Example of Continuous Function Without Cluster Points

This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

Continuity at a Point

Definition: Let $f : A \to \mathbb{R}$ be a function and let $c \in A$. We say that $f$ is continuous at the point $c$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that $\forall x \in A$ with $\mid x - c \mid < \delta$ then $\mid f(x) - f(c) \mid < \epsilon$. Another definition of continuity of $f$ at $c$ is $\forall x \in A \cap V_{\delta} (c)$ we have that $f(x) \in V_{\epsilon} (f(c))$.

The graphic below illustrates the definition a function being continuous at a point $c$ in its domain.

Therefore, a function $f$ is continuous at the point $c$ if for every such $\epsilon > 0$ and corresponding $\delta > 0$, when $x$ in in the domain of $A$ and is $\delta$-close $c$, then $f(x)$ is $\epsilon$-close to $f(c)$.

Continuity at Cluster and Isolated Points

If $c$ is a cluster point of the set $A$, then $f$ is continuous at $c$ if and only if $\lim_{x \to c} f(x) = f(c)$.

If $c$ is an isolated point of the set $A$, then $f$ is always continuous at $c$. Why? For all $\epsilon > 0$ there will exist a $\delta > 0$ such that if $x \in A$ and $\mid x - c \mid < \delta$ (notice how we aren't restricting $x$ to be $c$) then $\mid f(x) - f(c) \mid < \epsilon$. In other words, if we get choose $\epsilon$ small enough, then the corresponding $\delta$-neighbourhood around $c$ will only contain $c$ and thus the corresponding $\epsilon$-neighbourhood around $f(c)$ will only contain $f(c)$ and clearly $\mid f(c) - f(c) \mid = 0 < \epsilon$ since $\epsilon$ is positive.

Continuity on a Set

Definition: Let $f : A \to \mathbb{R}$ be a function and let $B \subseteq A$. Then $f$ is continuous on the set $B$ if $\forall b \in B$, $f$ is continuous at $b$.

For example, consider the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 5$ which can geometrically be represented as a horizontal line that crosses the $x$-axis at the point $(0, 5)$. Since $f(x) = 5$ for all $x \in \mathbb{R}$, it shouldn't be too hard to understand that $f$ is continuous on all of $\mathbb{R}$.

Another example if the function $f(x) = x$ which is the diagonal line that intersects the origin.

Yet another example is any function $p : \mathbb{R} \to \mathbb{R}$ that is an $n^{\mathrm{th}}$-degree polynomial defined by $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ where $a_0, a_1, ..., a_n \in \mathbb{R}$. $p$ will be continuous on all of $\mathbb{R}$ as well.

The last three examples depicted functions there were continuous on all of $\mathbb{R}$. This is not always the case though. For example, consider the function $f : (0, \infty) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$. The domain of this function is $\{ x \in \mathbb{R} : x > 0 \}$. We will show that $f$ is continuous on its entire domain.

Let $c$ be a member of the domain of $f$, i.e., $c \in D(f) = (0, \infty)$. Let $\epsilon > 0$ be given. We need to find $\delta > 0$ such that $\forall x \in D(f)$ where $\mid x - c \mid < \delta$ then $\biggr \rvert \frac{1}{x} - \frac{1}{c} \biggr \rvert < \epsilon$. Now:

(1)

\begin{align} \biggr \rvert \frac{1}{x} - \frac{1}{c} \biggr \rvert = \frac{\mid x - c \mid}{xc} \end{align}

We want $\frac{\mid x - c \mid}{xc} < \epsilon$. Let $\delta = \mathrm{min} \{ \frac{c}{2}, \frac{\epsilon c^2}{2} \}$. Then:

(2)

\begin{align} \frac{\mid x - c \mid}{xc} < \frac{2\mid x - c \mid}{c^2} < \frac{ 2\epsilon c^2}{2c^2} = \epsilon \end{align}

Therefore $f$ is continuous for all $c$ in its domain $D(f)$.

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Source: http://mathonline.wikidot.com/continuous-functions

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