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

Explain the answer.

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.

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    $\begingroup$ 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. $\endgroup$
    – user296602
    Commented Mar 6, 2018 at 21:32
  • $\begingroup$ Well, I just reread it and you are right, I'll try to rephrase it. $\endgroup$
    – Konstantinos Zafeiris
    Commented Mar 6, 2018 at 22:27
  • $\begingroup$ 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 $(0,1)$). $\endgroup$
    – AlkaKadri
    Commented Mar 6, 2018 at 22:51

2 Answers 2

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To be honest the phrasing of the question makes very little sense. Let me try to answer what I think you are asking:

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.

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  • $\begingroup$ 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 $(1)$) $\endgroup$
    – AlkaKadri
    Commented Mar 6, 2018 at 22:47
  • $\begingroup$ @AlkaKadri Of course you are right ! I will edit it . Thanks :) $\endgroup$
    – Nick A.
    Commented Mar 6, 2018 at 23:21
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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.

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