I need to find a bijective map from $A=[0,1)$ to $B=(0,1).$ Is there a standard method for coming up with such a function, or does one just try different functions until one fits the requirements?
I've considered some "variation" of the floor function, but not sure if $\left \lfloor{x}\right \rfloor$ is bijective. Now i'm thinking of maybe some trig function that takes elements of $A$ and maps them to $B$.
I know how to show that a function is injective and surjective, but how do I find such a function? Am I overthinking this or am I under thinking this?
Any help would be greatly appreciated!!!
Lastly, since we are to find a bijective map from $A$ to $B$, assuming one exists, is this equivalent to saying that these two sets have the same cardinality?
Thank you
Best Answer
Consider the function $\psi: A \rightarrow B$ given by $$ \psi(x) = \begin{cases} x & \text{ if}\; \; \; x \neq 0 \; \; \text{and}\; \; x \neq \frac{1}{n} \in \mathbb{N} \\ \frac{1}{2} & \text{if} \;\; x = 0 \\ \frac{1}{n+1} & \text{if } \; \; x= \frac{1}{n}, \; n \in \mathbb{N} \end{cases} $$ It is not difficult to verify that $\psi$ is bijective. The problem with constructing a bijection using our favorite elementary functions is that no continuous bijection between the sets exists. This is due to the differences in character (or topology) of the two sets. So unfortunately one must turn their attention to different functions.
The function above relies on the fact that we can get this bijection by taking a sequence in $(0,1)$, say $\{x_n\}$ and mapping $x_1$ to the extra point $0$, and then mapping $x_n$ to $x_{n-1}$ for all other $n$. Hopefully this trick helps with your future work.