Bounded function in $[0,1]$ without max and min.

Question

Do we have a function $f$ defined in $[0, 1]$, which is bounded but has no maximum and minimum ?

I do know that $\arctan x$ can give me a hint, which is bounded without max and min but that's in $\mathbb R$. Thanks!

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EDITED:
Firstly thank all the answers that I received.
I would like to ask a little additional question: could we find some edited or mixed trigonometric functions to satisfy this problem?
If yes, a example please!
The reason is, that I got stuck with the idea of $\arctan x$ and other possible trigonometric functions.

Thanks!

Answer 1

You cannot have a purely trigonometric function defined on $[0,1]$ that does not attain its max/min because it would be continuous and the extreme value theorem would apply.

$$ f(x)=\begin{cases}(1-x)\sin(\frac 1x) &; x\in(0,1]\\ 0 &; x=0 \end{cases} $$ This function is almost a trigonometric function as you specified. It has $\sup_{x\in[0,1]} f(x)=1$ and $\inf_{x\in[0,1]} f(x)=-1$ but those values are not attained.

Answer 2

$f(x)=x$ if $.25<x<.75$, and $f(x)=0.5$ otherwise

Answer 3

The function $f$ defined below has $\sup_{x\in[0,1]} f(x)=1$ and $\inf_{x\in[0,1]} f(x)=-1$ but doesn't attain those values. $$ f(x)=\begin{cases}\frac {(-1)^nn}{n+1} &; x=\frac 1n \text{ for $n\in\Bbb N$}\\ 0 &; \text{otherwise} \end{cases} $$

Answer 4

The key point is continuity, without that we can construct counterexample as

• $f(x)=x \quad x\in [0,1/2)$

• $f(x)=0 \quad x=1/2$

• $f(x)=1-x \quad x\in (1/2,1]$

Refer also to Extreme value theorem.