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Consider the Thomae's function

$$f(x)=\begin{cases} 0 \text{ ; when } x \text{ is irrational} \\\frac 1 q \text{ ; for } x=\frac p q \text{ irreducible fraction}\end{cases}$$

In the following proof a negative rational number has a negative numerator and a positive denominator.

Let $z$ be an irrational number, thus $f(z)=0$. I need to prove that $\lim_{x\rightarrow z} f(x)=0$. Thus, given any $\epsilon>0$ we need to prove that there exists $\delta >0$ such that $x \in (z-\delta,z+\delta) \implies |f(x)|<\epsilon$.

Take any $\delta_1>0$. Consider the minimum denominator,$m_1$ of all the rational numbers $\frac p q$ in the interval $(z-\delta_1,z+\delta_1$). There exists a finite number of rational numbers of the form $\frac{t_i}{m_1}$ in the interval $(z-\delta_1,z+\delta_1$). Let the closest one from $z$ be $\frac {t}{m_1}.$ Take $$\delta_2=\frac{|z-\frac {t}{m_1}|}{2}$$. Now let the minimum denominator in the interval ($z-\delta_2,z+\delta_2$) be $m_2$. Then $m_2>m_1$. Similarly we can find $\delta_3,\delta_4,\dots$ and the strictly increasing sequence ${m_1,m_2,\dots}$.

Let n be the minimum number so that $m_n> \left \lceil{\frac{1}{\epsilon}}\right \rceil $. Take $\delta = \delta_n.$

When x is rational and $x\in (z-\delta_n , z+\delta_n)$, $$|f(x)|=\frac1 j < \frac {1} {m_n}<\epsilon$$

When x is irrational and $x\in (z-\delta_n , z+\delta_n)$

$$f(x)=0< \epsilon$$

Thus proved.

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  • $\begingroup$ Instead of minimum denominator, maybe use the idea of "fix some $m$ and consider all rationals which may be rewritten in the form $k/m$ with $k$ an integer" etc. $\endgroup$ – coffeemath Nov 26 '14 at 6:48
  • $\begingroup$ @coffeemath can you please elaborate? $\endgroup$ – Soham Nov 26 '14 at 6:49
  • $\begingroup$ Lucyfer -- I just put up an "answer" explaining what I had in mind using fractions $k/m$ (reduced or not). $\endgroup$ – coffeemath Nov 26 '14 at 18:11
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I'm going to give you an overall review of your proof.

Let $z$ be an irrational number, thus $f(z)=0$. I need to prove that $\lim_{x\rightarrow z} f(x)=0$. Thus, given any $\epsilon>0$ we need to prove that there exists $\delta >0$ such that $x \in (z-\delta,z+\delta) \implies |f(x)|<\epsilon$.

Try to avoid using "thus" so often. It's a good way to indicate that you have come to a final conclusion. The term "then" is usually used to refer to a next step. Thank you for writing what you need to prove. This is great! If you wish to keep it in your proof I suggest using "we" instead. Or this can be omitted. It is assumed most of the time we will know from our theorem or such what we want to prove! Instead of the second thus you could say, "we will check that the definition of continuity holds" or we could restructure this sentence to read "by the definition of continuity, $f$ is continuous when yadadada". Also note, the above should read $\forall \varepsilon >0$, $\exists \delta$ such that $\forall z$ with $|x-z|<\delta$ then $|f(x)-f(z)|<\varepsilon$. In our case $f(x)=0$ so we may write this as $\forall z$ with $|x-z|<\delta$ then $|f(z)|<0$.

Take any $\delta_1>0$. Consider the minimum denominator,$m_1$ of all the rational numbers $\frac p q$ in the interval $(z-\delta_1,z+\delta_1$).

Okay, I'm imagining taking the smallest denominator. My immediate consideration is what if there are two rationals in the interval, say $\frac{1}{3}$ and $\frac{2}{3}$. Not an issue so far.

There exists a finite number of rational numbers of the form $\frac{t_i}{m}$ in the interval $(z-\delta_1,z+\delta_1$).

I think it is important to explain why there must be at least one rational in the interval. Also, does the above fraction mean to read $\frac{t_i}{m_1}$? As was written in the comments, maybe it is a good idea to write, " for some $m$ where $\frac{n}{m}$ is in this interval, there are finitely many elements $n_i$ with $i\in \{1,2,...,k\}$." Then in your next paragraph you can continue with taking the minimum distance from $x$ to this rational.

Let the closest one from $z$ be $\frac t m.$ Take $$\delta_2=\frac{|z-\frac tm|}{2}$$. Now let the minimum denominator in the interval ($z-\delta_2,z+\delta_2$) be $m_2$. Then $m_2>m_1$. Similarly we can find $\delta_3,\delta_4,\dots$ and the strictly increasing sequence ${m_1,m_2,\dots}$.

Maybe instead say that, set $\delta_2=\min\{\frac{|x-\frac{t_1}{m_1}|}{2},...,\frac{|x-\frac{t_j}{m_1}|}{2}\}$. That's completely optional. Just make sure you mention what closest means in the space we are talking about! I would suggest rephrasing that last part from "similarly we can find ... " to "We may continue this process ad infinitum, constructing the strictly increasing sequence of positive integers $\{m_n\}_{n=1}^\infty$."

Let n be the minimum number so that $m_n> \left \lceil{\frac{1}{\epsilon}}\right \rceil $. Take $\delta = \delta_n.$

Italicize that $n$!

When x is rational and $x\in (z-\delta_n , z+\delta_n)$, $$|f(x)|=\frac1 j < \frac {1} {m_n}<\epsilon$$

Same thing with the $x$! Also, where did $j$ come from? You could just say $q$ as in the top or explain.

When x is irrational and $x\in (z-\delta_n , z+\delta_n)$

You get the point with the $x$.

$$f(x)=0< \epsilon$$

Or we could write "then $|f(z)-f(x)|=|0-0|=0<\varepsilon$.

Thus proved.

Reiterate your conclusion. Therefore $f$ is continuous at $x$ irrational. Or something similar.

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  • $\begingroup$ Be warned, I am going to review your proof again. I am trying to understand what everyone else is saying as well. $\endgroup$ – Eoin Nov 26 '14 at 7:13
  • $\begingroup$ I edited the part $\frac{t}{m_1}$ as it was a typographical error. $\endgroup$ – Soham Nov 26 '14 at 7:37
  • $\begingroup$ @LucyferZedd Your proof does the job. haha $\endgroup$ – Eoin Nov 26 '14 at 7:38
  • $\begingroup$ After "In our case $f(x)=0$ it should be "..then $|f(z)|<\varepsilon.$ [Your 0 there is a typo?] $\endgroup$ – coffeemath Nov 26 '14 at 16:53
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The Thomae function has period $1$ so the irrational $z$ may without loss be assumed to lie in the interval $(0,1).$ Now let $x \in (0,1)$ be any rational $x=k/n$ where $1 \le k < n$ and $\gcd(k,n)=1,$ and use $d(x)$ to mean the "denominator of $x$" so that $f(x)=1/d(x)$ when $x$ is rational. For rational $x$ then we have $f(x) \ge 1/n$ if and only if $d(x) \le n.$

An intricate argument can be avoided if we look at the collection of all rational $x$ in $(0,1)$ for which $d(x) \le n.$ These rationals aren't evenly spaced, which is one thing which makes the proof intricate working directly with them. Instead we may define $m$ to be the l.c.m of the numbers $\{1,2,...,n \}$ and note that then all the rational $x$ in $(0,1)$ for which $f(x)\ge 1/n$ may be written in the form $k/m.$ The latter numbers are evenly spaced.

Example: the numbers for which $f(x) \ge 1/4$ are $\{1/4,1/3,1/2,2/3,3/4\}$ which are not evenly spaced, but when put over the g.c.d $12$ of $1,2,3,4$ become $\{3/12,4/12,6/12,8/12,9/12\}.$ Of course these numbers are still not evenly spaced (they're the same numbers), but there is no harm in throwing in all the numbers admitting denominator $12$ (reduced or not), so that we get the list $\{1/12,2/12, \cdots 11/12\}.$ Better also throw in $0/12$ and $12/12$. Now the irrational $z$ must fall between two adjacent fractions $k/12,(k+1)/12$. Then if $\delta=\min(c-k/12,(k+1)/12-c)$ we can be sure that when a rational $x$ lies within $\delta$ of $c$ it will have $f(x)<1/4.$

Replacing $1/4$ here by $1/n$ and using $m=\rm{lcm} (1,2,...,n)$ we have a similar argument to get $f(x)<1/n$ when $x$ is within $\delta$ of $z$, where this $\delta$ is similarly defined as the above, using adjacent terms of the form $k/m,(k+1)/m.$ Then one can put a "front end" on the argument by: Given $\varepsilon>0$ start by picking $n$ so that $1/n<\varepsilon,$ and continue as outlined.

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