[SOLVED]discontinuities of a function

dwsmith

Well-known member
Construct a function on $(0,1)$ that is continuous at all points except the rationals, is monotone increasing, is right continuous at all points on $(0,1)$, and such that $f(0) = 0$ and $f(1) = 1$.

$$f(x) = \sum_{n=1}^{\infty}\frac{a_{n+1} - a_n}{a_na_{n+1}}H_n(x-\mathbb{Q})$$
where $H_n$ is the Heaviside function and the rationals are of the form $\frac{1}{a_n}$.
Is this correct?

Opalg

MHB Oldtimer
Staff member
Construct a function on $(0,1)$ that is continuous at all points except the rationals, is monotone increasing, is right continuous at all points on $(0,1)$, and such that $f(0) = 0$ and $f(1) = 1$.

$$f(x) = \sum_{n=1}^{\infty}\frac{a_{n+1} - a_n}{a_na_{n+1}}H_n(x-\mathbb{Q})$$
where $H_n$ is the Heaviside function and the rationals are of the form $\frac{1}{a_n}$.
Is this correct?
I think that you have the right general idea, though I don't understand the actual formula that you are proposing (and to be blunt, I don't think it makes any sense at all). The key thing must be that there are countably many rationals in the unit interval. Let $\{r_1,r_2,r_3,\ldots\}$ be an enumeration of the rationals in $(0,1)$, and define $$f(x) = \sum_{n=1}^{\infty}2^{-n}H(x-r_n),$$ where $H$ is the Heaviside function.

chisigma

Well-known member
Construct a function on $(0,1)$ that is continuous at all points except the rationals...
I think that this concept would be a little better specified. Calling $\varphi(x)$ this function if $x_{0}$ is irrational, under Your hypothesis, for an $\varepsilon>0$ it exists a $\delta>0$ for which for all x for which $|x-x_{0}| < \delta$ is $|\varphi(x) - \varphi(x_{0})|< \varepsilon$. Now in any interval $(x,x_{0})$ there are infinite rational values of x so that, swapping the roles of $x_{0}$ and $x$ we conclude that $\varphi(x)$ is continuos also per rational values of x... contradiction!...

Kind regards

$\chi$ $\sigma$

Opalg

MHB Oldtimer
Staff member
I think that this concept would be a little better specified. Calling $\varphi(x)$ this function if $x_{0}$ is irrational, under Your hypothesis, for an $\varepsilon>0$ it exists a $\delta>0$ for which for all x for which $|x-x_{0}| < \delta$ is $|\varphi(x) - \varphi(x_{0})|< \varepsilon$. Now in any interval $(x,x_{0})$ there are infinite rational values of x so that, swapping the roles of $x_{0}$ and $x$ we conclude that $\varphi(x)$ is continuos also per rational values of x... contradiction!...
That argument is false, and it neatly illustrates the importance of order of choice in analysis proofs.

When you use the fact that $\varphi$ is continuous at $x_0$, you start with two given elements: the point $x_0$ and the number $\varepsilon>0$. You then obtain a number $\delta>0$ (which depends on $x_0$ and $\varepsilon$) and you choose a rational point $x$ such that $|x-x_0|<\delta$. Thus $x$ depends on $x_0$, $\varepsilon$ and $\delta$.

You then claim that "swapping the roles of $x_{0}$ and $x$ we conclude that $\varphi(x)$ is continuous" at $x$. But that is not allowed. The roles of $x_0$ and $x$ are not symmetrical, because $x$ depends on $x_0$. To prove that $\varphi(x)$ is continuous at $x$, you would need to start with $x$ and $\varepsilon$, and find a $\delta>0$ such that $|\varphi(x') - \varphi(x)|<\varepsilon$ whenever $|x'-x|<\delta.$ But you cannot use the $\delta$ from the previous paragraph, because that was chosen before $x$. To prove continuity at $x$, you have to start with $x$ before knowing $\delta$. The order of choice is wrong, and the "proof" fails.

In fact, it is quite possible to have a function that is discontinuous at every rational point and continuous at every irrational point. See the function that Wikipedia calls Thomae's function.

chisigma

Well-known member
... when you use the fact that $\varphi$ is continuous at $x_0$, you start with two given elements: the point $x_0$ and the number $\varepsilon>0$. You then obtain a number $\delta>0$ (which depends on $x_0$ and $\varepsilon$) and you choose a rational point $x$ such that $|x-x_0|<\delta$. Thus $x$ depends on $x_0$, $\varepsilon$ and $\delta$...
That is not what I said... the condition $|x-x_{0}|<\delta$ must be true for any x, rational and irrational...

Anyway the question is complex and requires to me adequate 'reflection time'...

Kind regards

$\chi$ $\sigma$

chisigma

Well-known member
Before to proceed in this discussion an important detail: I'm not a Mathematician and all what is about 'abstact logic',' patological functions', 'indeterminate forms' and 'removable singularties' is very far from how my brain works...

Now I'try to clarify a little, and with benefit for anyone, the question about the function that is contonous for irrationals and discontinuos for rationals. Let br f(x) such a functions and for clarity sake we will use the following [pure explicative...] pitcure... Let be $x_{0}$ an irrational and 'by definition' f(x) is continuos in $x=x_{0}$. That means that, given an $\varepsilon>0$ it exists a $\delta>0$ for which for any $|x-x_{0}|<\delta$ is $|f(x)-f(x_{0}| < \varepsilon$. Now we choose two rational values of x in $(x_{0},x_{0}+\delta)$ with $x_{2}-x_{1}= \sigma>0$. According with what we have said is $|f(x_{1})-f(x_{0}| < \varepsilon$ and $|f(x_{2})-f(x_{0}| < \varepsilon$ and that means that $|f(x_{2})-f(x_{1})|< 2\ \varepsilon$ for $|x_{2}-x_{1}| < \sigma$...

... in other words f(x) seems to be continous both in $x_{2}$ and $x_{1}$...

For the reasons I explained before this is my last post on this thread...

Kind regards

$\chi$ $\sigma$