real analysis - Find a bijection between 2 intervals

So I'm supposed to find a bijection between $[0,1) \longrightarrow [0,1]$. My attempt of solution is the function defined as

$$f(x):=\begin{cases}
2x, & \text{iff} \,\, x=\dfrac{1}{2^n} \\
x, &\text{otherwise}
\end{cases}$$

Using this we have that if $x=\frac{1}{2}$, then $f(1/2)=1$ and so we got that covered. And for $x=\frac{1}{4}$ we have $f(1/4)=1/2$ and so on. Is this correct? Is it possible to assume that there is a bijection, when $n$ goes to infinity?

I also got another question: define a bijection from $[0,1) \longrightarrow [0,1) \times [0,1) \times \ \dots \ \times [0,1)$
$(n-$times$)$. My solution is just to define the value of $f(x)$ as a vector, i.e take $x \in [0,1)$ and $$f(x):=(x,x,x,\dots , x)$$ Is this correct?

Thank you in advance!

Answer

For the first function you either compute the inverse, and show that is is the right and left inverse, or you should show that it is injective and surjective. (that is take to elements and show that they are mapped to different images, and show that ever element in $[0,1]$ has a pre-image)

The function is correct. But why you say assume there is a bijection? What do you mean by $n$ goes to infinity? You should take care that $f$ is well-defined which it is since $x= \frac{1}{2^n}$ is either true or false.

For the second function. Is $(0,\frac{1}{2}, \cdots, 0)$ in the image? You may also should say $n$-tuple instead of vector, since the codomain is not equipped with a vector space structure.