Q:

How is the function $x^+$ defined?

Can I define the function $x^+$ in the following way (here $\epsilon \ge 0)$
$$ x^+ = \lim_{\epsilon \to 0} (x + \epsilon)$$ where $\epsilon \ne 0$.
Same question for $x^- = \lim_{\epsilon\to0} x - \epsilon$.
I thought that these functions (if they are correct) can be used to define a bijective mapping from $[0,1] \to (0,1)$ in the following way
$$ f(x) = x^+ \mid 0 \le x \lt \frac{1}{2} $$
$$f(x) = x^- \mid \frac{1}{2} \lt x \le 1$$
$$ f(x) = 1/2 \mid x = \frac{1}{2} $$.

A:

Your definition of the functions is fine.  They are a complicated way of defining the identity function, because for all $x$ we have $x^+=x^-=x$.  The problem in your application is that $0^+=0$ and $1^-=1$, so you have not defined a mapping from $[0,1] \to (0,1)$.  The image of $[0,1]$ is again $[0,1]$.

