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

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- Oct 24th 2006, 06:59 AM #1

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## 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

- Oct 24th 2006, 07:33 AM #2

- Oct 24th 2006, 07:39 AM #3

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- Oct 24th 2006, 07:40 AM #4
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

- Oct 24th 2006, 08:12 AM #5

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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).

- Oct 24th 2006, 08:44 AM #6

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- Oct 24th 2006, 09:03 AM #7

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- Oct 24th 2006, 09:20 AM #8

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- Oct 24th 2006, 09:29 AM #9

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- Oct 24th 2006, 09:37 AM #10

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- Oct 24th 2006, 02:51 PM #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.

- Oct 25th 2006, 12:43 PM #12

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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).