f: [0,1) -> R

**srt12** · October 24th 2006, 06:59 AM

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

**earboth** · October 24th 2006, 07:33 AM

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

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

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

EB

**TD!** · October 24th 2006, 07:39 AM

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

**topsquark** · October 24th 2006, 07:40 AM

Originally Posted by earboth

Hi,

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

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

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

**ThePerfectHacker** · October 24th 2006, 08:12 AM

Originally Posted by topsquark

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

You are wrong topsquark.

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

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

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).

**CaptainBlack** · October 24th 2006, 08:44 AM

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

$\tan(\pi\ x/2)$

**srt12** · October 24th 2006, 09:03 AM

Originally Posted by CaptainBlack

$\tan(\pi\ x/2)$

I'm affraid it returns only positive values for arguments from [0,1).

**srt12** · October 24th 2006, 09:20 AM

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?

**ThePerfectHacker** · October 24th 2006, 09:29 AM

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.

**CaptainBlack** · October 24th 2006, 09:37 AM

Originally Posted by srt12

I'm affraid it returns only positive values for arguments from [0,1).

Opps.

RonL

**Plato** · October 24th 2006, 02:51 PM

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

Now define a function from (0,1] to (0,1):
$<br /> \begin{array}{l}<br /> f<img src=$ 0,1] \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}
" alt="
\begin{array}{l}
f

0,1] \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.

**ThePerfectHacker** · October 25th 2006, 12:43 PM

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

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

) will create a bijection,
$[0,1)\to \mathbb{R}^+$

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).

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