Preimage of a set under floor function

Question

How would I interpret the pre-image of $U = (\frac{1}{2},\frac{3}{2})$ under the function $g(x) = \left \lfloor{x}\right \rfloor$? Friend says it's $[1,2)$ but I just can't undrstand it...

Answer 1

System won't let me write a comment so I'll write an answer instead.

Think in the following steps might help:

1. What value can $g(x)$ take? --Integers.
2. Then, what value in $(1/2,3/2)$ can $g(x)$ take? --$1$.
3. What is $g^{-1}(1)$?

Hope this helps.

Answer 2

Well, this depends on what "pre-image" means.

The Pre-image of $U = g^{-1}(U) = \{x \in \mathbb R|g(x) \in U\}$.

Therefore $g: g^{-1}(U) \rightarrow U$. BUT Note that the image of the pre-image is not necessarily all of $U$.

It's possible that $g:g^{-1}(U) \rightarrow K \subsetneq U$ where $K = \{g(s)|s \in g^{-1}(U)\}$.

Which is precisely what happens here.

$S=g^{-1}(\frac 12, \frac 32) = \{x \in \mathbb R| \lfloor x \rfloor \in (\frac 12, \frac 32)\} =$

$ \{x \in \mathbb R| \frac 12 < \lfloor x \rfloor < \frac 32\} =$

$ \{x \in \mathbb R| \lfloor x \rfloor =1\} =$

$[1,2)$.

And that's all there is to it.

Now it is true that $g(S) = \{1\} \ne (\frac 12, \frac 32)$. But that is not relevent.

It is NOT true that $g(pre-image(U)) = U$. It is only nescessary that $g(pre-image(U)) \subset U$. It is nescessary that if $g(x) \in U$ then $x \in pre-image (U)$. But that is all.

So... yes, the question is misleading. There are not any $\frac 12 < x < 1$ nor any $1< y < \frac 32$ so that $g(k) = x$ or $g(k) = y$ and the pre-image of $(\frac 12, \frac 32)$ is the same pre-image of $\{1\}$ and the same pre-image of $(1,2)$. The points that are not mapped to don't affect anything and are totally irrelevent.