# $\epsilon$-$\delta$ proof that $f$ is continuous for $x\notin\mathbb Q$ but isn't for $x\in\mathbb Q$

I'm trying to give an $\epsilon$-$\delta$ proof that the following function $f$ is continuous for $x\notin\mathbb Q$ but isn't for $x\in\mathbb Q$.

Let $f:\mathbb{A\subset R\to R}, \mathbb{A=\{x\in R| x>0\}}$ be given by: $$f(x) = \begin{cases} 1/n,&x=m/n\in\mathbb Q \\ 0,&x\notin\mathbb Q \end{cases}$$

where $m/n$ is in the lowest terms.

Can anyone help me with this proof (I'd prefer an answer with an $\epsilon$-$\delta$ proof).

Thank you very much!

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Your function is not yet well-defined: for example, $1=1/1=2/2$, so your definition gives $f(1)=1$ and $f(1)=1/2$. –  Chris Eagle Jan 10 '13 at 21:02
Presumably you want to specify that $x=m/n$ in lowest terms. –  Brian M. Scott Jan 10 '13 at 21:03
@ChrisEagle This is actually from an exercise. I suppose it means the simplified fractions, but it didn't make that clear. I'll edit the question. Thanks! –  Gabriel Bianconi Jan 10 '13 at 21:03
i can't do either ... nice question +1 and would love to see the solution –  Santosh Linkha Jan 10 '13 at 21:08
@ChrisEagle Just as I mentioned at this other similar comment, this is not as constructive as it could be. We shouldn't really use minutae as trout to whack about posters when their real question is obvious. –  rschwieb Jan 10 '13 at 21:11

HINTS: To show that $f$ is continuous at an irrational $x$, show for any $n\in\Bbb Z^+$ you can choose $\delta>0$ small enough so that the interval $(x-\delta,x+\delta)$ contains no fraction with a denominator $\le n$ in lowest term. Use the fact that between two rationals with denominator $m$ there is a gap of at least $\frac1m$.

To show that $x$ is discontinuous at a rational $\frac{m}n$, let $\epsilon=\frac{m}n$ (or any smaller positive real), and just observe that every non-empty open interval in $\Bbb R$ contains an irrational number.

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For $x=\frac mn\in\mathbb Q$ select $\epsilon=\frac1{2n}>0$. Assume there is $\delta>0$ such that $|x-y|<\delta$ implies $|f(y)-f(x)|<\epsilon$. Then especially $f(y)>\frac{1}{2n}$ for such $y$, which means that alls such $y$ are rational and have denominator $<2n$. Even without knowing that the irrationals are dense, we can see that $y=\frac ab$ with $b<2n$ and $y\ne x$ implies $$|y-x|=\frac{|an-bm|}{bn}\ge \frac1{bn}>\frac1{2n^2}$$ because the numerator must be $\ge 1$. But of course there are (rational) numbers $y$ with $|y-x|\le \frac1{2n^2}$ (for example $y=x+\frac1{2n^2+1}$), hence no such $\delta$ exists, $f$ is not continuous.

Let $x\notin\mathbb Q$ and $\epsilon>0$ be given. We want a $\delta>0$ such that $|y-x|<\delta$ implies $|f(y)-f(x)|<\epsilon$. Select $n>\frac1\epsilon$. Then we want that $|y-x|<\delta$ implies that $y$ is either irrational or has a denominator $\ge n$. For each $k< n$, there are only finitely many rationals $\frac mk$ with denominator $k$ and $|\frac mk-x|<1$. Then $$\delta =\min\left\{\left|\frac mk-x\right|\colon k<n, \left|\frac mk-x\right|<1\right\}$$ will do the trick.

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We prove that for every $a\in(0,1)$ we have $$\lim_{x\to a}f(x)=0$$

from where we'll see it is only continuous at the irrational points. So, let's pick any $a\in(0,1)$, and let us be given $\epsilon >0$. Choose $n$ so that $1/n\leq \epsilon$.

First, we note that the only points where it might be false that $|f(x)-0|<\epsilon$ are $$\frac{1}{2};\frac{1}{3},\frac{2}{3};\frac{1}{4},\frac{3}{4};\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}; \cdots ;\frac{1}{n}, \cdots ,\frac{{n - 1}}{n}$$ If $a$ is rational, then $a$ could be one of those numbers. But, as many as there can be, the amountof these numbers are finite. Thus, for some $p/q$ in that list, the number $$|a-p/q|$$ is least. If $a$ is one of these numbers, consider $p/q\neq a$. Then, take $\delta$ as this distance. It will then be the case that if $0<|x-a|<\delta$, then $x$ will be none of the numbers

$$\frac{1}{2};\frac{1}{3},\frac{2}{3};\frac{1}{4},\frac{3}{4};\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}; \cdots ;\frac{1}{n}, \cdots ,\frac{{n - 1}}{n}$$

thus it will be true that $|f(x)-0|<\epsilon$.

You can find another interesting example here.

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