Theoretical Continuity

A continuous function couldn’t be more intuitive—its a function that is continuous! Visually, this means drawing the graph of the function without ever needing to lift up your pencil (or for the more hardcore among you, Hagoromo chalk). But that’s all hand-wavy. What does it mean, and what can we prove using it?

Definition 1. Let $K \subseteq \mathbb R$ . A function $f : K \to \mathbb R$ is continuous at $c \in K$ if for any $\epsilon > 0$ , there exists $\delta > 0$ such that for any $x \in K$ ,

$|x-c| < \delta \quad \Rightarrow \quad |f(x) - f(c)| < \epsilon.$

We say that $f$ is continuous on $K$ if $f$ is continuous at every $c \in K$ .

This definition is one of the more general forms we have, but has a convenient equivalence, known as sequential continuity.

Theorem 1. Let $K \subseteq \mathbb R$ and $f : K \to \mathbb R$ be a function. Then $f$ is continuous at $c$ if and only if for any sequence $\{c_n\}$ , $c_n \to c$ implies $f(c_n) \to f(c)$ . In the latter case, we say that $f$ is sequentially continuous at $c$ .

Proof. We first prove $(\Rightarrow)$ directly. Fix $\{c_n\}$ such that $c_n \to c$ . Fix $\epsilon > 0$ . Since $f$ is continuous at $c$ , there exists $\delta > 0$ such that

$|x - c| < \delta \quad \Rightarrow \quad |f(x) - f(c)| < \epsilon.$

Since $c_n \to c$ , there exists $N \in \mathbb N$ such that

$n > N \quad \Rightarrow \quad |c_n - c| < \delta \quad \Rightarrow \quad |f(c_n) - f(c)| <\epsilon.$

This implies $f(c_n) \to f(c)$ , as required.

We will prove $(\Leftarrow)$ by contrapositive (please let me know if there is a direct approach). Suppose $f$ is not continuous at $c$ . Unraveling definitions, this means there exists $\epsilon_0 > 0$ such that for any $\delta > 0$ , there exists $x \in K$ such that

$|x - c| < \delta \quad \text{and}\quad |f(x) - f(c)| \geq \epsilon_0$

We will now construct a sequence $\{c_n\}$ such that $c_n \to c$ and yet $f(c_n) \not\to f(c)$ . For any $n \in \mathbb N$ , $1/n > 0$ . Thus, for each $\delta_n = 1/n$ , choose $c_n \in K$ such that

$\displaystyle |c_n - c| < \delta_n = \frac 1n \quad \text{and}\quad |f(c_n) - f(c)| \geq \epsilon_0.$

It is clear that $c_n \to c$ . To show that $f(c_n) \not\to f(c)$ , we will prove by contradiction. Suppose for a contradiction that $f(c_n) \to f(c)$ . This means that for the positive number $\epsilon := \epsilon_0/2 > 0$ , there exists $N \in \mathbb N$ such that

$\displaystyle n > N \quad \Rightarrow \quad |f(c_n) - f(c)| < \epsilon = \frac{\epsilon_0}{2}.$

However, consider $n := N+1 > N$ . By our construction,

$\displaystyle \frac{\epsilon_0}{2} < \epsilon_0 \leq |f(c_n) - f(c)| < \frac{\epsilon_0}{2},$

a contradiction. Thus, $f(c_n) \not\to f(c)$ , as required.

Now, who cares about continuity? Well, continuity turns out to be essential to two incredibly useful theorems in calculus—the intermediate value theorem and the extreme value theorem. Now, these ideas could be generalised via topics in topology, but that will require more motivation for future discussion (topology basically expands our ideas of convergence beyond the real numbers).

Intermediate Value Theorem. Let $f : [a, b] \to \mathbb R$ be a function. If $f$ is continuous on $[a, b]$ and $f(a) < 0$ and $f(b) > 0$ , then there exist $c \in (a, b)$ such that $f(c) = 0$ .

Proof. Define the nonempty bounded set $K := \{x \in [a, b]: f(x) < 0\} \subseteq [a, b]$ and denote $c := \sup K < b$ . We claim that $f(c) = 0$ . Find a sequence $c_n^- \to c$ where $c_n^- \in K$ , so that $c_n^- \leq c$ . By sequential continuity, since each $f(c_n^-) \leq 0$ ,

$\displaystyle f(c) = \lim_{n \to \infty} f(c_n^-) \leq 0.$

On the other hand, for any $n \in \mathbb N^+$ , $c_n^+ := c + (b-c)/n \notin K$ so that $f(c_n^+) > 0$ . Since $(b-c)/n \to 0$ , $c_n^+ \to c + 0 = c$ . By sequential continuity, since each $f(c_n^+) \geq 0$ ,

$\displaystyle f(c) = \lim_{n \to \infty} f(c_n^+) \geq 0.$

Thus, we have $0 \leq f(c) \leq 0$ , which means $f(c) = 0$ , as required.

Corollary 1. Let $f : [a, b] \to \mathbb R$ be a function with $f(a) < f(b)$ . Suppose $f$ is continuous on $[a, b]$ . Then, for any $m \in (f(a), f(b))$ , there exists $c \in (a, b)$ such that $f(c) = m$ .

Proof. Fix $m \in (f(a),f(b))$ . Apply the vanilla intermediate value theorem to the function $f(\cdot) - m$ to find $c \in (a, b)$ such that $f(c) - m = 0 \iff f(c) = m$ .

Coupled with the Bolzano-Weierstrass theorem and several notions related to compactness in topology, we can prove the extreme value theorem.

Lemma 1. Let $f : [a, b] \to \mathbb R$ be a function. If $f$ is continuous on $[a, b]$ , then $f([a, b])$ is bounded.

Proof. If we can show that $f([a, b])$ is bounded above, then a symmetric argument shows that $f([a, b])$ is bounded below.

Suppose for a contradiction that $f([a, b])$ is not bounded above. This means that for any $n \in \mathbb N$ , there exists $x_n \in [a, b]$ such that $f(x_n) \geq n$ .

Now, the sequence $\{x_n\}$ is bounded and contained in $[a, b]$ . By the Bolzano-Weierstrass theorem, $\{x_n\}$ contains a convergent subsequence $\{x_{n_k}\}$ . Denote $x_{n_k} \to x \in [a, b]$ .

Since $f$ is continuous at $x$ , it is sequentially continuous at $x$ , so that $f(x_{n_k}) \to f(x)$ . This means that for any $\epsilon > 0$ , there exists $K \in \mathbb N$ such that

$k > K \quad \Rightarrow \quad |f(x_{n_k}) - f(x)| < \epsilon.$

Particularise to $\epsilon = 1$ so that

$k > K \quad \Rightarrow \quad f(x_{n_k}) < f(x) + 1.$

However, by the Archimedean property of $\mathbb R$ , there exists $N \in \mathbb N$ such that $N > f(x) + 1$ . This means that for the integer $M := \max\{K, N\} + 1$ ,

$f(x) + 1 < M \leq f(x_{n_M}) < f(x) + 1,$

a contradiction. Therefore, $f([a, b])$ must be bounded above. Applying the argument to the continuous function $-f$ yields $-f([a,b])$ being bounded above. This implies $f([a, b])$ is bounded below.

Extreme Value Theorem. Let $f : [a, b] \to \mathbb R$ be a function. If $f$ is continuous on $[a, b]$ , then there exist $c_1,c_2 \in [a, b]$ such that for any $x \in [a, b]$ ,

$f(c_1) \leq f(x) \leq f(c_2).$

Proof. Since $f$ is continuous on $[a, b]$ , $f([a, b])$ is bounded and obviously nonempty. Denote $M := \sup f([a, b])$ . This also means that for any $n$ , there exists $y_n \in f([a, b])$ such that $M-1/n < y_n \leq M$ . Thus, $y_n \to M$ .

For each $y_n$ , there exists $x_n \in [a, b]$ such that $y_n = f(x_n)$ . Thus, $\{x_n\}$ is a bounded sequence in $[a, b]$ that does not necessarily converge.

But by the Bolzano-Weierstrass theorem, $\{x_n\}$ contains a convergent subsequence $\{x_{n_k}\}$ with $x_{n_k} \to c_2 \in [a, b]$ .

By the sequential continuity of $f$ ,

$\displaystyle f(c_2) = \lim_{k \to \infty} f(n_k) = \lim_{k \to \infty} y_{n_k} = M.$

By definition of $M$ , for any $x \in [a, b]$ , $f(x) \leq M = f(c_2)$ , as required. For the lower bound, apply the previous result to the function $-f$ , so that there exists $c_1$ such that for any $x \in [a, b]$ ,

$-f(x) = (-f)(x) \leq (-f)(c_1) = -f(c_1)\quad \Rightarrow \quad f(x) \geq f(c_1).$

—Joel Kindiak, 16 Dec 24, 1948H

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Theoretical Continuity

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