# Can all Continuous Functions be Drawn without Lifting the Pen from the Paper?

This is part of a series on common misconceptions.

Is this true or false?

All continuous functions can be drawn without lifting the pen from the paper.

**Why some people say its true:** Because of the name "continuous," it seems to them that these functions can be drawn continuously without any interruption or break. Often students are taught in that way in their junior classes.

**Why some people say its false:** Well, people who have a very basic knowledge of calculus don't actually say this is false, since the idea that continuous functions are continuous in the aforesaid loose sense is sufficient for them to solve ordinary application-oriented sums.

The statement is \( \color{red} {\textbf{false}}\). There are continuous functions which cannot be drawn without lifting the pen from the paper.

Explanation:Let us first begin with the formal definition of a continuous function. Let \(A\subseteq R, f:A\rightarrow R\) and \(c\in A\). Then \(f\) is said to be continuous at \(c\) if given an \(\varepsilon >0\) there exists a \(\delta >0\) such that if \(x\in A\) and \(\left| x-c \right| <\delta \), then \(\left| f(x)-f(c) \right| <\varepsilon \). If \(f\) is not continuous at \(c\), then it is said to be discontinuous at \(c\). Now note that the definition does not make any mention of whether \(c\) is a limit point of \(A\) or not (limit points or cluster points are just points such that any arbitrarily small neighborhood of them contains infinitely many points in \(A\)). If \(c\) is so, then we have to check the following three conditions to testify the continuity of \(f\) at \(c\):

\(\hspace{.7cm} \text{ (i)}\) \(f\) must be defined at \(c\), so that \(f(c)\) is meaningful.

\(\quad \text{ (ii)}\) The limit of \(f\) at \(c\) must exist in \(R\).

\(\quad \text{(iii)}\) The limit value must equal the functional value at \(c\).However, if \(c\) is not a limit point of \(A\), we see that there exists a neighborhood \({ V }_{ \delta }\left( c \right) \) of \(c\) such that \(A\cap { V }_{ \delta }\left( c \right) =\left\{ c \right\}\). Thus we conclude that \(f\) is by default continuous at those points in its domain \(A\), which are not limit points of \(A\), i.e., which are "isolated points" of \(A\).

Example:Let \(f:N\rightarrow R\) be a function such that \(f\left( x \right) =x~ \forall x\in N\). Then \(f\) seems to be discontinuous but it is actually not. See that its domain consists of isolated points only.

**Cite as:**Can all Continuous Functions be Drawn without Lifting the Pen from the Paper?.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/can-continuous-functions-be-drawn-lifting-the-pen/