1. Some curious topics.

Could you help me with some topics?

1. How many continuus functions [0, 1] to R exists? (some people says that it is continuum! I can accept this!). Justify.

2. How many continuus functions R to R exists (some people says that it is also continuum. That i cannot accept! I bet that it is more than continuum.). Justify.

I belive that this tasks will be no problem for all you mathematics genius.

Thank you for any help.

2. Originally Posted by albi
Could you help me with some topics?

1. How many continuus functions [0, 1] to R exists? (some people says that it is continuum! I can accept this!). Justify.
Not going to prove this but just observe that $\displaystyle C^0_{[0,1]}$ (the
space of continuous functions on the unit interval) has a countable
basis, so in an appropriate sense is equivalent to the set of all sequences
of reals, the cardinality of which is $\displaystyle C$ (the cardinality of the continuum)

RonL

3. Thank you for your reply. You mean of course the Fourier expansion on the interval [0, 1]?

But what about the (0, 1) interval (or the functions from R to R). I bet that their number is bigger than continuum!.

As far as I know some functions on (0, 1) (continuus!) doesn't satisfy Dirichlett conditions and cannot be represented by the Fourier series expansion. So what about that?

Regards
Albi

4. Um, what happens when the function IS NOT countinous is the cardinality of the countinuum?

(I find this funny for some reason: just like there is not such thing as a biggest number, there is no such thing as the biggest cardinal).

5. Originally Posted by albi
Thank you for your reply. You mean of course the Fourier expansion on the interval [0, 1]?
They will do, as will the a set of orthogonal polynomials on the interval.

RonL

6. Ok, I was wrong.
Now I am sure that the number of continuus functions from R to R is also continuum! I get this result realising that the number of realvalued sequences is also continuum (what a surprise!).

If we take all functions from [0, 1] to R their number will be grater than continuum.

It will be $\displaystyle c^c$ which is grater than contiunuum cause that continuum is grater than two. (I don't know how to make gothic "c" in this "latex")

If you know some interesting proofs of the above theorems share with me

Ok it is 1.15 on my watch so it is a good time to go to bed. And tomorrow (or today maybe) is a very good day to have some exam .

Goodnight

7. I decided to refresh this thread. I know that doubling the posts is forbidden but I found the answer to this old question (this is what happens when one studies mathematics).

And I think it may be usful to someone...

The thing is relatively easy. We should note that $\displaystyle \mathbb{R}$ is separable. So there exists countable set $\displaystyle \{ a_n \}$ which is dense (this may be for instance $\displaystyle \mathbb{Q}$ )

Let A denote continuous functions $\displaystyle \mathbb{R} \rightarrow \mathbb{R}$ and B denote functions $\displaystyle \mathbb{Q} \rightarrow \mathbb{R}$

There exists injection: $\displaystyle \sigma: A \rightarrow B$.

Let $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$

We define $\displaystyle \sigma(f)$ the following way:

$\displaystyle \sigma(f) := f|_{\mathbb{Q}}$

$\displaystyle \sigma$ is injection because if there are functions $\displaystyle f, g$ which differ in some irrational x, $\displaystyle \sigma(f) = \sigma(g)$ implies (simply by taking limit and continuity) $\displaystyle f(x) = g(x)$ which is contradiction!

Hence $\displaystyle |A| \geq |B|$. But $\displaystyle |A| = c^\omega = c$

But B has subset of continuum cardinality (constant functions), hence $\displaystyle |B| = c$

Of couse this workes for any separable metric space!