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Let $f:\mathbb R$→$\mathbb R$ defined as - $f(x)=0$, if $x$ is irrational or $x=0$ and $f(x)=1/q$, if $x=p/q$, $p\in$$\mathbb Z$ ,$q\in$$\mathbb N$, $(p,q)=1$. What are the points of continuity of $f(x)$? I wanted to use simple left hand and right hand limits criteria but the uncertainity of $\epsilon>0$ (rational as well as rational) is restricting my thoughts to evaluate these.

Arthur
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Nitin Uniyal
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1 Answers1

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Hint: For a neighbourhood around an irrational number, the smaller that neighbourhood is, the larger the smallest denominator of any rational number in that neighbourhood becomes.

Arthur
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  • which can be seen for example by the continued fraction expansion for that irrational. – snulty Jun 25 '16 at 16:23
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    @snulty I was thinking more along the lines of "I want a neigbourhood of the irrational number $x$ so that any rational number in that neighbourhood has denominator larger than $n$. Since there are only finitely many rational numbers between $x-1$ and $x+1$ that has too small denominators for me, there is one that is closest to $x$. I choose the neighbourhood to be smaller than that." – Arthur Jun 25 '16 at 16:26
  • That's actually also a nice way to see it. I had't thought of that, thanks. – snulty Jun 25 '16 at 16:29
  • @Arthur...do you want to state that $f(x)$ is continuous at all irrationals and $0$? – Nitin Uniyal Jun 25 '16 at 16:52
  • @mathlover Yes, that isthe solution in this case. It is also discontinuous at all nonzero rationals. – Arthur Jun 25 '16 at 17:24
  • ok...thanks a lot.. – Nitin Uniyal Jun 25 '16 at 18:08