We consider $h:A\to\Bbb R$ where $A=(0,\infty).$
In Thomae's Function we consider:
$$
\begin{align}
h(x) = \begin{cases}
0 & \text{if $x$ is irrational}\\
\frac{1}{n} & \text{if $x = \frac{m}{n}$ where $\gcd(m,n) = 1$}
\end{cases}
\end{align}
$$
It is also to be noted that $x\in A.$
I was studying a proof about the fact that $h$ is continuous at the irrational points in $A$ from the book, "Real Analysis " by Robert G Bartle and Donald R Sherbert.
The proof went on as :
If $b$ is an irrational number and $\epsilon > 0,$ then (by the Archimedean Property) there is a natural number $n_0$ such that $1/n_0 < \epsilon.$ There are only a finite number of rationals with denominator less than $n_0$ in the interval $(b-1,b+1).$ (Why?) Hence
$\delta > 0$ can be chosen so small that the neighborhood $(b-\delta, b+\delta)$ contains no rational
numbers with denominator less than $n_0.$ It then follows that for $|x-b| < \delta; x \in A,$ we have $|h(x)-h(b)|=|h(x)|\leq 1/n_0\lt \epsilon.$ Thus $h$ is continuous at the irrational number $b.$
However, I have $3$ doubts regarding the proof.
Doubt 1: When they say,
There are only a finite number of rationals with denominator less than $n_0$ in the interval $(b-1,b+1).$
, how does they know that indeed there's a rational number(expressed in canonical form) whose denominator is less than $n_0$ is present in the interval $(b-1,b+1).$ I mean, it might so happen, that no rationals (expressed in the canonical form) in the interval $(b-1,b+1)$ are there, whose denominator is less than $n_0.$ All rationals in the canonical form, with denomiantor $m$ where $m\lt n_0$ maybe outside the interval $(b-1,b+1).$
My thoughts: I think the proof is not saying that there exists at least one rational number in canonical form whose denominator is less than $n_0,$ but they are saying instead, that even if there is some rationals whose denominator is less than $n_0$ in canonical form, then the number of such rationals are finite.
But I don't know whether this is the thing intended ? I need some confirmation.
Doubt 2: Secondly, again, in that same statement i.e
There are only a finite number of rationals with denominator less than $n_0$ in the interval $(b-1,b+1).$
it is asserted that, that if such rational numbers exist they must be finite.
My thoughts: We know that there are, finite number of natural numbers less than $n_0$ i.e $n_0-1$ natural numbers are there. Similarly, there exist a finite number of natural numbers in the interval $(b-1,b+1)$ say, $z$. Thus, the total number of rationals with dwnominator less than $n_0$ in the interval $(b-1,b+1)$ is atmost $z(n_0-1),$ which is finite. Hence, they made such a statement.
Am I correct in my thoughts/reasonings above ?
Doubt 3: How are the saying, $|h(x)-h(b)|=|h(x)|\leq 1/n_0\lt \epsilon,$ in their proof? I precisely don't get how did they write, $|h(x)|\leq 1/n_0$?
Best Answer
Looking at the question as well as the answer posted by the asker it is clear that the problem is in Doubt 2.
In particular one needs to prove that interval $(b-1,b+1)$ contains a finite number of rationals with denominators less than $n_0$.
To handle the situation where $n_0=1$ it is best to prove the following:
Lemma: Let $I$ be a bounded interval and let $m$ be a positive integer. Then $I$ contains a finite number of rationals with denominator less than or equal to $m$.
Let us consider the sets $$A_m=\{n/m\mid n\in\mathbb{Z}, \text{gcd} (n, m) =1\}, B_m=\{n/m\mid n\in\mathbb{Z} \} $$ Both these sets are infinite and $A_m\subseteq B_m$. Let us now observe that if $A$ is an infinite set with $A\subseteq B_m$ then $A$ is necessarily unbounded. This is because we then have infinitely many $n/m$ lying in $A$ and this means that we need integers $n$ with arbitrarily large absolute values and then set $A$ is unbounded.
Thus $A_m $ is unbounded but its intersection with a bounded interval $I$ is bounded and then $A_m\cap I\subseteq B_m$ is necessarily finite.
Further note that $A_{m_1}\cap A_{m_2}=\emptyset$ if $m_1\neq m_2$ and hence the set of all rationals lying in $I$ and having denominator less than or equal to $m$ is $$(A_1\cap I) \cup (A_2\cap I) \cup \dots\cup (A_m\cap I) $$ Since each of $A_i\cap I$ is finite their union above is also finite.