Does there exist a function f, defined by $\displaystyle

f:\left[ {0,1} \right) \to \mathbb{R}

$, which is bounded above and below but does not ever attain the upper bound nor the lower bound? Can anyone give an example of such a function?

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- Feb 1st 2009, 07:45 AMTriAngleContinuous functions
Does there exist a function f, defined by $\displaystyle

f:\left[ {0,1} \right) \to \mathbb{R}

$, which is bounded above and below but does not ever attain the upper bound nor the lower bound? Can anyone give an example of such a function? - Feb 1st 2009, 09:48 AMflyingsquirrel
Hello

Well $\displaystyle \begin{array}{rl} f:[0,1)&\rightarrow \mathbb{R}\\ x & \mapsto \sin x\end{array}$ is bounded above and below by 2 and -2 respectively and $\displaystyle f$ never attains these two bounds...

Another example : let $\displaystyle \begin{array}{rl} f:[0,1)&\rightarrow \mathbb{R}\\ x & \mapsto x\sin\left(\frac{1}{1-x}\right)\end{array}$. This function is bounded above and below by $\displaystyle 1$ and $\displaystyle -1$ respectively but $\displaystyle f(x)$ never equals $\displaystyle \pm1$ since $\displaystyle \left|x\sin\left(\tfrac{1}{1-x}\right)\right|\leq \left|x\right| \times 1<1$. - Feb 1st 2009, 10:48 AMTriAngle
That's what I was thinking. It seems like a simple solution should work, but the problem is presented in such a way as to make it seem like it should be harder.