• May 14th 2010, 05:50 PM
Bernhard
On page 252 of William Dunham's "Journey Through Genius" he is writing about Nineteenth Century mathematics and writes the following:

"As the nineteenth Century progressed, mathematical discoveries came to light indicating that these two classes of numbers [rationals and irrationals] did not carry the same weight. The discoveries often required very technical, very subtle reasoning. For instance a function was described that was continuous at each irrational point and discontinuous at each rational point; however it was also proved that no function exists that is continuous at each rational point and discontinuous at each irrational point ... ... "

[I thought for a moment that the reference might be to DIrichlet's Charactersitic Function of the Rationals where f(x) is defined as 1 if x is rational and 0 if x is not rational - but David Bressoud in A Radical Approach to Lebesgue's Theory of Integration states on page 45 that "Dirichlet's function is totally discontinous since it is discontinous at every point!]
• May 14th 2010, 05:55 PM
Drexel28
Quote:

Originally Posted by Bernhard
On page 252 of William Dunham's "Journey Through Genius" he is writing about Nineteenth Century mathematics and writes the following:

"As the nineteenth Century progressed, mathematical discoveries came to light indicating that these two classes of numbers [rationals and irrationals] did not carry the same weight. The discoveries often required very technical, very subtle reasoning. For instance a function was described that was continuous at each irrational point and discontinuous at each rational point; however it was also proved that no function exists that is continuous at each rational point and discontinuous at each irrational point ... ... "

[I thought for a moment that the reference might be to DIrichlet's Charactersitic Function of the Rationals where f(x) is defined as 1 if x is rational and 0 if x is not rational - but David Bressoud in A Radical Approach to Lebesgue's Theory of Integration states on page 45 that "Dirichlet's function is totally discontinous since it is discontinous at every point!]

The relevant theorem says that if $f:\mathbb{R}\to\mathbb{R}$ then it's set of discontinuities is a $F_{\sigma}$ and thus a meager set. Thus, since $\mathbb{Q}$ is a meager set so would the union of $\mathbb{Q}$ and the irrationals, which is $\mathbb{R}$. But isn't $\mathbb{R}$ a Baire space?
• May 14th 2010, 05:59 PM
Bernhard
Thanks, but can you help further ...

Can you recommend a good text that covers this theorem and its proof.

Is there such a text accessible to undergraduates?

Bernhard
• May 14th 2010, 06:03 PM
Drexel28
Quote:

Originally Posted by Bernhard

Can you recommend a good text that covers this theorem and its proof.

Is there such a text accessible to undergraduates?

Bernhard

Oh God, I don't know. I only tangentially know about this from other stuff. I would guess that Rudin's Real and Complex Analysis or Royden's book would be a good start. I would wait for someone more used to this subject to give you a good suggestion for a book.

What year undergraduate are you? Those two books are in general graduate texts.

Until then I managed to find the exact topic on wikipedia

See here, here, and here.
• May 14th 2010, 06:12 PM
Bernhard
Thanks for the Wikipedia refs

Also found a brief discussion on Thomae's Function in Stephen Abbotts undergraduater analysis book, "Undertanding Analysis"

I am not taking a formal course in maths but am a math hobbyist - but my level is about senior undergraduate.

Bernhard
• May 14th 2010, 09:01 PM
roninpro
Hello.

Consider the following function:

$
f(x)=\begin{cases}

\frac{1}{n} & \text{for } x=\frac{m}{n} \text{ rational, in lowest terms} \\

0 & \text{for } x \text{ irrational}
\end{cases}
$

You can can show that $f$ is continuous at the irrationals and discontinuous on the rationals.
• May 14th 2010, 09:20 PM
Drexel28
Quote:

Originally Posted by roninpro
Hello.

Consider the following function:

$
f(x)=\begin{cases}

\frac{1}{n} & \text{for } x=\frac{m}{n} \text{ rational, in lowest terms} \\

0 & \text{for } x \text{ irrational}
\end{cases}
$

You can can show that $f$ is continuous at the irrationals and discontinuous on the rationals.

Yes, that is Thomae's function. He is asking for a function whose set of continuities is the rationals and discontinuities is the irrationals.