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$f(x)$ is given by: $$f(x)=1/(1/[x]),\quad 0\leq x\leq 1,$$ where $[x]$ represents the largest integer less than or equal to $x$. How to prove that $f(x)$ is discontinuous at infinitely many points on $(0,1)$?

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Are you sure that you don’t mean $$f(x)=\frac1{\lfloor 1/x\rfloor}$$ for $0\le x<1$? Because the function that you’ve written makes no sense. – Brian M. Scott Oct 15 '12 at 2:45
$f$ isn't defined for any $x \in (0, 1)$ as $[x] = 0$ (usually I would use the notation $\lfloor x\rfloor$). – Michael Albanese Oct 15 '12 at 2:47
I have edited your post, please let me know if this really agree with what you want to ask. Since you have already received at least one good answer to each you've asked please consider accept your answers. – leo Oct 15 '12 at 4:24
Do you know about accepting answers to questions you post to this site? Please read up on it and consider doing it. – Gerry Myerson Oct 15 '12 at 4:25
up vote 2 down vote accepted

I assume you mean $$f(x) = \dfrac1{\lfloor 1/x \rfloor}$$ For $x \in \left(\dfrac1{n+1}, \dfrac1n \right]$, we have $\dfrac1x \in \left[n,n+1 \right)$. This means $f(x) = \dfrac1n$.

Do you now see the points of discontinuity of $f(x)$?

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