Continuous Functions

Definition

Let f \colon D \to \mathbb{R}, where x,c \in D. We say that f is continuous at c if

\forall \varepsilon > 0, \exists \delta > 0, |x-c| < \delta \implies |f(x) - f(c)| < \varepsilon.

In English, this means a function is continuous if x approaching c implies the difference between f(x) and f(c) is arbitrarily small.

If f is continuous at all points x \in D, we say that f is continuous on D. Note that this definition of continuity is a pointwise notion.

Theorem 2.2.1

Let f \colon D \to \mathbb{R} and let c \in D. Then the following three conditions are equivalent

(a) f is continuous at c.

(b) If (x_n) is any sequence in D such that (x_n) converges to c, then \lim f(x_n) = f(c).

(c) For every neighborhood V of f(c) there exists a neighborhood U of c such that f(U \cap D) \subseteq V.

Futhermore if c is an accumulation point of D, then the above are all equivalent to

(d) f has a limit at c and \lim_{x \to c} f(x) = f(c).

We will assume the function is continuous at c, and show it implies the rest of the properties.

(a) \implies (c)

Then there exists a neighborhood U of c such that U \cap D = \{c\}. It follows that, for any neighborhood V of f(c), f(U \cap D) = \{f(c)\} \subseteq V. Thus (a) implies (c).

Taking the negation of (a) and (b) in this Theorem, we have a useful Corollary for what it means for a function to be discontinous.

Corollary 2.2.2

Let f \colon D \to \mathbb{R} and c \in D. Then f is discontinous at c if and only if there exists a sequence (x_n) in D such that (x_n) converges to c but the sequence (f(x_n)) does not converge to f(c).

Example

Take the function f \colon \mathbb{R} \to \mathbb{R} where

f(x) = \begin{cases} 1 \text{ if $x$ is rational}\\ 0 \text{ if $x$ is irrational} \end{cases}

Claim: f is discontinous at every real number.

For any point c \in \mathbb{R}, take the neighborhood V of f(c) with \varepsilon < 1. The neighborhood U of c will always contain both rational and irrational points (Due to both being dense in \mathbb{R}), such that f(U \cap D) \not \subseteq V. Thus by Theorem 2.2.1, the function is not continous at c.

Theorem 2.2.3 (Algebraic Relations of Continuity)

Let f,g\; \colon D \to \mathbb{R} and c \in D. If f,g are both continous at the point c, then

(a) f + g and fg are continous at c, and

(b) f/g is continous at c if g(c) \not = 0.

Use convergence criterion and previous results.

Theorem 2.2.4

Let f \colon D \to \mathbb{R} and g \colon E \to \mathbb{R} such that f(D) \subseteq E.

If f is continuous at c \in D and g is continuous at f(c), then g \circ f \colon D \to \mathbb{R} is continuous at c.

We want to show that given any neighborhood V of g(f(c)), there exists a neighborhood U of c such that f(U \cap D) \subseteq V.

Since g is continous at f(c), we know that there exists a neighborhood W of f(c) such that g(W \cap E) \subseteq V. Similarly since f is continous at c, there exists a neighborhood U of c such that f(U \cap D) \subseteq W.

Theorem 2.2.5

A function f\colon D \to \mathbb{R} is continous on D if and only if for every open set U of \mathbb{R}, there exists an open subset V of \mathbb{R} such that V \cap D = f^{-1}(U).

WTS: f being continous on D implies V \cap D = f^{-1}(u) .

Assume f is continous on D. Let U be an open subset of \mathbb{R}. If c \in f^{-1}(U), then f(c) \in U. Since U is open, there exists a neighborhood U^\prime \subseteq U containing f(c). Since f is continous, there exists a neighborhood V(c) of c such that f(V(c) \cap D) \subseteq U^\prime.

Let V = \bigcup_{c \in f^{-1}(u)}V(c). Now since V(c) is open, we know V is open and V \cap D = f^{-1}(U). \square

WTS: V \cap D = f^{-1}(u) implies that f is continous on D.

Note that the pre-image of a set may not exist. In this case we would say f^{-1}(U) is the empty set, \varnothing.

$And when D is \mathbb{R}, we can write

Corollary 2.2.6

A function f \colon \mathbb{R} \to \mathbb{R} is continuous if and only if f^{-1}(G) is open in \mathbb{R}

Practice

Recap

In this section we proved

Theorem 2.2.1

Let f \colon D \to \mathbb{R} and let c \in D. Then the following three conditions are equivalent

(a) f is continuous at c.

(b) If (x_n) is any sequence in D such that (x_n) converges to c, then \lim f(x_n) = f(c).

(c) For every neighborhood V of f(c) there exists a neighborhood U of c such that f(U \cap D) \subseteq V.

Futhermore if c is an accumulation point of D, then the above are all equivalent to

(d) f has a limit at c and \lim_{x \to c} f(x) = f(c).

Corollary 2.2.2

Let f \colon D \to \mathbb{R} and c \in D. Then f is discontinous at c if and only if there exists a sequence (x_n) in D such that (x_n) converges to c but the sequence (f(x_n)) does not converge to f(c).

Theorem 2.2.3 (Algebraic Relations of Continuity)

Let f,g\; \colon D \to \mathbb{R} and c \in D. If f,g are both continous at the point c, then

(a) f + g and fg are continous at c, and

(b) f/g is continous at c if g(c) \not = 0.

Theorem 2.2.4

Let f \colon D \to \mathbb{R} and g \colon E \to \mathbb{R} such that f(D) \subseteq E.

If f is continuous at c \in D and g is continuous at f(c), then g \circ f \colon D \to \mathbb{R} is continuous at c.

Theorem 2.2.5

A function f\colon D \to \mathbb{R} is continous on D if and only if for every open set U of \mathbb{R}, there exists an open subset V of \mathbb{R} such that V \cap D = f^{-1}(U).

Corollary 2.2.6

A function f \colon \mathbb{R} \to \mathbb{R} is continuous if and only if f^{-1}(G) is open in \mathbb{R}