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Section 4.1 Definition of Continuity. Examples.

Subsection 4.1.1 Continuity at a Point

Let \(S \subseteq \R\) and let \(f \colon S \to \R\) be a function. We say that \(f\) is continuous at\(c \in S\) if for every \(\varepsilon \gt 0\) there exists \(\delta \gt 0\) such that whenever \(x \in S\) and \(|x-c| \lt \delta\text{,}\) we have
\begin{equation*} |f(x)-f(c)| \lt \varepsilon. \end{equation*}
In other words, \(f\) is continuous at \(c\) if the output \(f(x)\) can be made arbitrarily close to \(f(c)\) provided that the input \(x\) is sufficiently close to \(c\text{.}\)
If \(A \subseteq S\text{,}\) then \(f\) is continuous on\(A\) if it is continuous at every point of \(A\text{.}\) We say simply that \(f\) is continuous if it is continuous on its whole domain.
The definition only refers to points \(x\) that lie in the domain \(S\text{.}\) So continuity at \(c\) concerns the behavior of \(f\) on domain points near \(c\text{;}\) we do not need \(f\) to be defined on an entire open interval around \(c\text{.}\)

Subsection 4.1.2 Cluster Points and Isolated Points

A neighborhood of a point \(c \in \R\) is a subset of \(\R\) that contains an open interval around \(c\text{.}\) In particular, every open interval containing \(c\) is a neighborhood of \(c\text{.}\)
Let \(S \subseteq \R\text{.}\) A real number \(c\) is called a cluster point of \(S\) if for every \(\delta \gt 0\) there exists \(x \in S\) such that \(x \neq c\) and \(|x-c| \lt \delta\text{.}\) In other words, \(c\) is a cluster point of \(S\) if every neighborhood of \(c\) contains a point in \(S\) other than \(c\text{.}\) A point \(c \in S\) is an isolated point of \(S\) if it is not a cluster point of \(S\text{.}\) That is, there exists \(\delta \gt 0\) such that
\begin{equation*} S \cap (c-\delta,c+\delta)=\{c\}. \end{equation*}
Equivalently, some neighborhood of \(c\) contains no points of \(S\) other than \(c\text{.}\) A cluster point of \(S\) need not belong to \(S\text{,}\) but an isolated point of \(S\text{,}\) by definition, must belong to \(S\text{.}\)

Example 4.1.1.

  • Every point of an open interval \((a,b)\) is a cluster point of \((a,b)\text{.}\)
  • The point \(0\) is a cluster point of \((0,1)\) even though \(0 \notin (0,1)\text{.}\)
  • Every integer is an isolated point of \(\Z\text{.}\)
  • In the set \(\{0\} \cup \{1/n : n \in \N\}\text{,}\) the point \(0\) is a cluster point, while each point \(1/n\) is an isolated point.

Proof.
Let \(c\) be an isolated point of the domain \(S\) of a function \(f\text{.}\) Then there exists some \(\delta_0 \gt 0\) such that \(S \cap (c-\delta_0,c+\delta_0)=\{c\}\text{.}\) So whenever \(x \in S\) and \(|x-c| \lt \delta_0\text{,}\) necessarily \(x=c\text{.}\) Hence,
\begin{equation*} |f(x)-f(c)|=|f(c)-f(c)|=0 \end{equation*}
which is less than any positive \(\varepsilon\text{.}\) This shows that \(f\) is continuous at \(c\text{.}\)
A cluster point is also called a limit point. Here is why:

Proof.
First suppose that \(c\) is a cluster point of \(S\text{.}\) We construct an injective sequence \((x_n)\) in \(S\) such that \(x_n \to c\text{.}\) Since \(c\) is a cluster point, there exists \(x_1 \in S\) such that \(0 \lt |x_1-c| \lt 1\text{.}\) Now suppose that \(x_1,\dots,x_{n-1}\) have been chosen in \(S\text{,}\) all distinct, with \(0 \lt |x_k-c| \lt 1/k\) for \(k=1,\dots,n-1\text{.}\) Let
\begin{equation*} \eta=\min\{|x_1-c|,\dots,|x_{n-1}-c|\} \gt 0. \end{equation*}
Because \(c\) is a cluster point of \(S\text{,}\) there exists \(x_n \in S\) such that
\begin{equation*} 0 \lt |x_n-c| \lt \min\{1/n,\eta\}. \end{equation*}
Then \(x_n \neq x_k\) for \(k=1,\dots,n-1\text{,}\) because \(|x_n-c| \lt |x_k-c|\text{.}\) Thus, by induction, we obtain an injective sequence \((x_n)\) in \(S\) with \(|x_n-c| \lt 1/n\) for every \(n\text{.}\) Since \(1/n \to 0\text{,}\) it follows that \(|x_n-c| \to 0\text{,}\) and hence \(x_n \to c\text{.}\)
Conversely, suppose there is an injective sequence \((x_n)\) in \(S\) such that \(x_n \to c\text{.}\) Let \(\delta \gt 0\text{.}\) Since \(x_n \to c\text{,}\) \(|x_n-c| \lt \delta\) for all \(n\) sufficiently large. In particular, the inequality is satisfied by infinitely many terms of the sequence. At most one of them can be \(c\) by injectivity. Therefore, \(x_N \neq c\) for some sufficiently large \(N\) and so \(|x_N-c| \lt \delta\) as well. This shows that every neighborhood of \(c\) contains a point of \(S\) other than \(c\text{.}\) Therefore \(c\) is a cluster point of \(S\text{.}\)

Subsection 4.1.3 Examples of Continuous Functions

The simplest examples of continuous functions are the constant functions and the identity function \(\mathrm{id}(x)=x\text{.}\) We leave the justification of these statements as an exercise.

Proof.
Fix \(c \in \R\text{.}\) By the reverse triangle inequality,
\begin{equation*} \bigl||x|-|c|\bigr| \le |x-c| \quad (x \in \R). \end{equation*}
Therefore, for any \(\varepsilon \gt 0\text{,}\) if \(|x-c| \lt \varepsilon\text{,}\) then \(||x|-|c|| \lt \varepsilon\) as well. Since \(c\) is an arbitrary real number, this shows that the function \(x \mapsto |x|\) is continuous on \(\R\text{.}\)

Proof.
To prove the continuity of the function \(x \mapsto 1/x\text{,}\) we need, roughly speaking, to show that for any fixed \(c \neq 0\text{,}\)
\begin{equation*} \left|\frac{1}{x}-\frac{1}{c}\right| = \frac{|x-c|}{|x||c|} \end{equation*}
is small whenever \(|x-c|\) is small. Thus, we need to make sure that the quantity \(|xc|\text{,}\) and hence \(|x|\text{,}\) stays bounded away from \(0\) when \(x\) is close to \(c\text{.}\) This can be achieved because if \(|x-c| \lt |c|/2\text{,}\) then \(||x|-|c|| \lt |c|/2\text{,}\) and that guarantees \(|x| \gt |c|/2\text{.}\) Thus,
\begin{equation*} \left|\frac{1}{x}-\frac{1}{c}\right| = \frac{|x-c|}{|x||c|} \le \frac{2|x-c|}{|c|^2}. \end{equation*}
So for any \(\varepsilon \gt 0\text{,}\) if we choose \(0 \lt \delta \lt \min\{|c|/2, \varepsilon|c|^2/2\}\text{,}\) then the inequality above implies \(|1/x - 1/c| \lt \varepsilon\text{,}\) and we are done, since \(c \neq 0\) is arbitrary.

Example 4.1.6.

If the domain consists entirely of isolated points, then every function on that domain is continuous. For example, every function \(f \colon \Z \to \R\) is continuous, because each integer is an isolated point of \(\Z\text{.}\)
Later we will prove that sums, products, quotients (where the denominator is nonzero), and compositions of continuous functions are again continuous. From those general results it will follow that every polynomial is continuous on \(\R\text{,}\) every rational function is continuous on its domain, and standard functions such as \(\sin x\text{,}\) \(\cos x\text{,}\) \(e^x\text{,}\) and \(\ln x\) are continuous on their natural domains.
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