Section 4.3 Basic Properties

The definition of continuity is local and quantitative, but in practice it is often convenient to test continuity using sequences. Once that criterion is available, several basic permanence properties follow immediately: sums, products, quotients, and compositions of continuous functions are again continuous. These results will let us build large classes of examples from a small stock of simple continuous functions.

Subsection 4.3.1 Sequential Characterization

The following criterion replaces the \(\varepsilon\)-\(\delta\) definition with a statement about images of convergent sequences. It is often the most efficient way to prove that a function is continuous.

Proposition 4.3.1.

Let \(S \subseteq \R\text{,}\) let \(f \colon S \to \R\text{,}\) and let \(c \in S\text{.}\) Then \(f\) is continuous at \(c\) if and only if for every sequence \((x_n)\) in \(S\) with \(x_n \to c\text{,}\) we have

\begin{equation*} f(x_n) \to f(c). \end{equation*}

Proof.

First suppose that \(f\) is continuous at \(c\text{,}\) and let \((x_n)\) be a sequence in \(S\) such that \(x_n \to c\text{.}\) Let \(\varepsilon \gt 0\) be given. By continuity at \(c\text{,}\) there exists \(\delta \gt 0\) such that \(|f(x)-f(c)| \lt \varepsilon\) whenever \(x \in S\) and \(|x-c| \lt \delta\text{.}\) Since \(x_n \to c\text{,}\) \(|x_n-c| \lt \delta\) for all but finitely many \(n\text{.}\) Hence \(|f(x_n)-f(c)| \lt \varepsilon\) for all but finitely many \(n\text{,}\) so \(f(x_n) \to f(c)\text{.}\)

Conversely, assume \(f(x_n) \to f(c)\) for every sequence \((x_n)\) in \(S\) that converges to \(c\) but \(f\) is not continuous at \(c\text{.}\) Then there exists \(\varepsilon_0 \gt 0\) such that for every \(n \in \N\) there exists \(x_n \in S\) with \(|x_n-c| \lt 1/n\) and

\begin{equation*} |f(x_n)-f(c)| \ge \varepsilon_0. \end{equation*}

Clearly, \(x_n \to c\text{,}\) but \(f(x_n)\) does not converge to \(f(c)\text{,}\) contradicting the hypothesis. Therefore \(f\) must be continuous at \(c\text{.}\)

We will use this sequential characterization of continuity repeatedly. Once continuity is translated into a statement about convergent sequences, the usual algebraic properties of limits give the corresponding algebraic properties of continuous functions.

Subsection 4.3.2 Algebra of Continuous Functions

Proposition 4.3.2.

Let \(f,g \colon S \to \R\) be continuous at \(c \in S\text{,}\) then so are \(f+g\) and \(fg\text{.}\) Moreover, if \(g(c) \neq 0\text{,}\) then \(1/g\) and hence \(f/g\) are also continuous at \(c\text{.}\)

Proof.

Pick a sequence \((x_n)\) in \(S\) that converges to \(c\text{.}\) Since \(f\) and \(g\) are continuous at \(c\text{,}\) the sequential characterization gives

\begin{equation*} f(x_n) \to f(c) \quad\text{and}\quad g(x_n) \to g(c). \end{equation*}

Therefore Proposition 2.1.6 implies that

\begin{equation*} (f+g)(x_n)=f(x_n)+g(x_n) \to f(c)+g(c)=(f+g)(c) \end{equation*}

and

\begin{equation*} (fg)(x_n)=f(x_n)g(x_n) \to f(c)g(c)=(fg)(c). \end{equation*}

Hence \(f+g\) and \(fg\) are continuous at \(c\text{.}\)

Now assume \(g(c) \neq 0\text{.}\) Since \(g(x_n) \to g(c)\text{,}\) we have \(g(x_n) \neq 0\) for all sufficiently large \(n\text{.}\) Thus, by dropping finitely many terms if necessary, we may assume \(g(x_n) \neq 0\) for all \(n\text{.}\) Then the reciprocal part of Proposition 2.1.6 gives

\begin{equation*} \frac{1}{g(x_n)} \to \frac{1}{g(c)}. \end{equation*}

Thus \(1/g\) is continuous at \(c\) by the sequential characterization of continuity. Applying the product result to \(f\) and \(1/g\text{,}\) we conclude that \(f/g=f(1/g)\) is continuous at \(c\text{.}\)

Proposition 4.3.3.

Let \(S,T \subseteq \R\text{,}\) let \(f \colon S \to T\text{,}\) and let \(g \colon T \to \R\text{.}\) If \(f\) is continuous at \(c \in S\) and \(g\) is continuous at \(f(c)\text{,}\) then the composition \(g \circ f\) is continuous at \(c\text{.}\)

Proof.

Let \((x_n)\) be a sequence in \(S\) with \(x_n \to c\text{.}\) Since \(f\) is continuous at \(c\text{,}\) we have \(f(x_n) \to f(c)\text{.}\) The sequence \((f(x_n))\) lies in \(T\text{,}\) and because \(g\) is continuous at \(f(c)\text{,}\) it follows that

\begin{equation*} (g \circ f)(x_n)=g(f(x_n)) \to g(f(c))=(g \circ f)(c). \end{equation*}

Therefore \(g \circ f\) is continuous at \(c\text{.}\)

Corollary 4.3.4.

Every polynomial is continuous on \(\R\text{,}\) and every rational function is continuous on its domain.

Proof.

Constant functions and the identity function are continuous. By repeated application of the preceding proposition, finite sums and products of continuous functions are continuous, so every polynomial is continuous on \(\R\text{.}\) A rational function is a quotient \(p/q\) of two polynomials, and the quotient part of the same proposition shows that \(p/q\) is continuous at every point where \(q \neq 0\text{.}\)

Subsection 4.3.3 Some Consequences

Continuity does more than preserve limits under algebraic operations. It also imposes rigidity: a continuous function is determined by its values on a dense subset, and a nonzero value at one point forces the function to keep the same sign nearby.

Proposition 4.3.5.

Let \(S \subseteq \R\text{,}\) let \(D \subseteq S\) be dense in \(S\text{,}\) and let \(f \colon S \to \R\) be continuous on \(S\text{.}\) If \(f(x)=0\) for every \(x \in D\text{,}\) then \(f(x)=0\) for every \(x \in S\text{.}\)

Proof.

Fix \(c \in S\text{.}\) Since \(D\) is dense in \(S\text{,}\) for each \(n \in \N\) we may choose \(d_n \in D\) such that \(|d_n-c| \lt 1/n\text{.}\) Then \(d_n \to c\text{.}\) Because \(f\) is continuous at \(c\text{,}\) the sequential characterization gives \(f(d_n) \to f(c)\text{.}\) But every \(d_n\) lies in \(D\text{,}\) so \(f(d_n)=0\) for all \(n\text{.}\) Hence the constant sequence \((f(d_n))\) converges to \(0\text{,}\) and therefore \(f(c)=0\text{.}\) Since \(c\) was arbitrary, \(f=0\) on \(S\text{.}\)

Corollary 4.3.6.

Let \(S \subseteq \R\) and let \(D \subseteq S\) be dense in \(S\text{.}\) If \(f,g \colon S \to \R\) are continuous on \(S\) and \(f(x)=g(x)\) for every \(x \in D\text{,}\) then \(f=g\) on \(S\text{.}\)

Proof.

The function \(f-g\) is continuous on \(S\) and vanishes on \(D\text{.}\) By the proposition, \(f-g=0\) on \(S\text{.}\) Therefore \(f=g\) on \(S\text{.}\)

Proposition 4.3.7.

Let \(S \subseteq \R\text{,}\) let \(f \colon S \to \R\text{,}\) and let \(c \in S\text{.}\) If \(f\) is continuous at \(c\) and \(f(c) \neq 0\text{,}\) then there exists \(\delta \gt 0\) such that \(f(x)\) has the same sign as \(f(c)\) whenever \(x \in S\) and \(|x-c| \lt \delta\text{.}\)

Proof.

Let \(\varepsilon = |f(c)|/2\text{.}\) Since \(f\) is continuous at \(c\text{,}\) there exists \(\delta \gt 0\) such that

\begin{equation*} |f(x)-f(c)| \lt \frac{|f(c)|}{2} \end{equation*}

whenever \(x \in S\) and \(|x-c| \lt \delta\text{.}\) If \(f(c) \gt 0\text{,}\) then

\begin{equation*} f(x) \gt f(c)-\frac{|f(c)|}{2}=\frac{f(c)}{2}\gt 0. \end{equation*}

If \(f(c) \lt 0\text{,}\) then

\begin{equation*} f(x) \lt f(c)+\frac{|f(c)|}{2}=\frac{f(c)}{2}\lt 0. \end{equation*}

In either case, \(f(x)\) has the same sign as \(f(c)\text{.}\)

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