Topology: Question on the inverse set

Question

Excerpt from "Topology Without Tears" (Sidney Morris).

In the above example, how is it that $f^{-1}((1,3)) = (2,3]$ ? Here is my understanding, kindly correct the misconceptions. The inverse for $(2,4]$ is not defined. The inverse is as below. $$f^{-1}(y)=\begin{cases} y+1 & \text { if } y \le 2\\ 2y-5 & \text{ if } y \gt 4\\ \end{cases}$$ So, if I have to find out for example, $f^{-1}(2\frac{1}{2})$, how do I do it? When does $f^{-1}(y)$ give me $3$ (to justify the $3$ in $(2,3]$ ) ?

Answer 1

The inverse image $f^{-1}(S)$ refers to the set $$\{x \in \Bbb{R} : f(x) \in S\}$$ This would mean that $$f^{-1}(\{2\frac{1}{2}\}) = \emptyset$$ We also have, $$f^{-1}(1, 3) = \{x \in S : 1 < f(x) < 3\},$$ which is true precisely for $2 < x \le 3$. There's no requirement that there be some $x$ such that $f(x) = 2.5$; just so long as it's less than $3$.

Answer 2

• If $x>3$ we have that $f(x) = \frac{1}{2}(x+5) > \frac{1}{2} \cdot 8 = 4$ and so $f(x) \notin (1,3)$.

• If $2< x \le 3$ we have that $f(x)=x-1 \in (1,2] \subseteq (1,3)$ and

• If $x \le 2$, $f(x)=x-1 \le 1$,so $f(x) \notin (1,3)$

Hence $f^{-1}((1,3) = \{x: f(x) \in (1,3) \} = (2,3]$, as we covered all options for $x$.

Inverse image is not "image under a (non-existent) inverse function". For example: if $f: \mathbb{R} \to \mathbb{R}$ is the function that is constant with value $2$, then $f^{-1}(\{2\}) = \mathbb{R}$ and $f^{-1}(\{1\}) = \emptyset$. We are talking about inverse images of sets, not of points.