Talk:Thomae Function is Continuous at Irrational Numbers

Help. I'm not clear on something.

Let $\Q$ be ordered in the following way:


 * $\dfrac {p_1} {q_1} \prec \dfrac {p_2} {q_2} \iff \begin {cases} q_1 < q_2 & : q_1 \ne q_2 \\ p_1 < p_2 & : q_1 = q_2 \end {cases}$

and so we can denote $\Q$ with this ordering as $\struct {\Q, \prec}$

So $\dfrac 9 {10} \prec \dfrac {10} {13}$ in this cooked-up ordering.

But... this is not how the rational numbers are actually ordered


 * No it's not. Why is that a problem? --prime mover (talk) 19:08, 9 April 2021 (UTC)

The whole point is to find an open interval:
 * $C = \openint a b$

which contains $x$ and no rational numbers whose denominators are less than $q$.

Is it legit to completely re-order the rational numbers to accomplish this task? (If so, how/why?)


 * Yes of course it is. In maths, you can define whatever you want. In particular, given a set, you can define whatever ordering you want on that set. If you treat $\dfrac p q$ as an ordered pair $\tuple {p, q}$, what we have effectively defined is a lexicographical ordering on $\Q$.


 * Don't confuse $\prec$ (which is the ordering defined here) and $<$ (which is the conventional ordering). --prime mover (talk) 19:09, 9 April 2021 (UTC)
 * Got it. The lightbulb just turned on. We first select q which caps the denominator and then the supremum and infimum guarantee no rational numbers with denominators less than $q$.


 * If $x = \sqrt 2$ and $q = 5$ then the open interval defined by the current proof is $C = \openint {\dfrac 7 5} {\dfrac 3 2}$
 * If $x = \sqrt 2$ and $q = 6$ then the open interval defined by the current proof is $C = \openint {\dfrac 7 5} {\dfrac 3 2}$
 * If $x = \sqrt 2$ and $q = 7$ then the open interval defined by the current proof is $C = \openint {\dfrac 7 5} {\dfrac {10} 7}$
 * Thanks for the response! --Robkahn131 (talk) 20:56, 9 April 2021 (UTC)

An alternate (and perhaps more intuitive?) approach would be to use convergents of continued fractions

For example, if our irrational number was $\sqrt 2$, then using convergents of continued fractions, our open intervals would be:


 * $C_1 = \openint {1} {\dfrac 3 2}$
 * $C_2 = \openint {\dfrac 7 5} {\dfrac 3 2}$
 * $C_3 = \openint {\dfrac 7 5} {\dfrac {17} {12} }$
 * $\cdots$

And from Denominators of Simple Continued Fraction are Strictly Increasing, we know we can find an open interval which contains $x$ and from Definition:Best Rational Approximation, also contains no rational numbers whose denominators are less than $q$.

--Robkahn131 (talk) 16:48, 9 April 2021 (UTC)


 * If you want to craft a different proof of this theorem using this technique, then feel free to have a go. As long as you don't replace the proof that's currently in place, which is simple and elegant. Of course, we'd probably have to split this proof up into multiple pages, so we don't have to pointlessly duplicate the rational number proof. --prime mover (talk) 19:08, 9 April 2021 (UTC)


 * I think it is max and not min - specific example to illustrate...


 * If $x = \sqrt 2$ and $q = 5$ then the open interval defined by the current proof is $C = \openint {\dfrac 7 5} {\dfrac 3 2}$
 * $\forall y \in C: \map {D_M} y \le \dfrac 1 q$
 * $\delta = \min \set {\size {\sqrt 2 - \dfrac 7 5}, \size {\dfrac 3 2 - \sqrt 2} } = \sqrt 2 - \dfrac 7 5$ (roughly $0.014$)
 * $\delta = \max \set {\size {\sqrt 2 - \dfrac 7 5}, \size {\dfrac 3 2 - \sqrt 2} } = \dfrac 3 2 - \sqrt 2$ (roughly $0.086$)


 * What if $y = \dfrac {149} {100}$?


 * We know $\size {y - x} \lt \delta$


 * Therefore $\size {\dfrac {149} {100} - \sqrt 2} =0.076 \lt 0.086$


 * --Robkahn131 (talk) 11:08, 10 April 2021 (UTC)


 * You want a $\delta$ such that $\size {y - x} < \delta$ guarantees that $\size {\map {D_M} y - \map {D_M} x} < \epsilon$.


 * Suppose $x$ is really close to $a$ but quite some distance from $b$. You set $\delta = b - x$.


 * Now suppose $y < a$ but $x - y < \delta$, because $\delta$ is some way bigger than $x - a$.


 * Then you have a $y$ for which $\map {D_M} y$ is not necessarily within the bound that is guaranteed it will be, by dint of $y$ being within $\openint a b$.


 * So, let's start again with your example.


 * What if $y = \dfrac {136} {100}$


 * With your big $\delta$ you know that $\size {y - x} < \delta$ because $1.414 - 1.36 < 0.086$


 * But $1.36$ is outside $\openint a b$.


 * I don't know (haven't checked) whether $\sqrt 2 - 0.086$ actually contains any $y$ such that $\size {\map {D_M} y - \map {D_M} x} > \epsilon$ but you can't guarantee that.


 * Short answer: you need to constrain $\delta$ to guarantee that the open $\delta$-ball is entirely within the interval $\openint a b$.


 * And you don't do that by selecting the maximum. --prime mover (talk) 13:24, 10 April 2021 (UTC)


 * Your answer clears it up for me - I agree - it is min and not max. And as for a counterexample, let $y = \dfrac 4 3$
 * $\size {\dfrac 4 3 - \sqrt 2} =0.081 \lt 0.086$ BUT...
 * $\size {\map {D_M} y - \map {D_M} x} =\dfrac 1 3 > \dfrac 1 5$


 * Thanks for the help! --Robkahn131 (talk) 16:01, 10 April 2021 (UTC)

Missing Link?
In the proof where:
 * Let $a$ be the supremum of the set:
 * $\set {z \in S: z < x}$


 * Let $b$ be the infimum of the set:
 * $\set {z \in S: x < z}$


 * Then we have that the open interval:
 * $C = \openint a b$
 * contains $x$ and no rational numbers whose denominators are less than $q$.

what is it that establishes:
 * $a < x < b$?

I assume that it is some argument invoking the Archimedean Principle to establish that $a$ and $b$ are rational. --Leigh.Samphier (talk) 05:34, 6 June 2021 (UTC)