# Range $R=[0,1]$ of function with domain $(0,1)$

Is there a function g continuous in the domain $(0,1)$ with range $R=[0,1]$.

This is a first semester calculus question, therefore I am curious about the depth someone has to reach to prove this one. I believe that simple stating that for any function f that in domain $(0,1)$ $$\lim_{x\to z^-} f(x)= \lim_{x\to z^+} f(x) =f(z)$$ is true $\forall z\in (0,1)$ because the Range has no discontinuities is not enough.

• I don't understand your question: you haven't written a statement to prove yet. You just said "let $g$ satisfy two properties" without a conclusion.
– user296602
Commented Mar 6, 2018 at 21:32
• Well, I just reread it and you are right, I'll try to rephrase it. Commented Mar 6, 2018 at 22:27
• The answers here already seem to address the question, but just as an instructional moment.. If you had the thought that for continuous functions, the pre-image of closed sets must be closed, the pre-image of the range is the domain, but $[0,1]$ is closed whereas $(0,1)$ is open, keep in mind that it's being relatively open/closed that matters (i.e., $(0,1)$ is relatively closed in $(0,1)$). Commented Mar 6, 2018 at 22:51

1) Yes there exists such function. Take for example $f(x)=\frac{1}{2}(1+\sin(2πx))$.
2) Continuity of $g$ doesn't guarantee that a function has that property. Indeed $g(x)=1$ doesn't have that property.
• Super small comment, but in order to have $\text{Ran} (f) = [0,1]$ on $(0,1)$, you probably want $f(x) = \frac{1}{2} \cdot \left(1 + \sin(2\pi x) \right)$ (i.e., half of what you have for $(1)$) Commented Mar 6, 2018 at 22:47
The simplest continuous function with bounded range familiar to all students of Calc 1 is $f(x) = \sin x$. Its domain is the line and its range is $[-1,1]$. The function $g(x) = \frac 12 (\sin x + 1)$ is continuous, its domain is the line, and its range is $[0,1]$.
To get a continuous function on $(0,1)$ whose range is $[0,1]$ just boost the periods up a bit. Try $g(x) = \frac 12 (\sin 100x + 1)$, for instance.