I simply want to know exactly how this particular function is bijective.
Since they defined our function $g$, I would like to assume the set A to be written as $A=\ \left\{a_{\ 1}{,}\ a_{\ 2}{,}\ a_{\ 3}….\right\}$
so therefore from the defined function we can assume that $B=\ \left\{a_{\ 1}\right\}$, so if we $A-B$, we are left with $A – B =\ \left\{a_{\ 2}{,}\ a_{\ 3}{,}\ a_{\ 4}….\right\}$
Now if we pick $a_{\ 1}$ from set $B$ and plug it into our function g we should be getting $a_{\ 2}$
since $g\left(a_{\ n}\right)=\ a_{n+1\ }$ for $a_{n\ \ }\in \ B$
at the same time if we choose $\left\{a_{2\ }\right\}\in \ A-B$ into $g$ we would get
$a_{\ 2}$
So now $a_{\ 2}$ (in the range of the function) has two pre-image. I fail to understand how this function can be considered a bijection? (my apologies if my question is too elementary, I simply want to understand this theorem)
Best Answer
You misunderstood the construction.
They define $f(ℤ_+) = B$, so $B = \{f(n) \,|\, n ∈ ℤ_+ \}$. Since $f(n)$ is written as $a_n$, we have $B = \{a_1, a_2, a_3, ...\}$ and $A-B = \{ x ∈ A \,|\, \not \exists n ∈ ℤ_+: f(n) a_n = x \}$. So $A - B$ is everything that is left when we take out the indexed elements.
So your first conclusion ($A = \{a_1, a_2, ... \}$) is not correct.
I don't know how you get the conclusion that $g(a_2) = a_2$. The definition of $g$ and the schematic image says $g(a_2) = a_3$ ($n = 2, n+1=3$).
The range of $g$ is $A - \{a_1\}$.