$f(x)$ 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 $(0,1)$?
How to prove that $f(x)$ is discontinuous at infinitely many points on $(0,1)$?
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analysis
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4Are 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. – 2012-10-15
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0$f$ isn't defined for any $x \in (0, 1)$ as $[x] = 0$ (usually I would use the notation $\lfloor x\rfloor$). – 2012-10-15
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1I 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](http://meta.math.stackexchange.com/q/3286/8271). – 2012-10-15
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0Do you know about accepting answers to questions you post to this site? Please read up on it and consider doing it. – 2012-10-15
1 Answers
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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 $f(x)$?