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)


 * I don't know. I just thought the fact that $\sup \set {z \in S: z < x}$ and $\inf \set {z \in S: x < z}$ are themselves elements of $S$, and thence rational, was intuitively obvious.


 * If not, then we have this:


 * If $x$ is irrational, and $S$ is a countable set of $\Q$ which goes from $-\infty$ to $+\infty$, with finite gaps between each element, then some of $S$ are going to be less than $x$ and the rest of them are going to be greater than $x$. The fact that $\set {z \in S: z < x}$ has a supremum and $\set {z \in S: x < z}$ has an infimum follows from the fact that $S$ intersected with a finite interval $\closedint {x - a} {x + a}$ for some real $a$ is itself finite, I suppose, but I do not have the patience to prove that latter fact. From that we have that $\sup \set {z \in S: z < x} \in \set {z \in S: z < x}$ and $\inf \set {z \in S: x < z} \in \set {z \in S: x < z}$ which are cases of some results somewhere on concerning suprema and infima of finite sets which again I can't be bothered to look up. (I expect my old mathematics teacher (who is still active, is familiar with  and knows how to get hold of me) is going to be thundering his way towards my front door with the whippiest cane he can find. He was never fond of my attitude: the only reason I do maths is because I'm too bone idle to think.)


 * So, granted that $a$ and $b$ exist, we know that $x$ is irrational and that $a$ and $b$ are rational.


 * So $a < x$ and $x < b$.


 * What precisely is missing? --prime mover (talk) 06:58, 6 June 2021 (UTC)
 * In my honest opinion:
 * If x is irrational, and S is a countable set of Q which goes from −∞ to +∞, with finite gaps between each element, then some of S are going to be less than x and the rest of them are going to be greater than x. The fact that {z∈S:z<x} has a supremum and {z∈S:x<z} has an infimum follows from the fact that S intersected with a finite interval [x−a..x+a] for some real a is itself finite. $\dots$ From that we have that $\sup \set {z \in S: z < x} \in \set {z \in S: z < x}$ and $\inf \set {z \in S: x < z} \in \set {z \in S: x < z}$ which are cases of some results somewhere on concerning suprema and infima of finite sets $\dots$
 * --Leigh.Samphier (talk) 08:02, 6 June 2021 (UTC)