I need to the detriment the cardinality of a group and prove why.
The group:
A = {The group of all real numbers in the range between 0 and 1, where the digits can only be ones and zeroes, and after each zero comes a one}
At first I thought this is א, and was planning to prove it using the good old technique of picking one where there is zero, and zero where there is one.
But then I came up with the following function:
f:A ---> Z
f(x) = {
if the first digit after the point is zero, conversion of the binary digits after the point in x (excluding the first zero) into a decimal number
if the first digit after the point is one, conversion of the binary digits after the point in x into a decimal number, multiplied by minus one
}
This function seems to be reversible, and since |Z| is א0, |A| should be א0 as well. on the other hand, if proving using the method I intended to use, the result is that |A| is א.
Which one is it, and why?
The cardinality of your set is that of the continuum, because it's basically the set of all infinite sequences composed of the strings $1$ and $01$.
As mentioned in the comments, the problem with your proof idea is that $f(x)$ is not always an integer. If you would like to interpret $f(x)$ as simply an infinite binary sequence, then that's fine, but then the injectivity of $f$ gives you no grounds on which to claim $A$ is countable.