#### 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!]

#### Drexel28

MHF Hall of Honor
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 $$\displaystyle f:\mathbb{R}\to\mathbb{R}$$ then it's set of discontinuities is a $$\displaystyle F_{\sigma}$$ and thus a meager set. Thus, since $$\displaystyle \mathbb{Q}$$ is a meager set so would the union of $$\displaystyle \mathbb{Q}$$ and the irrationals, which is $$\displaystyle \mathbb{R}$$. But isn't $$\displaystyle \mathbb{R}$$ a Baire space?

#### 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

#### Drexel28

MHF Hall of Honor

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.

#### 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

#### roninpro

Hello.

Consider the following function:

$$\displaystyle f(x)=\begin{cases} \frac{1}{n} & \text{for } x=\frac{m}{n} \text{ rational, in lowest terms} \\ 0 & \text{for } x \text{ irrational} \end{cases}$$

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

#### Drexel28

MHF Hall of Honor
Hello.

Consider the following function:

$$\displaystyle f(x)=\begin{cases} \frac{1}{n} & \text{for } x=\frac{m}{n} \text{ rational, in lowest terms} \\ 0 & \text{for } x \text{ irrational} \end{cases}$$

You can can show that $$\displaystyle f$$ 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.

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