- Thread starter
- #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?

- Thread starter dwsmith
- Start date

- Thread starter
- #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?

- Moderator
- #2

- Feb 7, 2012

- 2,786

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.

$$

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?

- Feb 13, 2012

- 1,704

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!...Construct a function on $(0,1)$ that is continuous at all points except the rationals...

Kind regards

$\chi$ $\sigma$

- Moderator
- #4

- Feb 7, 2012

- 2,786

That argument is false, and it neatly illustrates the importance ofI 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!...

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

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.

- Feb 13, 2012

- 1,704

That is not what I said... the condition $|x-x_{0}|<\delta$ must be true for any x, rational and irrational...... 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 arationalpoint $x$ such that $|x-x_0|<\delta$. Thus $x$ depends on $x_0$, $\varepsilon$ and $\delta$...

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

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

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

... 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

$\chi$ $\sigma$