Cardinality of continuous real functions

[evinda]

Hi! (Wave)

Find the cardinal number of $C(\mathbb{R}, \mathbb{R})$ of the continuous real functions of a real variable and show that $C(\mathbb{R}, \mathbb{R})$ is not equinumerous with the set $\mathbb{R}^{\mathbb{R}}$ of all the real functions of a real variable. That's what I have tried: We have the following: Obvioulsy it holds that $Card(\mathbb{R}) \leq Card(C(\mathbb{R}, \mathbb{R})) \rightarrow c \leq Card(C(\mathbb{R}, \mathbb{R})) $. We will show that $Card(C(\mathbb{R}, \mathbb{R})) \leq c$.

In order to do this we pick the function $h: \mathbb{R}^2 \to \mathbb{R}$ so that $\langle n,m \rangle \mapsto \sum_{n=0}^k \frac{2n}{3^m}, k \in \mathbb{R}$ and want to show that it is bijective.It doesn't seem right to me... (Tmi) But it is entirely wrong?

 

[Evgeny.Makarov]

Let's start by fixing an obvious mistake. You wrote one thing in your post, and then you wrote the opposite thing. Perhaps you redefined a notation. Unless this is fixed, I don't see the reason to go forward.

 

[evinda]

Let's start by fixing an obvious mistake. You wrote one thing in your post, and then you wrote the opposite thing. Perhaps you redefined a notation. Unless this is fixed, I don't see the reason to go forward.

I am sorry... I fixed the mistake... (Tmi)

 

[Evgeny.Makarov]

That's what I have tried: We have the following:

Ah, so now you have tried nothing? Well, this violates rule 11 of the forum, and if this happens again, you will be issued a formal infraction and banned forever. Kidding. (Smile)

Obvioulsy it holds that $Card(\mathbb{R}) \leq Card(C(\mathbb{R}, \mathbb{R}))$

Yes, it is obvious, but still: why?

We will show that $Card(C(\mathbb{R}, \mathbb{R})) \leq c$.

In order to do this we pick the function $h: \mathbb{R}^2 \to \mathbb{R}$ so that $\langle n,m \rangle \mapsto \sum_{n=0}^k \frac{2n}{3^m}, k \in \mathbb{R}$ and want to show that it is bijective.

I am not familiar with this approach. Usually you show first that a continuous function is completely determined by its values in rational points.

 

[evinda]

Ah, so now you have tried nothing? Well, this violates rule 11 of the forum, and if this happens again, you will be issued a formal infraction and banned forever.

(Sweating)

Kidding. (Smile)

(Whew) (Tongueout)

Yes, it is obvious, but still: why?

Could we consider for example a function $f: \mathbb{R} \to C(\mathbb{R}, \mathbb{R})$ such that $x \mapsto e^{x+y}$ where $y \in \mathbb{R}$ and show that it is injective?

I am not familiar with this approach. Usually you show first that a continuous function is completely determined by its values in rational points.

So do we have to find an injective function that maps a continuous function to a real number? (Thinking)

That what I wrote doesn't make sense, right? (Tmi)

 

[Evgeny.Makarov]

Could we consider for example a function $f: \mathbb{R} \to C(\mathbb{R}, \mathbb{R})$ such that $x \mapsto e^{x+y}$ where $y \in \mathbb{R}$ and show that it is injective?

That works. An even simpler example is a mapping $y\mapsto f(x)=y$ that returns constant functions.

So do we have to find an injective function that maps a continuous function to a real number?

Well, yes, but you need to go through rational numbers. Once you prove that a continuous function is determined by its values in rational points, you need to show that $\Bbb R^{\Bbb Q}\sim\Bbb R$.