The Timpson Lyceum · Real Analysis (2020)

Lecture 9 (Real Analysis 2020)

Overview

This lecture opens Chapter 3 by defining what it means for a function to be continuous, building directly on the sequence-limit machinery from Chapter 2. Where Chapter 2 studied limits of sequences, the central object now is the continuous function defined on an arbitrary subset of the real numbers. We adopt the sequential definition of continuity and use it to prove specific functions continuous, to exhibit functions that fail continuity at a point or everywhere, and finally to show that sums, products, quotients, and compositions of continuous functions are again continuous. These last results give us a fast way to certify that polynomials and rational functions are continuous.

Key Concepts

Real-valued function on a subset. We study functions $f : D \to \mathbb{R}$ where $D \subseteq \mathbb{R}$ . Unlike a sequence (whose domain is $\mathbb{N}$ ), the domain $D$ can be any subset of $\mathbb{R}$ . The function uses all of $D$ , but need not hit all of $\mathbb{R}$ — its range is only a subset of $\mathbb{R}$ .
Continuity at a point (sequential definition). $f : D \to \mathbb{R}$ is continuous at $x_0 \in D$ provided that for every sequence $(x_n)$ in $D$ with $x_n \to x_0$ , we have $f(x_n) \to f(x_0)$ . In words: the image of any sequence converging to $x_0$ must converge to the image of $x_0$ : $x_n \to x_0 \;\Longrightarrow\; \lim_{n\to\infty} f(x_n) = f(x_0).$
Continuous function. $f$ is a continuous function if it is continuous at every point of $D$ . Distinguish carefully: a function may be continuous at some points and not others. Only when it is continuous at all points of its domain do we call it a continuous function.
Graphical intuition. A continuous function can be drawn without lifting the pencil — no holes, no jumps, no breaks. A jump in the graph signals failure of continuity at that point.
Polynomials are continuous. Using the limit laws for sequences (limit of a sum/difference is the sum/difference of limits; limit of a product is the product of limits; limit of a constant is the constant), every polynomial $p : \mathbb{R} \to \mathbb{R}$ satisfies $x_n \to a \Rightarrow p(x_n) \to p(a)$ , so polynomials are continuous on all of $\mathbb{R}$ .
Disproving continuity at a point. To show $f$ is not continuous at $x_0$ , exhibit a single counterexample: one sequence $x_n \to x_0$ for which $\lim f(x_n) \neq f(x_0)$ .
Algebra of continuous functions. For $f, g : D \to \mathbb{R}$ continuous (same domain):
- $(f+g)(x) = f(x) + g(x)$ is continuous;
- $(fg)(x) = f(x)\,g(x)$ is continuous;
- $\left(\tfrac{f}{g}\right)(x) = \tfrac{f(x)}{g(x)}$ is continuous on the set where $g(x) \neq 0$ (the restriction is required so the quotient is always defined).
Each proof copies the additive proof, swapping in the corresponding sequence limit law.
Corollary — rational functions. A rational function is a quotient of two polynomials. Since polynomials are continuous and the quotient of continuous functions is continuous wherever the denominator is nonzero, every rational function is continuous on its natural domain (all points where the denominator $\neq 0$ ).
Function composition. Given $f : D \to \mathbb{R}$ and $g : U \to \mathbb{R}$ with $f(D) \subseteq U$ , the composition $g \circ f : D \to \mathbb{R}$ is defined by $(g \circ f)(x) = g(f(x))$ — apply $f$ first, then $g$ (" $g$ after $f$ "). The containment $f(D) \subseteq U$ is exactly what makes every output of $f$ a legal input to $g$ . Order matters: in general $g \circ f \neq f \circ g$ .
Composition preserves continuity. If $f : D \to \mathbb{R}$ and $g : U \to \mathbb{R}$ are continuous and $f(D) \subseteq U$ , then $g \circ f$ is continuous.
Two instructive non-examples.
- A jump function — e.g. $f(x) = 1$ for $x \ge 0$ and $f(x) = 2$ for $x < 0$ — is continuous everywhere except $0$ . The sequence $x_n = -\tfrac{1}{n} \to 0$ has $f(x_n) \equiv 2 \to 2 \neq 1 = f(0)$ .
- The Dirichlet function, $f(x) = 1$ if $x \in \mathbb{Q}$ and $f(x) = 0$ if $x \notin \mathbb{Q}$ , is continuous nowhere. Because both $\mathbb{Q}$ and its complement are dense in $\mathbb{R}$ , every point has both a sequence of rationals and a sequence of irrationals converging to it, and their images ( $1$ 's vs. $0$ 's) disagree.

Worked Example

Claim. $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = x^2 - 2x + 4$ , is continuous.

Let $a \in \mathbb{R}$ be arbitrary and let $(a_n)$ be any sequence in $\mathbb{R}$ with $\lim_{n\to\infty} a_n = a$ . We must show $\lim_{n\to\infty} f(a_n) = f(a)$ . Applying the sequence limit laws (sum/difference, product, constant-multiple, and constant rules):

\begin{aligned} \lim_{n\to\infty} f(a_n) &= \lim_{n\to\infty}\bigl(a_n^2 - 2a_n + 4\bigr) \\ &= \lim_{n\to\infty} a_n^2 \;-\; 2\lim_{n\to\infty} a_n \;+\; \lim_{n\to\infty} 4 \\ &= \Bigl(\lim_{n\to\infty} a_n\Bigr)^2 - 2\Bigl(\lim_{n\to\infty} a_n\Bigr) + 4 \\ &= a^2 - 2a + 4 \\ &= f(a). \end{aligned}

Since $a$ and $(a_n)$ were arbitrary, $f$ is continuous at every point, hence a continuous function. $\blacksquare$

Sketch — closure under addition. For continuous $f, g : D \to \mathbb{R}$ , take any $a \in D$ and $(a_n)$ in $D$ with $a_n \to a$ . Then $\lim_{n\to\infty}(f+g)(a_n) = \lim_{n\to\infty}\bigl(f(a_n)+g(a_n)\bigr) = \lim_{n\to\infty} f(a_n) + \lim_{n\to\infty} g(a_n) = f(a) + g(a) = (f+g)(a),$ so $f+g$ is continuous. The product and quotient cases are identical with the matching limit law (the quotient additionally requiring $g \neq 0$ on $D$ ).

Summary

A function $f : D \to \mathbb{R}$ is continuous at $x_0$ iff $x_n \to x_0$ forces $f(x_n) \to f(x_0)$ for every sequence in $D$ ; it is a continuous function when this holds at every point of $D$ .
To prove continuity, take an arbitrary point and an arbitrary convergent sequence and push the limit through using the Chapter 2 limit laws; to disprove it at a point, produce one bad sequence.
Polynomials are continuous on all of $\mathbb{R}$ , and rational functions are continuous wherever their denominator is nonzero.
Sums, products, quotients (where the denominator never vanishes), and compositions ( $f(D) \subseteq U$ ) of continuous functions are continuous.
Composition is $(g \circ f)(x) = g(f(x))$ , applied right-to-left, and is generally not commutative.
Classic non-examples: a jump function fails continuity at the jump only, while the Dirichlet function (1 on rationals, 0 on irrationals) is continuous nowhere because both sets are dense in $\mathbb{R}$ .