaches $1$ as $x$ approaches $1$ from below; it is not attained on $[0,1)$. Since this error does not approach zero, convergence is not uniform. Each $x^n$ is a continuous polynomial, whereas $f$ is discontinuous at $1$. This also gives a second proof: a uniform limit of continuous functions is continuous.

3. Prove that x squared is uniformly continuous on [0,1] and is not uniformly continuous on the whole real line.

   A hint

   Factor $x^2-y^2$. On the bounded interval you can control the second factor uniformly; on the line, choose nearby points whose sizes grow.

   Work through the solution

   If $x,y\in[0,1]$, then $|x^2-y^2|=|x-y||x+y|\le2|x-y|$. Given $\varepsilon>0$, choosing $\delta=\varepsilon/2$ gives the required implication for every pair in the interval.

   On $\mathbb R$, take $x_n=n$ and $y_n=n+1/n$ for $n\ge1$. Their distance is $1/n\to0$, but

   $$|y_n^2-x_n^2|=2+\frac1{n^2}>2.$$

   In particular, with $\varepsilon=1$, no single $\delta>0$ can work for all real pairs: choose $n$ so large that $1/n<\delta$. The function is continuous at every real point, but the allowed neighborhood cannot be chosen independently of the point.

4. For f_n(x)=sum from k=0 to n of x to the kth power on [0,1/2], prove the sequence is uniformly Cauchy, identify its limit, and give a uniform error bound.

   A hint

   Estimate all tails using the same geometric series with ratio 1/2. The bound must be independent of x.

   Work through the solution

   For integers $m>n\ge0$ and $0\le x\le1/2$,

   $$0\le f_m(x)-f_n(x)=\sum_{k=n+1}^{m}x^k\le\sum_{k=n+1}^{\infty}2^{-k}=2^{-n}.$$

   Given $\varepsilon>0$, choose an integer $N\ge0$ with $2^{-N}<\varepsilon$. For all $m,n\ge N$, interchange their roles if needed to obtain $\sup_x|f_m(x)-f_n(x)|\le2^{-N}<\varepsilon$. This is the uniform Cauchy condition.

   The finite geometric identity gives $f_n(x)=(1-x^{n+1})/(1-x)$ because $x\ne1$. Hence the pointwise limit is $f(x)=1/(1-x)$, and

   $$|f(x)-f_n(x)|=\frac{x^{n+1}}{1-x}\le2^{-n}.$$

   Equality holds at $x=1/2$, so the supremum error is exactly $2^{-n}$. It tends to zero, proving uniform convergence with an explicit rate.

   If this was the sticking point:

   • Calculus — Review finite geometric sums and geometric-series tails.
   • Proofs & Mathematical Writing — Track the order of quantifiers in the uniform Cauchy condition.

Already comfortable with this? Skip the sequence review if epsilon arguments and completeness proofs are reliable. Skip the full selection if you can produce the stated counterexamples and state sufficient hypotheses for interchanging a limit with an integral or derivative, not just compute sample limits.

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Reading choices revised 2026-10-17.

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