Continuity is the next property of curves that make them nice. Like with limits continuity is defined at a point and we say the function is continuous on an interval if the function is continuous at each point in that interval.
Our first step in understanding continuity is to consider the examples in Figure 1.8.1 which shows examples of points on curves where the function is continuous, and Figure 1.8.2 which shows examples of points on curves where the function is discontinuous.
all points
all points
all points
\([-2,0) \cup (0,2]\)
\([0,1) \cup (1,2]\)
\([0,2) \cup (2,4]\)
\([-3,-1) \cup (-1,1) \cup (1,4]\)
\((0,1) \cup (1,2) \cup \ldots \)
\((-\infty,0) \cup (0,\infty)\)
Figure1.8.1.The functions are continuous at the given points.
\(x=0\)
\(x=1\)
\(x=2\)
\(x \in \{ -1, 1\}\)
\(x \in \{ 0, 1, 2, 3, \ldots \}\)
\(x=0\)
Figure1.8.2.The functions are discontinuous at the given points.
Activity1.8.1.Describe Definining Traits of Continuity.
The goal of this activity is to determine traits of continuous functions.
Based on the example in Figure 1.8.1, Figure 1.8.2, and Figure 1.8.3 describe conditions for continuity. Note it may be easier to describe discontinuity.
(c)
Can the limit exist at a point that is discontinuous?
Figure1.8.3.Determine where these are continuous.
Activity1.8.2.Common description of continuity.
(a)
Draw a function that is continuous on \([0,3]\text{.}\)
(b)
Draw a function that has a discontinuity on \([0,3]\)
(c)
Draw a function that has a different type of discontinuity on \([0,3]\)
(d)
What do these experiments hint is true about drawing continuous functions.
(e)
Review the Cantor function (day 1 materials). How does this compare to your guess above?
Checkpoint1.8.4.Intervals of discontinuity.
From the examples above it may appear the discontinuities occur solely at a single point or on an interval where the function is undefined. However, there exists functions that are discontinuous on an interval where they are defined.
\begin{equation*}
sp(x) = \begin{cases} 0 \amp \text{ if } x \text{ is irrational} \\ 1 \amp \text{ if } x \text{ is rational} \end{cases}
\end{equation*}
\begin{equation*}
confetti(x) = \begin{cases} 1 \amp \text{ if } x \text{ is irrational} \\ \frac{1}{p} \amp \text{ if } x=\frac{n}{p^k} n,k \in \Z^+, p \text{ is prime} \\ 0 \amp \text{ otherwise} \end{cases}
\end{equation*}
Show that the salt and pepper (\(sp(x)\)) function is discontinuous everywhere.
Note, the confetti function is also discontinuous everywhere.
