1. 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....

2. Well by that logic it would then be impossible to construct a bijection between $\displaystyle \mathbb{Z}$ and $\displaystyle \mathbb{Q}$.

.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 $\displaystyle A=\{\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4 },\frac{1}{5},\ldots\} \subseteq [0,1]$ and set $\displaystyle B=\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5 },\ldots\} \subseteq [0,1)$. You can construct bijection $\displaystyle g:A\rightarrow B$ by setting $\displaystyle g\left(\frac{1}{i}\right)=\frac{1}{i+1}$ (verify it is a bijection).

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

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

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