Proof for: A fiunction f : R --> R is continous if and only if f^-1((a,b)) is a countable union of open intervals for every (a,b) subset of R

Question

I am reading this book Probability theory in Finance by Sean Dineen, In chapter they proof this proposition. I can;t understand this and also don't understand what he means by " a countable union of open intervals".

Answer 1

You may read the book "General Topology", written by John Kelley, to acquire some background knowledge about topology. The proposition you stated is correct and is based on the following well-known facts.

Fact 1: Let $X$ and $Y$ be topological spaces and let $f:X\rightarrow Y$ be a map. Let $\mathcal{SB}$ be a subbase for the topology on $Y$. Then $f$ is continuous iff $f^{-1}(A)$ is open for each $A\in\mathcal{SB}$.

(This is because the inverse image operator $A\mapsto f^{-1}(A)$ preserves all set-theoretic operations, including arbitrary union, aribrary intersection, taking complement)

Fact 2: $\{(a,b)\mid a<b\}$ is a base for the usual topology on $\mathbb{R}$.

Fact 3: Every open set on $\mathbb{R}$ can be written as disjoint union of at most countably many open intervals.