# Thread: f: [0,1) -> R

1. ## f: [0,1) -> R

Hi there,

I wonder if you can think of any bijective function that maps [0,1) interval into R?

In case of (0,1) I've done the following: y=tan(pi*x - pi/2), but I have no idea how to deal with [0,1).

Thanks in advance for any help,
Lucas

2. Originally Posted by srt12
Hi there,

I wonder if you can think of any bijective function that maps [0,1) interval into R?...
Hi,

try this one: $\displaystyle f(x)=\frac{\sqrt{x-x^2}}{x-\frac{1}{2}}$

Unfortunately this function maps $\displaystyle [0,1]\mapsto \mathbb{R}$

But may be you find a restriction so you can use your interval.

EB

3. Or: try mapping [0,1) to (0,1), use the composition to map [0,1) (via (0,1) to R.

4. Originally Posted by earboth
Hi,

try this one: $\displaystyle f(x)=\frac{\sqrt{x-x^2}}{x-\frac{1}{2}}$

Unfortunately this function maps $\displaystyle [0,1]\mapsto \mathbb{R}$

But may be you find a restriction so you can use your interval.

EB
Actually, the domain of f does not include x = 1/2, so this isn't quite true.

My thought is that this mapping is impossible. The interval [0, infinity) is half closed, whereas the "interval" R is either closed or open, depending on how you look at it. I'm not positive I'm remembering this correctly, but I believe there is a topological theorem that says such functions (mapping from different types of intervals) is not possible, for a continuous function anyway.

Naturally, someone please correct me if I'm wrong.

-Dan

5. Originally Posted by topsquark

Naturally, someone please correct me if I'm wrong.
You are wrong topsquark.

The idea is that $\displaystyle [0,1)$ is the countinuum as does $\displaystyle \mathbb{R}$. To see a prove of this using a variation of the Diagnol argument look here.

Now, by the countinuum hypothesis since $\displaystyle |[0,1)|>\aleph_0$ and $\displaystyle |[0,1)|\leq \mathbb{R}$ we must have,
$\displaystyle |[0,1)|=|\mathbb{R}|$.

The countinuum hypothesis provides us that there does exists a bijection between the two sets. However, whether is is possible to explicity state one, even via Axiom of Choice might not be possible (but I think yes).

6. Originally Posted by srt12
Hi there,

I wonder if you can think of any bijective function that maps [0,1) interval into R?

In case of (0,1) I've done the following: y=tg(pi*x - pi/2), but I have no idea how to deal with [0,1).

Thanks in advance for any help,
Lucas
$\displaystyle \tan(\pi\ x/2)$

7. Originally Posted by CaptainBlack
$\displaystyle \tan(\pi\ x/2)$
I'm affraid it returns only positive values for arguments from [0,1).

8. Originally Posted by ThePerfectHacker
The countinuum hypothesis provides us that there does exists a bijection between the two sets. However, whether is is possible to explicity state one, even via Axiom of Choice might not be possible (but I think yes).
Thanks a lot, hope it will do that I just show that there exist such a bijection.
And please correct me, if I'm wrong, but continuum hypothesis cannot be neither proved nor disproved, right?

9. Originally Posted by srt12
Thanks a lot, hope it will do that I just show that there exist such a bijection.
And please correct me, if I'm wrong, but continuum hypothesis cannot be neither proved nor disproved, right?
Yes it is completely independant from ZFC.

10. Originally Posted by srt12
I'm affraid it returns only positive values for arguments from [0,1).
Opps.

RonL

11. If we can biject (0,1] with R, then certainly we can [0,1).
$\displaystyle (0,1] = \bigcup\limits_{k = 1}^\infty {\left( {\frac{1}{{k + 1}},\frac{1}{k}} \right]}$

Now define a function from (0,1] to (0,1):
$\displaystyle \begin{array}{l} f0,1] \to (0,1) \\ x \in (0,1] \Rightarrow \left[ {\exists n} \right]\left( {x \in \left( {\frac{1}{{n + 1}},\frac{1}{n}} \right]} \right) \\ f\left( x \right) = x + 1 - \frac{1}{n} - \frac{1}{{n + 1}} \\ \end{array}$

Show that that is a bijection. Then finish the problem with you function.

12. I do not know what Plato was attempting to do but when I saw that union and I had a revealation.

If you can biject,
$\displaystyle f:A\to B$ and $\displaystyle g:B\to C$ then you can biject,
$\displaystyle A\to C$

Thus, you divide your interval as,
$\displaystyle [0,1),[1,2),[2,3),...$
And you can create a bijection between any adjacent to each other.
That means, the union of these bijective maps (there are $\displaystyle \aleph_o$ many thus the set is well-founded ) will create a bijection,
$\displaystyle [0,1)\to \mathbb{R}^+$

I am too lazy to work on a procedure but I got the idea from seeing what Plato wrote. (I hope I did not say the same thing).