Examples of continuous functions $f:(0,1)\to[0,1]$

Question

Hi so the question is asking if if there is a continuous function over $(0,1)$ which will result in the image $[0,1]$. Thank you so much! Explanations are greatly appreciated!

Answer 1

you may consider the function $f(x)=\frac{\cos{(100x)+1}}{2}$. Surely it is continuous and reaches its maximum and minimum several times in $(0,1)$.

Hope I helped.

Answer 2

HINT: Let $f$ be any bijection from $(0, 1)$ to $(-1, 2)$. This "spills over" the interval $[0, 1]$, so what you want to do is take the parts where it spills over and "fold them back." Do you see how to do this?

Answer 3

Consider the piecewise function: \begin{cases} 2x, \qquad &\text{ for } 0 < x < \frac{1}{2}, \\ 1, &\text{ otherwise. } \end{cases}

EDIT: Oops, thought we wanted the image to be $(0, 1]$. To get the appropriate answer, consider: \begin{cases} 4x, \qquad &\text{ for } 0 < x < \frac{1}{4}, \\ 1 - 4(x - \frac{1}{4}) = 2 - 4x, &\text{ for } \frac{1}{4} < x \leq \frac{1}{2}, \\ 0, &\text{ otherwise.} \end{cases}

Answer 4

Assuming the image desired is $[0,1]$ (including the endpoints) an example continous function on $(0,1)$ would be $$ f(x) = \left\{ \begin{array}{cl} 0 & x < \frac14 \\ 2x-\frac12 & \frac14 \leq x \leq \frac34 \\ 1 & x > \frac34 \end{array}\right. $$ This function is continuous and has the desired image.

A much tougher question is whether there can be a function that is infinitely differentiable that maps this open interval onto its closure. One of the other answers shows that even this is possible.

Answer 5

Assuming you mean the range to be $[0,1]$ as the title says, $f(x)=\frac{1+ sin(2 \pi x)}{2}$ has $f((0,1))=[0,1]$.

Answer 6

\begin{align*} f(x)=\begin{cases} 0, & \text{ if $x<1/4$,} \\ 2(x-1/4), & \text{if $1/4 \leq x \leq 3/4$,} \\ 1, & \text{ if $x>3/4$.} \end{cases} \end{align*}

EDIT: Similar to previous answer. If you want to get creative with a similar idea...

\begin{align*} f(x)=\begin{cases} sin(2(x-1/4)\pi), & \text{if $1/4 \leq x \leq 3/4$,} \\ 0, & \text{ else.} \end{cases} \end{align*}

Answer 7

There are many examples of such functions. Take for example $f:(0,1) \to [0,1]$, $x \mapsto \sin^2(4 \pi x)$.

You cannot have a continuous bijection, however. Then you would have a continuous bijection $f^{-1}:[0,1] \to (0,1)$ but the preimage of the open $(0,1)$ is $[0,1]$, absurd.

Answer 8

Draw a zigzag line in the $xy$-plane as follows. Start at $(0,0);$ go in a straight line to $(\frac13,1);$ then straight to $(\frac23,0);$ then straight to $(1,1).$ This is the graph of a continuous function on $[0,1].$ (You didn't say it had to be differentiable, did you?) Erase the endpoints $(0,0$ and $(1,1)$ and it's a continuous function on $(0,1).$ What's the image?