# Bijection

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• September 11th 2009, 02:08 AM
felixmcgrady
Bijection
.to construct a bijection

between [0,1] and [0,1)......................

I just think it is impossible , cuz the second one is one element less than the first....
• September 11th 2009, 03:56 AM
putnam120
Well by that logic it would then be impossible to construct a bijection between $\mathbb{Z}$ and $\mathbb{Q}$.
• September 11th 2009, 04:28 AM
Taluivren
Quote:

Originally Posted by felixmcgrady
.to construct a bijection

between [0,1] and [0,1)......................

I just think it is impossible , cuz the second one is one element less than the first....

Hi,
It must be possible.

If you know Cantor-Bernstein Theorem, it is easy, because there obviously exist an injective map from [0,1) to [0,1] (the identity map) and an injective map from [0,1] to [0,1) (x -> x/2).

If you don't know the theorem, consider set $A=\{\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4 },\frac{1}{5},\ldots\} \subseteq [0,1]$ and set $B=\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5 },\ldots\} \subseteq [0,1)$. You can construct bijection $g:A\rightarrow B$ by setting $g\left(\frac{1}{i}\right)=\frac{1}{i+1}$ (verify it is a bijection).

Next, we see that $[0,1]\smallsetminus A = [0,1)\smallsetminus B$, so the identity on $[0,1]\smallsetminus A$ is a bijection from $[0,1]\smallsetminus A$ to $[0,1)\smallsetminus B$.

Now we're ready to construct a bijection $f:[0,1]\rightarrow [0,1)$ as follows:

$f(x) = \begin{cases} g(x), & \mbox{if } x \in A \\ x, & \mbox{if } x\in [0,1]\smallsetminus A \end{cases}$