Do not understand: "A real function $f$ is continuous if $f^{-1}(O)$ is open for each open set $O$"

Question

I am looking at Measure, Integral and Probability book and that is what they say (about the title). Although, I fail to understand this.

If we take a function $f(x) = x^2$ and restrict its domain to $x \geq 0$. Then the inverse is $f^{-1}(x) = \sqrt{x}$. This function's output will be open for union of open intervals.

But why are we concerning ourselves with the inverse of a function to check its continuity? This definition feels like cheating to me. Because if the function $f$ is continuous, then $f^{-1}$ will be continuous, why does this have anything to do with open intervals? And so a proper definition of continuity has to be used, in that a function is differentiable at every point in the domain.

Answer 1

If $f$ is a function from $A$ to $B$ and $C \subset B$ then $f^{-1}(C)$ is defined as $\{a\in A:f(a)\in C\}$. This does not require the existence of an inverse for $f$.

Answer 2

That statement has nothing to do with inverse functions. It's about reverse images: if $f\colon A\longrightarrow B$ is a function and $C\subset B$, then$$f^{-1}(C)=\{a\in A\mid f(a)\in C\}.$$