real analysis - Prove $[0, 1]$ and $(0, 1]$ are equinumerous by use of bijection

What my thought was that the bijection was a piecewise function:

$f(x) = 1/n+1$ if $x = 1/n$ for some $n \in \Bbb N$ and $x=1/n$ if $x \neq1/n$ for some $n \in \Bbb N$.

However, the textbook says this is incorrect. I don't actually see why. Was it a matter of brackets or the open variable $(0,1]$, which is what I tend to think.