Image of $f$ in $f(x)=\lfloor x\rfloor$ out of bounds for intervals?

Question

Edit 1. This all being worked on with the real numbers $\mathbb{R}$

Given a function $f(x) = \lfloor x\rfloor$ (Floor function). Find the image of B, $f^{-1}(B)$ if $B = [0,1)$

For easier cases such as $B=\{0,1\}$, it's obvious that the image of B would be $f^{-1}(B) = \{x\in\mathbb{R}\colon 0\le x < 2\}$. However, I can't seem to figure out the one with a interval.

What I'm confused about is that, for example, $0.5$ doesn't have a solution in $f(x)$. Am I supposed to write that the image of $B$ does not exist in such case?

Edit 1 Here's a badly drawn image illustrating how I see this, the $2.5$ value cannot be traced back to A, because of the function transforming every number to an integer.

Answer 1

Normally the for $f(x)=[x]$ we have $f(x)$ is the greatest integer value of x. s.t $f(1.5)=1$ & $f(-1.5)=2$. So the inverse image $f^{-1}(B)=[0,1)$.

Some clarification for the asker

If $B=[x,x+1)$ where $x\ge 0$ then $f^{-1}B=x$ if $B=[x,x+1)$ where $x< 0$ then $f^{-1}B=x$. Now is it clear to you.

Again if you take $f(x)=x-[x]$ then $f^{-1}(B)=\Bbb R$