Well by that logic it would then be impossible to construct a bijection between and .
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 and set . You can construct bijection by setting (verify it is a bijection).
Next, we see that , so the identity on is a bijection from to .
Now we're ready to construct a bijection as follows: